CHAPTER ONE
Electric Charge, Forces in Nature and Electrostatics
1.1 Introduction to Electricity and Magnetism
Electricity and magnetism are closely related branches of physics concerned with electric charges, electric and magnetic fields, electric currents, electromagnetic induction, and electromagnetic radiation. Together, they form the subject known as electromagnetism.
Electrical and magnetic phenomena are encountered in both nature and technology. Lightning is an electrical discharge in the atmosphere. Electric motors convert electrical energy into mechanical energy through magnetic interactions. Generators produce electrical energy through electromagnetic induction, while communication systems depend heavily on electromagnetic waves.
The study of electricity begins with the concept of electric charge. Charges at rest produce electrostatic effects, whereas charges in motion constitute electric currents and are associated with magnetic fields.
Electrostatics is the branch of physics concerned with electric charges at rest and the forces, fields, potentials and energies associated with them.
Although electrostatics is introduced separately from magnetism, later chapters will show that electricity and magnetism are different aspects of a unified electromagnetic interaction.
1.2 Fundamental Forces in Nature
Interactions between physical objects are described in modern physics in terms of four fundamental forces or interactions. Many forces encountered in everyday mechanics can ultimately be traced to one or more of these fundamental interactions.
| Fundamental Interaction | Main Role | Relative Range |
|---|---|---|
| Gravitational interaction | Acts between bodies possessing mass-energy and governs much astronomical motion. | Infinite |
| Electromagnetic interaction | Acts between electrically charged particles and underlies electricity, magnetism and many atomic and molecular interactions. | Infinite |
| Strong nuclear interaction | Binds quarks and is responsible for the residual nuclear force that helps bind protons and neutrons in nuclei. | Very short |
| Weak nuclear interaction | Responsible for processes such as beta decay and important particle transformations. | Very short |
The electromagnetic interaction is particularly important in this course. It is responsible for electric forces between charged particles and magnetic effects associated with moving charges.
Familiar contact forces such as normal reaction, friction and tension arise ultimately from electromagnetic interactions between atoms and molecules, even though they are treated macroscopically as separate forces in elementary mechanics.
1.3 Electric Charge
Electric charge is a fundamental property of matter that determines how particles and bodies participate in electromagnetic interactions.
Electric charge is represented by the symbol q or Q. Its SI unit is the coulomb (C).
At the atomic level, ordinary matter contains positively charged protons, negatively charged electrons and electrically neutral neutrons. The magnitudes of the charges of the proton and electron are equal:
Thus, approximately,
and
An electrically neutral body contains equal amounts of positive and negative charge. A charged body has an imbalance between these charges.
In ordinary electrostatic processes, charging a body usually involves the transfer or redistribution of electrons. Protons are bound within atomic nuclei and normally do not move from one body to another during common charging processes.
1.4 Positive and Negative Charges
There are two kinds of electric charge: positive and negative.
Their interaction follows a simple qualitative rule:
Thus, two positively charged bodies repel each other, and two negatively charged bodies also repel each other. A positively charged body and a negatively charged body attract each other.
| Combination | Interaction |
|---|---|
| Positive and positive | Repulsion |
| Negative and negative | Repulsion |
| Positive and negative | Attraction |
A body becomes negatively charged when it gains excess electrons. It becomes positively charged when it loses electrons.
1.5 Fundamental Properties of Electric Charge
Electric charge possesses several important properties that form the foundation of electrostatics.
Additivity of Charge:
Electric charge is additive. The total charge of a system is the algebraic sum of the individual charges:
The signs of the charges must be included when calculating the net charge.
Question: A system contains charges of +5 μC, -8 μC and +2 μC. Determine the net charge.
Formula:
Solution:
Answer: The net charge is -1 μC.
Conservation of Charge:
The total electric charge of an isolated system remains constant. Charge may be transferred between bodies, but the net amount of charge is conserved.
Quantisation of Charge:
Observable charge occurs in discrete multiples of the elementary charge e. For ordinary bodies,
where n is an integer.
1.6 Quantisation of Electric Charge
Quantisation of electric charge means that the net electric charge on an isolated ordinary body occurs in integral multiples of the elementary charge.
where:
- Q = total charge,
- n = integer number of elementary charges,
- e=1.602×10-19C.
Question: How many electrons correspond in magnitude to a charge of 3.2×10-18 C?
Given:
Required: Number of electrons, n.
Formula:
Solution:
Answer: The charge corresponds to approximately 20 electrons.
Question: A body gains 5×1012 electrons. Calculate the charge acquired by the body.
Formula:
Solution:
Equivalently,
Answer: The body acquires a charge of approximately -8.01×10-7 C.
1.7 Conservation of Electric Charge
The algebraic sum of electric charge in an isolated system remains constant.
If two initially isolated bodies interact and exchange electrons, the charge lost by one body is gained by the other. The distribution of charge changes, but the total charge does not.
Question: Two isolated bodies initially have charges +8 μC and -3 μC. After interacting, the first body has +2 μC. Find the final charge on the second body.
Initial Total Charge:
Let the final charge of the second body be q.
Answer: The second body has a final charge of +3 μC.
1.8 Conductors and Insulators
Materials differ greatly in the ease with which electric charges can move through them. A basic distinction is made between conductors and insulators.
A conductor is a material in which electric charge can move relatively freely through the material.
Metals such as copper, aluminium and silver are good conductors because some of their electrons are mobile throughout the material.
An insulator is a material in which electric charges are not free to move easily throughout the material.
Examples include glass, dry rubber, plastics and ceramics.
| Property | Conductors | Insulators |
|---|---|---|
| Charge mobility | Charges can move relatively freely. | Charges are largely bound to atoms or molecules. |
| Examples | Copper, aluminium, silver | Glass, rubber, plastic |
| Charge placed on material | Can redistribute over the conductor. | Tends to remain localized near where it is deposited. |
Semiconductors:
Semiconductors have electrical behaviour intermediate between good conductors and good insulators, and their conductivity can be strongly controlled by temperature, impurities, light and electric fields. Silicon is an important example.
1.9 Charging by Friction
When two different materials are rubbed together, electrons may transfer from one material to the other because the materials have different tendencies to retain electrons.
The material that gains electrons becomes negatively charged, while the material that loses electrons becomes positively charged.
When suitable insulating materials are rubbed together, one may acquire excess electrons while the other develops an equal deficiency of electrons. The charges produced are equal in magnitude and opposite in sign if the two-body system remains isolated.
Friction does not create electric charge. It facilitates the transfer of electrons already present in the materials.
1.10 Charging by Conduction
Charging by conduction is the process of transferring electric charge to or from a body through direct physical contact with a charged body.
Suppose a negatively charged conductor touches an initially neutral conductor. Some excess electrons can move onto the neutral conductor. After separation, the second conductor may retain a negative charge.
Similarly, contact with a positively charged conductor can remove electrons from another conductor.
Charging by conduction involves direct contact and actual transfer of charge between the interacting bodies.
1.11 Charging by Induction
Charging by induction is the process of giving a conductor a net charge without direct contact with the external charged body used to produce the charge separation.
Consider a negatively charged rod brought close to an initially neutral conductor. The rod repels mobile electrons within the conductor toward the farther side. The near side therefore has an induced positive region, while the far side has an excess of electrons.
If the conductor is then connected temporarily to the Earth while the charged rod remains nearby, some electrons may flow to Earth. If the Earth connection is removed first and the charged rod is subsequently taken away, the conductor is left with a net positive charge.
Using a positively charged inducing body reverses the directions of charge movement and can leave the conductor negatively charged.
Grounding:
Grounding is the process of electrically connecting a body to the Earth so that charge can flow between the body and the Earth.
Because the Earth acts as an enormous charge reservoir, it can accept or supply electrons with negligible change in its overall electrical state.
| Method | Contact Required? | Main Process |
|---|---|---|
| Friction | Yes | Electron transfer between unlike materials. |
| Conduction | Yes | Charge transfer through direct contact. |
| Induction | No contact with inducing body | Charge redistribution followed, when net charging is desired, by grounding and separation. |
1.12 Polarisation
A charged object can attract a neutral body even when there is no net transfer of charge. This is explained by polarisation.
Electric polarisation is the separation or displacement of positive and negative charge within a material under the influence of an electric field.
In a neutral atom or molecule, the centres of positive and negative charge may shift slightly in opposite directions when an external electric field is applied. The material remains electrically neutral overall, but its charge distribution becomes polarised.
Because the oppositely charged region can be closer to the external charge than the like-charged region, the attractive interaction may exceed the repulsive interaction, producing a net attraction.
1.13 Distribution of Charge on Conductors
When a conductor reaches electrostatic equilibrium, its free charges are no longer undergoing net motion. Several important properties then follow.
- Excess charge on an isolated conductor resides on its surface.
- The electric field inside the conducting material is zero under electrostatic equilibrium.
- The conductor is an equipotential body.
- The electric field immediately outside the surface is perpendicular to the surface.
- Surface charge density tends to be greater at regions of greater curvature, such as sharp points.
The concentration of charge near sharp points can produce very strong local electric fields and can lead to ionisation of the surrounding air.
1.14 Electrostatic Discharge
Electrostatic discharge is a rapid transfer of accumulated electric charge between bodies when the electric field becomes sufficiently large or a conductive path is established.
A small spark produced after walking across a dry surface and touching a metal object is an everyday example. Lightning is a much larger atmospheric electrical discharge.
Electrostatic discharge can be useful in some technologies but may also damage sensitive electronic components. Proper grounding and antistatic procedures are therefore important in electronic environments.
1.15 Applications of Electrostatics
Electrostatic principles have important scientific, industrial and technological applications.
Electrostatic Precipitators:
Electrostatic precipitators remove fine particles from industrial exhaust gases. Particles are electrically charged and attracted toward oppositely charged or grounded collecting surfaces.
Photocopiers and Laser Printers:
Controlled electrostatic charge patterns are used to attract toner particles to selected regions before the toner is transferred and fixed onto paper.
Electrostatic Painting:
Paint droplets can be electrically charged so that they are attracted toward an oppositely charged or grounded object. This can improve coverage and reduce paint wastage.
Lightning Protection:
Lightning protection systems provide a low-resistance conducting path through which large electrical currents can be directed safely toward the ground.
Chapter Summary
- Electromagnetism deals with electric and magnetic interactions and their relationship.
- There are four fundamental interactions in nature: gravitational, electromagnetic, strong and weak interactions.
- Electric charge is a fundamental property of matter and is measured in coulombs.
- Electric charges occur as positive and negative charges; like charges repel and unlike charges attract.
- The elementary charge has magnitude approximately 1.602×10-19 C.
- Electric charge is additive, conserved and quantised.
- For ordinary charged bodies, Q=ne.
- Conductors permit relatively free movement of charge, while insulators strongly restrict such movement.
- Objects may be charged by friction, conduction or induction.
- Charging by induction does not require contact with the inducing charged body.
- Polarisation involves separation or displacement of positive and negative charge within a material.
- At electrostatic equilibrium, the electric field inside conducting material is zero and excess charge resides on the surface of an isolated conductor.
- Electrostatic discharge occurs when accumulated charge is transferred rapidly between bodies.
- Electrostatics is applied in printing, pollution control, painting and electrical protection systems.
CHAPTER TWO
Coulomb's Law and Electric Fields
2.1 Introduction to Electrostatic Force
Charged bodies can exert forces on one another even when they are not in physical contact. These electrostatic forces are manifestations of the electromagnetic interaction.
The magnitude of the force between two stationary point charges depends on the magnitudes of the charges, their separation and the electrical properties of the surrounding medium.
For point charges in vacuum, this relationship is expressed quantitatively by Coulomb's law.
2.2 Coulomb's Law
The magnitude of the electrostatic force between two stationary point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance separating them.
For point charges q1 and q2 separated by distance r in vacuum,
where
The electrostatic constant is approximately
and the permittivity of free space is
Thus Coulomb's law may also be written as
Coulomb force varies inversely as the square of separation. Doubling the separation reduces the force to one-quarter; tripling it reduces the force to one-ninth.
Question: Two point charges of +3 μC and +5 μC are separated by 0.20 m in vacuum. Calculate the magnitude of the electrostatic force between them.
Given:
Formula:
Solution:
Answer: The magnitude of the force is approximately 3.37 N. Since both charges are positive, the force is repulsive.
Question: Charges of +4 μC and -2 μC are separated by 0.30 m. Determine the force between them.
Formula:
Solution:
Answer: The force has magnitude approximately 0.799 N and is attractive.
2.3 Direction and Vector Nature of Coulomb Force
Electric force is a vector quantity. The force between two point charges acts along the straight line joining them.
A vector form of Coulomb's law may be written as
where r̂12 is a unit vector specifying the chosen direction from one charge toward the other.
The sign of q1q2 determines whether the force follows or opposes the chosen unit-vector direction.
The forces exerted by two charges on one another satisfy Newton's third law:
2.4 Comparison of Electric and Gravitational Forces
Coulomb's law and Newton's law of universal gravitation have similar mathematical forms.
| Electric Force | Gravitational Force |
|---|---|
| F=k|q1q2|/r2 | F=Gm1m2/r2 |
| Depends on electric charge. | Depends on mass. |
| May be attractive or repulsive. | Attractive for ordinary positive masses. |
| Can be extremely strong at atomic scales. | Much weaker between elementary particles. |
| Obeys an inverse-square dependence. | Obeys an inverse-square dependence. |
Despite their mathematical similarity, the two interactions have very different physical properties. In ordinary matter, positive and negative charges can largely cancel, whereas gravitational mass does not ordinarily occur in positive and negative forms that cancel one another.
2.5 Principle of Superposition
When more than two charges are present, the net electric force on any one charge is found by calculating the force due to each of the other charges separately and then adding the individual force vectors.
The resultant electrostatic force on a charge due to several other charges is the vector sum of the individual forces exerted on it by those charges.
Because force is a vector, forces acting in different directions must be resolved into components before they are combined.
Question: A charge +2 μC is located midway between two charges +4 μC and +6 μC. The distance from the middle charge to each outer charge is 0.20 m. Find the net force on the middle charge.
The +4 μC charge repels the middle charge in one direction, while the +6 μC charge repels it in the opposite direction.
Force Due to +4 μC:
Force Due to +6 μC:
Since the forces oppose each other,
Answer: The net force is approximately 0.899 N toward the +4 μC charge, because the stronger repulsion from the +6 μC charge pushes the middle charge in that direction.
2.6 Electric Field
The idea of an electric field provides a convenient way of describing how a source charge influences the space around it.
An electric field is a region of space in which an electric charge experiences an electric force.
The electric field at a point is the electric force per unit positive test charge placed at that point, provided the test charge is sufficiently small not to disturb the original charge distribution.
The SI unit of electric field is
It is also equivalent to volts per metre, V/m, a relationship that becomes clearer when electric potential is studied.
The direction of the electric field at a point is defined as the direction of the force that would act on a positive test charge placed there.
The electric field exists because of the source charges. A test charge is conceptually used to detect the field. Once the field is known, the force on a charge q is obtained from F=qE.
2.7 Electric Field Due to a Point Charge
Consider a source charge Q and a small positive test charge q0 at distance r.
From Coulomb's law,
Since
the magnitude of the field is
Equivalently,
For a positive source charge, the field points radially outward. For a negative source charge, the field points radially inward.
Question: Calculate the electric field magnitude 0.30 m from a point charge of +5 μC.
Given:
Formula:
Solution:
Answer: The electric field is approximately 4.99×105 N/C, directed radially away from the positive charge.
Question: A charge of -2 μC is placed in an electric field of magnitude 3×104 N/C. Determine the magnitude and direction of the electric force.
Formula:
Magnitude:
Answer: The force has magnitude 0.06 N. Since the charge is negative, the force acts opposite to the direction of the electric field.
2.8 Superposition of Electric Fields
Electric fields obey the principle of superposition. If several charges produce electric fields at the same point, the resultant field is the vector sum of the individual fields.
For fields along the same straight line, signs may be assigned according to a chosen positive direction. For two-dimensional arrangements, the fields should be resolved into Cartesian components.
Question: Two equal positive charges are placed at equal distances on opposite sides of a point P. Determine the net electric field at P.
Each positive charge produces an electric field directed away from itself. Because the charges and distances are equal, the two fields have equal magnitudes but opposite directions.
Answer: The net electric field at the midpoint is zero.
2.9 Electric Field Lines
An electric field line is an imaginary curve drawn so that its tangent at any point gives the direction of the electric field at that point.
Electric field diagrams follow several useful rules:
- Field lines originate on positive charges and terminate on negative charges, or extend to or from infinity when appropriate.
- The tangent to a field line gives the local field direction.
- Field lines are more densely packed where the field is stronger.
- Electric field lines never cross.
- For a point charge, the lines are radial.
- In electrostatic equilibrium, field lines meet a conductor's surface perpendicularly.
Positive Point Charge:
The field lines of an isolated positive point charge extend radially outward.
Negative Point Charge:
The field lines of an isolated negative point charge point radially inward.
Electric field lines are not physical strings or paths permanently present in space. They are graphical tools used to represent the direction and relative strength of an electric field.
2.10 Electric Field of Two-Charge Systems
The electric field of two charges is found by applying the superposition principle at every point of interest.
Two Like Charges:
For two charges of the same sign, field lines bend away from the region between the charges because each charge repels a positive test charge.
If the two charges are equal, the electric field is zero at the midpoint because the two field contributions are equal and opposite.
Two Unlike Charges:
For a positive and a negative charge, field lines are directed from the positive charge toward the negative charge.
An electric dipole consists of two equal charges of opposite sign separated by a small distance.
The detailed properties of electric dipoles, including dipole moment, torque and potential energy, will be developed in Chapter 4.
2.11 Continuous Charge Distributions
Coulomb's law is naturally expressed for point charges. However, many physical objects contain charge distributed continuously along a line, over a surface or throughout a volume.
A continuous distribution can be treated as being composed of very small charge elements dq. Each element produces a small electric field contribution dE, and the total field is obtained by integration.
The mathematical difficulty depends strongly on the geometry and symmetry of the charge distribution.
2.12 Linear, Surface and Volume Charge Densities
Charge density describes how electric charge is distributed in space.
Linear Charge Density:
For charge distributed along a line,
For a uniform distribution over length L,
The SI unit is C/m.
Surface Charge Density:
For charge distributed over a surface,
For a uniform distribution over area A,
The SI unit is C/m2.
Volume Charge Density:
For charge distributed throughout a volume,
For a uniform distribution throughout volume V,
The SI unit is C/m3.
| Distribution | Charge Density | Uniform Form | SI Unit |
|---|---|---|---|
| Line | λ=dQ/dl | λ=Q/L | C/m |
| Surface | σ=dQ/dA | σ=Q/A | C/m2 |
| Volume | ρ=dQ/dV | ρ=Q/V | C/m3 |
Question: A uniformly charged wire of length 2.0 m carries a total charge of 6 μC. Find its linear charge density.
Formula:
Solution:
Answer: The linear charge density is 3×10-6 C/m, or 3 μC/m.
2.13 Electric Field Due to Continuous Charge Distributions
For a small charge element dq, the magnitude of its electric field contribution at a point a distance r away is
Because electric field is a vector, the direction of each contribution must also be considered:
The total field is then
Depending on the distribution,
or
Symmetry can greatly simplify such calculations. Chapter 3 will develop Gauss's law, which provides a particularly efficient method for highly symmetric charge distributions.
Question: A circular ring carries charge uniformly around its circumference. Explain qualitatively why the electric field at the exact centre is zero.
Reasoning: Every small charge element on the ring has a corresponding element on the opposite side. The two elements produce electric fields of equal magnitude but opposite direction at the centre.
Answer: By symmetry and superposition, the resultant electric field at the centre of a uniformly charged ring is zero.
2.14 Motion of a Charged Particle in an Electric Field
When a particle of charge q is placed in an electric field E, it experiences force
If the electric force is the only significant force, Newton's second law gives
Therefore,
In a uniform electric field, the force is constant, so the charged particle undergoes constant acceleration. A positive charge accelerates in the direction of the field, while a negative charge accelerates in the opposite direction.
Question: A particle of mass 2×10-6 kg and charge +4×10-6 C is placed in a uniform electric field of 5×103 N/C. Neglect other forces and calculate its acceleration.
Force:
Acceleration:
Answer: The acceleration is 1.0×104 m/s2 in the direction of the electric field.
Question: An electron is placed in a uniform electric field of magnitude 2.0×104 N/C. Determine the magnitude of the electric force on the electron.
Given:
Formula:
Solution:
Answer: The force magnitude is approximately 3.20×10-15 N. Because an electron is negatively charged, its force is directed opposite to the electric field.
2.15 Applications of Electric Fields
Electric fields provide a powerful framework for understanding and controlling charged particles. They are important in many physical systems and technologies.
Charged-Particle Control:
Electric fields can accelerate or deflect charged particles. This principle is used in particle-beam systems and many laboratory instruments.
Electrostatic Precipitation:
Electric fields can exert forces on charged pollutant particles and drive them toward collecting surfaces.
Printing and Imaging:
Controlled electric fields guide charged toner particles toward selected regions in electrostatic printing processes.
Capacitive Systems:
Electric fields between conductors are central to the operation of capacitors, which store electrical energy. Capacitance and energy storage will be developed in Chapter 4.
Coulomb's law describes the direct force between charges, while the electric-field concept separates the problem into two stages: source charges create a field, and another charge placed in that field experiences a force. This viewpoint becomes especially powerful when dealing with many charges and continuous charge distributions.
Chapter Summary
- Coulomb's law gives the magnitude of the force between stationary point charges as F=k|q1q2|/r2.
- The electrostatic constant is approximately 8.99×109 N m2/C2.
- Electrostatic force acts along the line joining two point charges and may be attractive or repulsive.
- Coulomb's law is an inverse-square law.
- The principle of superposition states that the resultant force or field is the vector sum of the individual contributions.
- Electric field strength is defined by E=F/q0.
- The electric field due to a point charge has magnitude E=k|Q|/r2.
- The field of a positive point charge is directed outward, while that of a negative point charge is directed inward.
- The force on a charge in an electric field is F=qE.
- Electric field lines show the direction and relative strength of an electric field.
- An electric dipole consists of equal and opposite charges separated by a distance.
- Continuous charge distributions may be described using linear, surface and volume charge densities.
- The field of a continuous distribution can be determined by integrating the contributions from small charge elements.
- A charged particle in a uniform electric field experiences constant electric force and, when other forces are negligible, constant acceleration.
- The electric-field concept provides the foundation for Gauss's law, electric potential, capacitance and later topics in electromagnetism.
CHAPTER THREE
Electric Flux, Gauss's Law and Electric Potential
3.1 Introduction to Electric Flux
In the previous chapter, electric field was introduced as a vector quantity that describes the force per unit positive test charge at a point. For many charge distributions, especially those possessing a high degree of symmetry, calculating the electric field directly from Coulomb's law can become unnecessarily difficult.
A powerful alternative is obtained by considering how much electric field passes through a surface. This idea is described by electric flux. Electric flux, together with Gauss's law, provides one of the most important methods for calculating electric fields in electrostatics.
Electric flux is a measure of the amount of electric field passing through a surface.
Electric flux depends on the strength of the electric field, the area of the surface and the orientation of the surface relative to the field.
3.2 Area Vectors and Electric Flux
To describe the orientation of a surface, an area vector is introduced. For a flat surface of area A, the area vector A has magnitude equal to the area and is directed perpendicular to the surface.
For an open surface, one of the two possible normal directions is chosen according to the problem. For a closed surface, the conventional area vector points outward from the enclosed volume.
where n̂ is a unit vector normal to the surface.
3.3 Electric Flux Through a Flat Surface
For a uniform electric field passing through a flat surface, the electric flux is
or
where θ is the angle between the electric field vector and the area vector.
The angle in the flux equation is measured between the electric field and the normal to the surface, not between the field and the surface itself.
Three important cases follow immediately.
| Orientation | Angle | Flux |
|---|---|---|
| Field perpendicular to surface | θ=0° | ΦE=EA, maximum magnitude |
| Field at an angle | 0°<θ<90° | ΦE=EAcosθ |
| Field parallel to surface | θ=90° | ΦE=0 |
The SI unit of electric flux is
Question: A uniform electric field of magnitude 400 N/C passes through a flat surface of area 0.50 m2. The field makes an angle of 60° with the area vector. Find the electric flux.
Given:
Formula:
Solution:
Answer: The electric flux is 100 N m2/C.
Question: A uniform field of 250 N/C is parallel to a rectangular surface of area 0.80 m2. Determine the electric flux.
Reasoning: If the field is parallel to the surface, it is perpendicular to the area vector. Thus θ=90°.
Answer: The electric flux is zero.
3.4 Flux Through Closed Surfaces
For a curved or closed surface, the electric field and surface orientation may vary from point to point. The surface is therefore divided conceptually into very small area elements dA.
The infinitesimal flux through one element is
The total flux through a closed surface is
The circle on the integral sign indicates integration over a closed surface.
Flux leaving a closed surface is taken as positive, while flux entering the surface is negative because the outward area vector points opposite to the incoming field direction.
3.5 Gauss's Law
The net electric flux through any closed surface is equal to the total charge enclosed by the surface divided by the permittivity of free space.
The charge Qenc is the algebraic sum of all charges enclosed by the Gaussian surface.
Charges outside the surface may produce electric fields at points on the surface, but their total contribution to the net flux through the closed surface is zero.
Gauss's law is always valid, but it is most useful for calculating electric fields when the charge distribution possesses enough symmetry for the electric field to be taken outside the surface integral.
Question: A closed surface encloses a charge of 5 μC. Determine the net electric flux through the surface.
Given:
Formula:
Solution:
Answer: The net electric flux is approximately 5.65×105 N m2/C.
3.6 Gaussian Surfaces and Symmetry
A Gaussian surface is an imaginary closed surface chosen to exploit the symmetry of a charge distribution when applying Gauss's law.
The most useful Gaussian surfaces correspond to the symmetry of the physical charge distribution:
- spherical surfaces for spherical symmetry;
- cylindrical surfaces for cylindrical symmetry;
- pillbox-shaped surfaces for planar symmetry.
A suitable Gaussian surface allows the magnitude of E to be constant over relevant portions of the surface and makes the angle between E and dA simple.
3.7 Electric Field of a Spherically Symmetric Charge Distribution
Consider a point charge Q at the centre of a spherical Gaussian surface of radius r.
Because of spherical symmetry, the electric field has the same magnitude everywhere on the sphere and is radial. Thus,
Since the surface area of a sphere is 4πr2,
Hence,
This is the same result obtained from Coulomb's law.
Question: Use Gauss's law to determine the electric field 0.20 m from a point charge of 4 μC.
Formula:
Solution:
Answer: The electric field is approximately 8.99×105 N/C, directed radially outward.
Field of a Charged Conducting Sphere:
For a conducting sphere in electrostatic equilibrium, all excess charge resides on the outer surface. Therefore:
- inside the conducting material, E=0;
- outside the sphere, the field is the same as that of a point charge located at the centre.
Thus for r>R,
while for points inside the conductor,
3.8 Electric Field of an Infinite Line of Charge
Consider an infinitely long straight line carrying uniform linear charge density λ.
By cylindrical symmetry, the electric field points radially outward or inward and has constant magnitude at every point on a cylindrical Gaussian surface of radius r and length L.
Flux passes only through the curved surface because the field is parallel to the flat end surfaces.
The enclosed charge is
From Gauss's law,
Therefore,
Question: An infinitely long line has linear charge density 2.0 μC/m. Find the electric field at a distance of 0.10 m from the line.
Formula:
Solution:
Answer: The electric field is approximately 3.60×105 N/C, directed radially away from the positively charged line.
3.9 Electric Field of an Infinite Plane Sheet of Charge
Consider an infinite plane sheet with uniform surface charge density σ. By symmetry, the electric field is perpendicular to the plane and has equal magnitude on both sides.
Choose a cylindrical pillbox Gaussian surface passing through the plane. Flux passes through the two flat ends but not through the curved side.
The enclosed charge is
Using Gauss's law,
Thus,
An important feature is that, for an ideal infinite sheet, the field magnitude does not depend on distance from the sheet.
Question: An infinite sheet has surface charge density 5×10-8 C/m2. Determine the electric field magnitude on either side.
Formula:
Solution:
Answer: The field magnitude is approximately 2.82×103 N/C.
3.10 Conductors in Electrostatic Equilibrium
Gauss's law helps explain important properties of conductors under electrostatic equilibrium.
Electric Field Inside a Conductor:
If a non-zero electric field existed inside the conducting material, free charges would continue to move. Therefore, at electrostatic equilibrium,
within the conductor.
Excess Charge Resides on the Surface:
Consider a Gaussian surface entirely within the conducting material. Since E=0 everywhere on the Gaussian surface,
Hence, by Gauss's law,
Thus there can be no net excess charge in the bulk of the conductor; excess charge resides on its surface.
Electric Field Just Outside a Conductor:
Immediately outside the surface of a conductor in electrostatic equilibrium, the field is perpendicular to the surface. For a surface charge density σ,
just outside the surface in vacuum, provided the local surface can be treated appropriately.
3.11 Electric Potential and Potential Difference
Electric field is a vector quantity. In many problems, a scalar quantity called electric potential provides a simpler way to describe electrostatic systems.
Electric potential at a point is the electric potential energy per unit positive test charge at that point.
The SI unit is the volt:
The electric potential difference between two points is the change in electric potential energy per unit charge in moving a test charge between those points.
For motion from point A to point B,
The negative sign indicates that electric potential decreases in the direction of the electric field.
Question: A charge of 3 C gains 24 J of electric potential energy when moved between two points. Find the potential difference.
Formula:
Solution:
Answer: The potential difference is 8 V.
3.12 Potential Due to a Point Charge
Taking the electric potential to be zero at infinity, the potential at distance r from a point charge Q is
Equivalently,
Unlike electric field, electric potential is a scalar. Therefore, the sign of Q directly determines the sign of the potential.
- A positive point charge produces positive potential.
- A negative point charge produces negative potential.
Question: Calculate the electric potential 0.25 m from a point charge of +4 μC.
Formula:
Solution:
Answer: The electric potential is approximately 1.44×105 V.
Question: Find the potential at a point 0.50 m from a charge of -3 μC.
Solution:
Answer: The potential is approximately -5.39×104 V.
3.13 Electric Potential Due to Several Charges
Because electric potential is scalar, the total potential due to several point charges is obtained by ordinary algebraic addition:
This is often easier than adding electric field vectors.
Question: A point is 0.20 m from a charge of +2 μC and 0.30 m from a charge of -3 μC. Find the electric potential at the point.
Formula:
Solution:
Answer: The net electric potential is 0 V.
Electric potential is scalar, while electric field is vector. Therefore, zero potential at a point does not necessarily imply zero electric field at that point.
3.14 Relationship Between Electric Field and Potential
The electric field points in the direction in which electric potential decreases most rapidly.
In one dimension,
In vector form,
For a uniform electric field along the x-direction,
when displacement is taken along the field direction.
Question: Two parallel points are separated by 0.040 m along the direction of a uniform electric field of magnitude 500 N/C. Find the change in potential moving in the direction of the field.
Formula:
Solution:
Answer: The electric potential decreases by 20 V.
3.15 Equipotential Surfaces
An equipotential surface is a surface on which the electric potential has the same value at every point.
No net work is required to move a test charge along an equipotential surface because the potential difference is zero:
Hence,
Electric field lines intersect equipotential surfaces at right angles. If there were a component of electric field along the surface, a charge moving along that surface would experience work, contradicting the definition of an equipotential.
For a point charge, equipotential surfaces are concentric spheres centred on the charge.
3.16 Potential Energy of Electric Charges
The potential energy of a charge q placed at a point of electric potential V is
For two point charges,
The sign of the potential energy depends on the signs of the charges.
- Like charges give positive potential energy.
- Unlike charges give negative potential energy.
Question: Calculate the electrostatic potential energy of two charges +2 μC and +5 μC separated by 0.30 m.
Formula:
Solution:
Answer: The potential energy is approximately 0.300 J.
Question: A charge of -4 μC moves through a potential difference of +50 V. Find the change in its electric potential energy.
Formula:
Solution:
Answer: The potential energy decreases by 2.0×10-4 J.
3.17 Applications of Electric Potential
Electric potential and potential difference are central to the behaviour of electrical systems. Potential difference provides the energy-per-unit-charge description that is used in batteries, circuits, capacitors and electric power systems.
A battery maintains a potential difference between its terminals. When connected in a complete circuit, electric charges can move through components and exchange energy.
Electric potential is also useful in accelerating charged particles. If a particle of charge q moves through a potential difference ΔV, the change in electric potential energy is
When electric potential energy decreases and no other energy-transfer mechanisms are important, kinetic energy increases.
Question: An electron is accelerated from rest through a potential difference of 200 V. Find the kinetic energy gained in joules. Use e=1.602×10-19C.
Required: Kinetic energy gained.
Principle: The magnitude of the energy gained is
Solution:
Answer: The electron gains approximately 3.20×10-17 J of kinetic energy.
Chapter Summary
- Electric flux measures the amount of electric field passing through a surface.
- For a uniform field through a flat surface, ΦE=EAcosθ.
- For a closed surface, electric flux is ∮E·dA.
- Gauss's law states that ∮E·dA=Qenc/ε0.
- Gaussian surfaces are chosen to exploit symmetry in spherical, cylindrical or planar charge distributions.
- The field of a point charge or spherically symmetric external distribution is E=Q/(4πε0r2).
- The electric field of an infinite line of charge is E=λ/(2πε0r).
- The electric field of an infinite plane sheet is E=σ/(2ε0).
- Inside a conductor in electrostatic equilibrium, the electric field is zero.
- Electric potential is potential energy per unit charge.
- The potential due to a point charge is V=kQ/r.
- Electric potential is a scalar and obeys algebraic superposition.
- Potential difference is related to electric field by VB-VA=-∫ABE·dl.
- In one dimension, E=-dV/dx.
- Equipotential surfaces are perpendicular to electric field lines.
- The potential energy of a charge at potential V is U=qV.
- For two point charges, U=kq1q2/r.
CHAPTER FOUR
Electric Dipoles, Capacitance and Dielectrics
4.1 Introduction to Electric Dipoles
An electric dipole is one of the simplest charge systems beyond a single point charge. It consists of equal positive and negative charges separated by a finite distance. Dipoles are important because many molecules and electrical systems possess dipole-like behaviour.
An electric dipole consists of two equal charges of opposite sign separated by a distance.
If the charges are +q and -q, separated by a displacement vector from the negative charge toward the positive charge, the system is characterized by its electric dipole moment.
4.2 Electric Dipole Moment
The electric dipole moment is a vector equal to the magnitude of either charge multiplied by the displacement vector from the negative charge to the positive charge.
Its magnitude is
The SI unit of dipole moment is
The direction of p is from the negative charge toward the positive charge.
Question: Charges of magnitude 3 μC are separated by 5 mm. Find the dipole moment.
Given:
Formula:
Solution:
Answer: The dipole moment is 1.5×10-8 C m.
4.3 Electric Field of a Dipole
The electric field of a dipole is obtained by vector addition of the fields due to the positive and negative charges.
At points very far from the dipole compared with the charge separation, useful approximate expressions can be obtained.
Field on the Axial Line:
For a point far from the dipole on its axis,
The field is directed along the dipole moment on the side beyond the positive charge.
Field on the Equatorial Line:
For a point far from the dipole on the perpendicular bisector,
Its direction is opposite to the dipole moment.
Far from a dipole, the electric field decreases approximately as 1/r3, faster than the 1/r2 field of a single point charge.
4.4 Torque on a Dipole in an Electric Field
When an electric dipole is placed in a uniform electric field, the positive and negative charges experience forces equal in magnitude and opposite in direction.
The resultant force is zero in a perfectly uniform field, but the two forces form a couple that can rotate the dipole.
where θ is the angle between the dipole moment and the electric field.
In vector form,
Question: A dipole of moment 2×10-8 C m is placed in a uniform electric field of 5×104 N/C. The angle between the dipole moment and the field is 30°. Find the torque magnitude.
Formula:
Solution:
Answer: The torque magnitude is 5×10-4 N m.
The torque tends to rotate the dipole until its dipole moment aligns with the field.
4.5 Potential Energy of an Electric Dipole
The potential energy of a dipole in a uniform electric field is
Thus,
The potential energy is minimum when the dipole aligns with the field:
Then,
The potential energy is maximum when the dipole is opposite to the field:
giving
Question: A dipole moment of 4×10-9 C m is aligned with a uniform electric field of 2×105 N/C. Find its potential energy.
Given: Since the dipole is aligned with the field, θ=0°.
Formula:
Solution:
Answer: The potential energy is -8×10-4 J.
4.6 Introduction to Capacitance
A capacitor is an electrical device designed to store separated electric charge and electrical energy. It usually consists of two conductors separated by an insulating material or vacuum.
A capacitor is a system of two conductors separated by an insulating region and arranged so that equal and opposite charges can be stored on the conductors.
Capacitance is the ratio of the magnitude of charge stored on either conductor to the potential difference between the conductors.
The SI unit of capacitance is the farad:
Common practical units include:
Question: A capacitor stores a charge of 12 μC when the potential difference is 6 V. Find its capacitance.
Formula:
Solution:
Answer: The capacitance is 2 μF.
4.7 The Parallel-Plate Capacitor
A parallel-plate capacitor consists of two large conducting plates of area A separated by a small distance d.
If the plates carry equal and opposite surface charge densities +σ and -σ, the electric field between them, neglecting edge effects, is
Since
then
The potential difference between the plates is
Therefore,
Since C=Q/V,
Question: A parallel-plate capacitor has plate area 0.020 m2 and separation 1.0 mm. Find its capacitance in air, approximated as vacuum.
Given:
Formula:
Solution:
Therefore,
Answer: The capacitance is approximately 1.77×10-10 F, or 177 pF.
4.8 Factors Affecting Capacitance
For a parallel-plate capacitor,
This expression shows that capacitance depends on geometry and the material between the plates.
Plate Area:
Capacitance is directly proportional to plate area:
Increasing the plate area increases the amount of charge that can be stored for a given potential difference.
Plate Separation:
Capacitance is inversely proportional to separation:
Reducing the separation increases capacitance.
Dielectric Material:
Introducing an insulating dielectric between the plates generally increases the capacitance. This effect will be examined later in the chapter.
4.9 Combination of Capacitors in Parallel
Capacitors are connected in parallel when both terminals of each capacitor are connected to the same pair of nodes.
Therefore, each capacitor has the same potential difference:
The total charge supplied is
Since Qi=CiV,
Thus,
Question: Capacitors of 2 μF, 3 μF and 5 μF are connected in parallel. Find the equivalent capacitance.
Formula:
Solution:
Answer: The equivalent capacitance is 10 μF.
Question: The three capacitors in the previous example are connected across 12 V. Find the total charge stored.
Formula:
Solution:
Answer: The total charge is 120 μC.
4.10 Combination of Capacitors in Series
Capacitors are connected in series when they are joined end-to-end so that the same magnitude of charge appears on each capacitor.
The total potential difference is
Since
then
Therefore,
For two capacitors,
Question: Capacitors of 6 μF and 3 μF are connected in series. Find the equivalent capacitance.
Formula:
Solution:
Answer: The equivalent capacitance is 2 μF.
The equivalent capacitance of capacitors in series is always smaller than the smallest individual capacitance.
Question: The 6 μF and 3 μF capacitors are connected in series across 12 V. Find the charge on each capacitor.
Equivalent Capacitance:
Total Charge:
In series, each capacitor carries the same magnitude of charge.
Answer: Each capacitor carries 24 μC.
4.11 Energy Stored in a Capacitor
Work must be done to transfer charge from one capacitor plate to the other. This work is stored as electric potential energy.
The energy stored in a capacitor is
Using Q=CV, this becomes
Using V=Q/C, it may also be written as
Question: A 20 μF capacitor is charged to 100 V. Find the energy stored.
Formula:
Solution:
Answer: The energy stored is 0.10 J.
4.12 Energy Density in an Electric Field
The energy stored by a capacitor may be regarded as energy stored in its electric field.
For a parallel-plate capacitor in vacuum,
and
Then,
Thus,
Since Ad is the volume between the plates,
The electric energy stored per unit volume in vacuum is
The SI unit of energy density is J/m3.
Question: Determine the energy density in an electric field of magnitude 2×105 N/C in vacuum.
Formula:
Solution:
Answer: The energy density is approximately 0.177 J/m3.
4.13 Dielectrics
A dielectric is an insulating material that can become polarised when placed in an electric field.
Examples include glass, mica, paper, ceramics, certain plastics and air under ordinary conditions.
Although free charges do not move through a dielectric as readily as they do in a conductor, the positive and negative charges within atoms or molecules can shift slightly under an external electric field.
The response of a dielectric is described in part by its relative permittivity, also called dielectric constant:
where ε is the permittivity of the material.
4.14 Polarisation of Dielectrics
When a dielectric is placed in an external electric field, bound positive and negative charges within the material shift slightly in opposite directions, or permanent molecular dipoles tend to align partially with the field.
This creates induced surface charges called bound charges. Their electric field generally opposes the applied electric field.
Dielectric polarisation is the development or alignment of electric dipoles within an insulating material under the influence of an external electric field.
Polarisation therefore reduces the net electric field inside the dielectric compared with the field that would exist in vacuum for the same free-charge arrangement.
4.15 Capacitors with Dielectrics
If the space between the plates of a capacitor is completely filled with a dielectric of relative permittivity κ, the capacitance becomes
where C0 is the capacitance without the dielectric.
For a parallel-plate capacitor,
Question: A parallel-plate capacitor has capacitance 4 μF in vacuum. A dielectric of constant κ=5 completely fills the space between the plates. Find the new capacitance.
Formula:
Solution:
Answer: The new capacitance is 20 μF.
Isolated Capacitor with Dielectric:
If a charged capacitor is disconnected from the battery before a dielectric is inserted, the charge remains constant.
Since C increases by a factor κ,
decreases by the same factor:
The electric field between the plates also decreases:
The stored energy becomes
so it decreases by a factor κ.
Battery-Connected Capacitor with Dielectric:
If the battery remains connected during insertion, the potential difference remains constant:
Because C increases,
also increases by a factor κ.
The stored energy
also increases by the factor κ.
| Quantity | Isolated Capacitor | Battery Connected |
|---|---|---|
| Charge Q | Constant | Increases by factor κ |
| Capacitance C | Increases by factor κ | Increases by factor κ |
| Voltage V | Decreases by factor κ | Constant |
| Stored energy U | Decreases by factor κ | Increases by factor κ |
Question: A 6 μF capacitor is charged to 12 V and disconnected from the battery. A dielectric with κ=3 is then inserted completely. Find the new capacitance and voltage.
Initial Charge:
New Capacitance:
New Voltage:
Answer: The new capacitance is 18 μF, and the new voltage is 4 V.
4.16 Dielectric Strength and Breakdown
Every insulating material can withstand only a limited electric field before it becomes electrically conducting.
Dielectric strength is the maximum electric field magnitude that an insulating material can withstand without electrical breakdown.
If the electric field exceeds this value, the material may ionise or otherwise lose its insulating behaviour. A sudden current can then flow through the material.
Dielectric breakdown is the process by which an insulating material becomes conducting because the applied electric field exceeds its dielectric strength.
Air breakdown is responsible for sparks and contributes to lightning formation when the local electric field becomes sufficiently large.
In capacitor design, the applied voltage must remain safely below the breakdown voltage:
Question: An insulating material has dielectric strength 3×106 V/m and thickness 2 mm. Estimate the maximum voltage it can withstand without breakdown.
Given:
Formula:
Solution:
Answer: The approximate maximum voltage is 6000 V.
4.17 Applications of Capacitors and Dielectrics
Capacitors are widely used because they can store electric charge and energy, respond to changing voltages and influence the timing and frequency characteristics of electrical circuits.
Energy Storage:
Capacitors can store energy temporarily and release it rapidly. This principle is used in systems requiring short bursts of energy.
Electronic Timing:
Capacitors combined with resistors can produce characteristic charging and discharging times. These properties are important in timing and signal-processing circuits.
Filtering and Smoothing:
Capacitors are used in power supplies and electronic circuits to reduce unwanted voltage fluctuations and help produce smoother electrical signals.
Signal Coupling:
Capacitors can allow changing signals to pass between sections of a circuit while blocking steady direct-current components under appropriate operating conditions.
Sensing:
Capacitance changes when plate area, separation or dielectric properties change. This principle is used in capacitive touch sensors, proximity detectors and other measurement systems.
Question: A capacitor of 100 μF stores 0.50 J of energy. Find the voltage across it.
Given:
Formula:
Solution:
Answer: The capacitor voltage is 100 V.
Electric dipoles explain how separated positive and negative charges respond to an electric field. Dielectrics behave through microscopic dipole formation or alignment, while capacitors exploit electric fields and dielectrics to store electrical energy. These ideas provide the foundation for the study of electric current and direct-current circuits.
Chapter Summary
- An electric dipole consists of equal and opposite charges separated by a distance.
- The electric dipole moment is p=qd and points from negative to positive charge.
- A dipole in a uniform electric field experiences torque τ=p×E.
- The potential energy of a dipole is U=-pEcosθ.
- A capacitor stores equal and opposite charges on separated conductors.
- Capacitance is defined by C=Q/V.
- The capacitance of a parallel-plate capacitor in vacuum is C=ε0A/d.
- Capacitance increases with plate area and decreases with plate separation.
- For capacitors in parallel, Ceq=ΣCi.
- For capacitors in series, 1/Ceq=Σ(1/Ci).
- The energy stored in a capacitor is U=(1/2)QV=(1/2)CV2=Q2/(2C).
- The energy density of an electric field in vacuum is uE=(1/2)ε0E2.
- A dielectric is an insulating material that can become polarised in an electric field.
- Inserting a dielectric completely into a capacitor increases capacitance by the relative permittivity κ.
- For a fully filled parallel-plate capacitor, C=κε0A/d.
- Dielectric breakdown occurs when an electric field exceeds the insulating capability of a material.
- Capacitors are used for energy storage, filtering, timing, sensing and signal processing.
CHAPTER FIVE
Electric Current, Resistance and Direct-Current Circuits
5.1 Introduction to Electric Current
Electrostatics deals primarily with electric charges at rest. Many practical electrical systems, however, depend on the controlled motion of charge. When electric charges move through a conductor or another medium, they constitute an electric current.
Electric current is central to the operation of electrical circuits, electronic devices, heating systems, batteries, motors and communication equipment. To analyse such systems quantitatively, we must understand current, potential difference, resistance, electromotive force, electrical power and the relationships among them.
Direct current, abbreviated DC, is electric current whose direction does not reverse with time. Its magnitude may be constant or may vary while maintaining the same direction.
5.2 Electric Current and Charge Flow
Electric current is the rate at which electric charge passes through a chosen cross-section of a conductor or other conducting path.
If a charge ΔQ passes through a cross-section in time Δt, the average current is
The instantaneous current is
The SI unit of current is the ampere.
Question: A charge of 24 C passes through a wire in 8 s. Determine the average current.
Given:
Required: Current I.
Formula:
Solution:
Answer: The average current is 3 A.
Question: A current of 2.5 A flows for 4 minutes. How much charge passes through the circuit?
Given:
Formula:
Solution:
Answer: The charge transferred is 600 C.
5.3 Conventional Current and Electron Flow
The conventional direction of current is defined as the direction in which positive charge would move. This convention was established before the microscopic nature of electric conduction was fully understood.
In metallic conductors, the mobile charge carriers are primarily electrons. Since electrons are negatively charged, their average drift is opposite to the direction of conventional current.
| Quantity | Direction in a Metallic Conductor |
|---|---|
| Conventional current | From higher electric potential toward lower potential through the external resistor network. |
| Electron drift | Opposite to conventional current. |
Circuit analysis normally uses conventional current. Unless otherwise stated, all current directions in this course follow that convention.
5.4 Current Density and Drift Velocity
Current describes the total rate of charge flow through a conductor. Current density describes how that current is distributed over cross-sectional area.
Current density is the electric current per unit cross-sectional area measured perpendicular to the direction of charge flow.
For a uniform current distribution,
The SI unit is
In vector form, J points in the direction of conventional current.
Drift Velocity:
Charge carriers in a conductor already possess microscopic motion, but an applied electric field produces a small average directed velocity called the drift velocity.
For charge carriers of number density n, charge magnitude q, cross-sectional area A, and drift speed vd,
Therefore,
Question: A current of 4 A flows uniformly through a wire of cross-sectional area 2×10-6 m2. Find the current density.
Formula:
Solution:
Answer: The current density is 2×106 A/m2.
Question: A metal contains 8.0×1028 conduction electrons per cubic metre. A wire of cross-sectional area 1.0×10-6 m2 carries 2.0 A. Estimate the drift speed. Use e=1.602×10-19C.
Formula:
Solution:
Answer: The drift speed is approximately 1.56×10-4 m/s.
The drift speed of individual electrons in a conductor can be quite small. This does not mean that an electrical effect travels through a circuit at the drift speed. Changes in the electric field propagate through the circuit much more rapidly.
5.5 Potential Difference and Electromotive Force
Potential difference describes energy transferred per unit charge between two points:
A source such as a cell or battery supplies energy to charges and maintains a potential difference between its terminals.
Electromotive force, abbreviated emf, is the energy supplied by a source per unit charge passing through the source.
Although called an electromotive force, emf is not a force. It is an energy-per-charge quantity and is measured in volts.
Question: A battery supplies 48 J of energy while moving 4 C of charge through itself. Find its emf.
Formula:
Solution:
Answer: The emf of the battery is 12 V.
5.6 Electrical Resistance
Electrical resistance is a measure of the opposition that a material or component offers to electric current.
For a component with potential difference V and current I, resistance is defined under the relevant operating condition as
The SI unit of resistance is the ohm:
A larger resistance generally means that a smaller current flows for a given potential difference.
5.7 Ohm's Law
For an ohmic conductor maintained at constant physical conditions, particularly temperature, the current through the conductor is directly proportional to the potential difference across it.
Hence,
An ohmic conductor has an approximately straight-line current-voltage characteristic through the origin over the range in which its resistance remains constant.
Question: A resistor of 8 Ω is connected across a 24 V supply. Find the current.
Given:
Formula:
Solution:
Answer: The current is 3 A.
Question: A current of 0.25 A flows through a resistor when the potential difference is 5 V. Find the resistance.
Formula:
Solution:
Answer: The resistance is 20 Ω.
Non-Ohmic Devices:
Not all electrical devices obey Ohm's law. A device is non-ohmic when the ratio V/I is not constant over the operating range.
Examples include semiconductor diodes and filament lamps whose resistance changes significantly with operating conditions.
5.8 Resistivity and Conductivity
Resistance depends partly on the dimensions of a conductor, while resistivity characterizes the electrical nature of the material itself.
Resistivity is a material property that measures how strongly the material opposes electric current.
For a uniform conductor,
so
The SI unit of resistivity is
Conductivity is the reciprocal of resistivity.
Its SI unit is siemens per metre:
5.9 Dependence of Resistance on Dimensions
For a uniform wire,
This relation reveals two important geometric dependences:
and
A longer wire has greater resistance because charge carriers must move through more material. A thicker wire has lower resistance because the larger cross-sectional area provides more conducting pathways.
Question: A wire has resistivity 1.7×10-8 Ωm, length 5 m and cross-sectional area 2×10-6 m2. Find its resistance.
Formula:
Solution:
Answer: The resistance is 0.0425 Ω.
Question: A wire has resistance 12 Ω. Another wire of the same material and cross-sectional area has three times the length. What is its resistance, assuming the same temperature?
Principle:
Solution:
Answer: The resistance is 36 Ω.
5.10 Temperature Dependence of Resistance
The resistance of a material generally depends on temperature. For many metals over a moderate temperature range,
where R0 is the resistance at reference temperature T0 and α is the temperature coefficient of resistance.
For ordinary metallic conductors, resistance generally increases with increasing temperature because lattice vibrations increase and impede electron motion.
Question: A metal resistor has resistance 10 Ω at 20°C. Its temperature coefficient is 0.004 °C-1. Estimate the resistance at 70°C.
Formula:
Solution:
Answer: The estimated resistance is 12 Ω.
5.11 Resistors in Series
Resistors are in series when they are connected end-to-end so that the same current passes through each resistor.
The total potential difference is
Using V=IR,
Therefore,
Question: Resistors of 4 Ω, 6 Ω and 10 Ω are connected in series across a 40 V supply. Find the total current.
Equivalent Resistance:
Current:
Answer: The circuit current is 2 A.
Question: For the circuit in Worked Example 11, find the voltage across the 6 Ω resistor.
Formula:
Solution:
Answer: The potential difference across the 6 Ω resistor is 12 V.
5.12 Resistors in Parallel
Resistors are in parallel when their corresponding terminals are connected to the same pair of nodes. Each branch therefore has the same potential difference.
The total current is the sum of the branch currents:
Using Ohm's law,
Hence,
For two resistors,
The equivalent resistance of resistors connected in parallel is always less than the smallest individual branch resistance.
Question: Resistors of 6 Ω and 3 Ω are connected in parallel. Find their equivalent resistance.
Formula:
Solution:
Answer: The equivalent resistance is 2 Ω.
Question: The 6 Ω and 3 Ω resistors are connected in parallel across 12 V. Find the branch currents and total current.
Current Through 6 Ω:
Current Through 3 Ω:
Total Current:
Answer: The branch currents are 2 A and 4 A, while the total current is 6 A.
5.13 Mixed Resistor Networks
Many practical circuits contain combinations of series and parallel resistors. Such circuits are simplified in stages by identifying groups that are unmistakably in series or parallel.
Question: A 4 Ω resistor is connected in series with a parallel combination of 6 Ω and 3 Ω. Find the total resistance.
Parallel Section:
Series Combination:
Answer: The total resistance is 6 Ω.
Question: The circuit in Worked Example 15 is connected to an 18 V ideal source. Find the total current and the voltage across the parallel section.
Total Current:
Voltage Across 4 Ω:
Voltage Across Parallel Section:
Answer: Total current = 3 A; voltage across the parallel combination = 6 V.
5.14 Electrical Power
Electrical power is the rate at which electrical energy is transferred or converted.
If a charge Q moves through potential difference V, the energy transferred is
Since I=Q/t,
For a resistor obeying Ohm's law,
Question: An electrical device operates at 12 V and draws 3 A. Find the power.
Formula:
Solution:
Answer: The power is 36 W.
Question: A 10 Ω resistor carries 2 A. Determine the power dissipated.
Formula:
Solution:
Answer: The resistor dissipates 40 W.
5.15 Electrical Energy and Heating Effect
When current flows through a resistance, electrical energy is commonly converted into thermal energy. For a constant current,
Therefore,
For a resistor,
or
This conversion of electrical energy into thermal energy is known as Joule heating.
Question: A 1000 W electric heater operates for 15 minutes. Determine the electrical energy converted.
Given:
Formula:
Solution:
Answer: The energy converted is 9.0×105 J.
Kilowatt-Hour:
Electrical energy supplied to homes and industries is often measured in kilowatt-hours rather than joules.
Question: A 2 kW appliance operates for 3 h. Determine the energy consumed in kilowatt-hours and joules.
Energy in kWh:
Energy in Joules:
Answer: The energy consumed is 6 kWh, equivalent to 2.16×107 J.
5.16 Cells, Batteries and Internal Resistance
A real cell or battery does not behave exactly like an ideal source. The materials within the source oppose current flow and can be represented by an internal resistance.
Internal resistance is the effective resistance within a source of emf that causes part of the source's energy per unit charge to be dissipated internally when current flows.
A practical cell may be modelled as an ideal emf ε in series with internal resistance r.
If the external load resistance is R, the total circuit resistance is
The current is therefore
Question: A cell has emf 12 V and internal resistance 1 Ω. It is connected to an external resistance of 5 Ω. Find the current.
Formula:
Solution:
Answer: The circuit current is 2 A.
5.17 Terminal Potential Difference
When a cell supplies current, some potential is lost across its internal resistance. The potential difference measured across its external terminals is therefore
This quantity is called the terminal potential difference.
Since the external load has resistance R,
Thus,
Question: A battery has emf 10 V and internal resistance 0.5 Ω. It supplies a current of 4 A. Find its terminal potential difference.
Formula:
Solution:
Answer: The terminal potential difference is 8 V.
If no current is drawn from a cell, I=0, so there is no internal voltage drop and the terminal potential difference equals the emf.
5.18 Measuring Current and Voltage
Ammeter:
An ammeter measures current and is connected in series with the component whose current is required.
An ideal ammeter has zero resistance so that its insertion does not significantly alter the circuit current.
Voltmeter:
A voltmeter measures potential difference and is connected in parallel across the two points of interest.
An ideal voltmeter has infinite resistance so that it draws no current from the circuit.
| Instrument | Measures | Connection | Ideal Resistance |
|---|---|---|---|
| Ammeter | Current | Series | Zero |
| Voltmeter | Potential difference | Parallel | Infinite |
5.19 Applications of DC Circuits
Direct-current circuits are fundamental in battery-powered electronics, automotive systems, control circuits, portable devices and many laboratory instruments.
Understanding current, resistance and power allows engineers and scientists to select conductors appropriately, control current levels, divide voltages, estimate energy consumption and protect equipment against excessive heating.
Question: A 24 V source is connected to a 12 Ω resistor. Determine the current, power dissipated and energy converted in 5 minutes.
Current:
Power:
Time:
Energy:
Answer: Current = 2 A; power = 48 W; energy converted = 1.44×104 J.
Chapter Summary
- Electric current is the rate of flow of electric charge and is defined by I=dQ/dt.
- The SI unit of current is the ampere, equivalent to one coulomb per second.
- Conventional current is opposite to electron drift in metallic conductors.
- Current density is J=I/A for uniform current distribution.
- For charge carriers, I=nqAvd.
- Electromotive force is energy supplied per unit charge and is measured in volts.
- Resistance measures opposition to current flow.
- Ohm's law for an ohmic conductor is V=IR.
- Resistance of a uniform conductor is R=ρL/A.
- Conductivity is the reciprocal of resistivity.
- For many metals, resistance increases approximately linearly with temperature over a limited range.
- Series resistances add directly.
- For parallel resistors, reciprocals of resistance add.
- Electrical power is P=VI=I2R=V2/R for an ohmic resistor.
- Electrical energy is E=Pt.
- A real source can be modelled by an emf and an internal resistance.
- For a source delivering current, terminal potential difference is V=ε-Ir.
- An ammeter is connected in series, while a voltmeter is connected in parallel.
CHAPTER SIX
Kirchhoff's Laws and Analysis of Direct-Current Circuits
6.1 Introduction to Circuit Analysis
Simple circuits can often be reduced by combining resistors in series and parallel. More complicated networks cannot always be simplified by these methods alone. In such cases, a systematic analysis can be performed using Kirchhoff's laws.
Kirchhoff's circuit laws are direct consequences of two fundamental conservation principles. The junction rule follows from conservation of electric charge, while the loop rule follows from conservation of energy.
These laws make it possible to determine unknown currents and potential differences in circuits containing several branches, resistors and sources.
6.2 Nodes, Branches and Loops
Before applying Kirchhoff's laws, three circuit terms should be clearly understood.
A node is a point in a circuit at which two or more conducting branches meet.
A branch is a conducting path between two nodes and may contain one or more circuit elements.
A loop is any closed conducting path that begins and ends at the same point without requiring the path to reverse direction.
Correct identification of branches and independent loops is essential in multi-loop circuit analysis.
6.3 Kirchhoff's Junction Rule
At any junction in a steady-state electrical circuit, the total current entering the junction equals the total current leaving it.
Equivalently, if currents entering a junction are treated as positive and currents leaving as negative,
Question: Currents of 5 A and 3 A enter a junction. One branch carries 2 A away from the junction. Find the current leaving through the remaining branch.
Formula:
Solution:
Answer: The unknown current is 6 A leaving the junction.
6.4 Conservation of Charge in Circuits
The junction rule follows from conservation of electric charge. In a steady-state circuit, charge does not continuously accumulate at an ordinary junction. Therefore, whatever amount of charge arrives per unit time must leave per unit time.
Since current is charge flow rate,
the conservation of charge requires that the algebraic sum of currents at a junction be zero.
Kirchhoff's junction rule applies most directly to steady-state lumped circuits. If charge were accumulating at a junction, the current entering and leaving would not remain balanced.
6.5 Kirchhoff's Loop Rule
The algebraic sum of all potential changes around any closed loop in a circuit is zero.
As one moves around a loop, the potential may rise across a source and fall across resistive elements. After returning to the starting point, the net change in electric potential must be zero.
6.6 Conservation of Energy in Circuits
The loop rule is a consequence of energy conservation. A charge returning to its original point after travelling around a complete circuit loop must return to the same electric potential.
A source supplies energy per unit charge, while resistors convert electrical energy into thermal energy or other forms. Thus, in a simple loop,
For a source of emf ε and a resistor R carrying current I,
Therefore,
for an ideal source and single external resistor.
6.7 Sign Conventions for Loop Equations
Correct signs are essential when applying Kirchhoff's loop rule. The following conventions provide a consistent method.
Moving Through a Resistor:
If the chosen loop direction moves through a resistor in the same direction as the assumed current, the potential change is
If the loop direction moves through the resistor opposite to the assumed current, the potential change is
Moving Through a Source:
Moving from the negative terminal to the positive terminal of an ideal source gives a potential rise:
Moving from the positive terminal to the negative terminal gives a potential fall:
| Circuit Element | Traversal | Potential Change |
|---|---|---|
| Resistor | With assumed current | -IR |
| Resistor | Against assumed current | +IR |
| Source | Negative to positive terminal | +ε |
| Source | Positive to negative terminal | -ε |
A current direction may be assumed before solving the equations. If the calculated current is negative, the actual current flows opposite to the assumed direction. The negative result is not an error.
6.8 Single-Loop Circuits
Consider a simple loop containing an ideal source ε and two series resistors R1 and R2.
Applying the loop rule,
Therefore,
Question: A 12 V battery is connected in series with resistors of 2 Ω and 4 Ω. Find the current using Kirchhoff's loop rule.
Loop Equation:
Solution:
Answer: The current is 2 A.
6.9 Multi-Loop Circuit Analysis
A multi-loop circuit usually contains more than one source or more than one branch, making simple series-parallel reduction insufficient.
A systematic approach is:
- assign current variables to the required branches;
- choose convenient current directions;
- apply the junction rule where branches meet;
- select enough independent loops;
- write a loop equation for each independent loop;
- solve the resulting simultaneous equations.
The number of independent equations must equal the number of unknown currents.
6.10 Two-Loop Networks
Consider a circuit with two loops sharing a resistor. Let current I1 circulate in the left loop and current I2 circulate in the right loop. If both mesh currents pass through the shared resistor in opposite directions, the actual current through that resistor is the difference between the mesh currents.
for the chosen orientation.
Question: Two loops share a 2 Ω resistor. The left loop contains a 10 V source and an additional 3 Ω resistor. The right loop contains a 4 V source and an additional 4 Ω resistor. Assume clockwise mesh currents I1 in the left loop and I2 in the right loop, with the two mesh currents opposing each other in the shared resistor. Determine the mesh currents.
Left Loop:
Right Loop:
Solve Simultaneously:
Multiply the first equation by 3:
Add the second equation:
Substitute into the second equation:
Answer: The mesh currents are approximately I1=2.62 A and I2=1.54 A.
The current in the shared resistor is
6.11 Multiple Sources in a Circuit
When more than one source appears in a loop, the algebraic contribution of each emf depends on the direction in which the loop passes through that source.
Question: A single loop contains a 12 V source and a 5 V source connected in opposition, together with total resistance 7 Ω. Find the current.
Net Emf:
Loop Equation:
Answer: The current is 1 A, driven in the direction of the stronger source.
Question: Two sources of 6 V and 9 V are connected in series aiding one another with a total resistance of 5 Ω. Find the current.
Net Emf:
Current:
Answer: The current is 3 A.
6.12 Cells Connected in Series
Cells may be connected in series to increase the total emf available to a circuit.
For n identical cells, each of emf ε and internal resistance r, connected in series aiding,
and
If connected to an external resistance R, the current is
Question: Three identical cells, each of emf 1.5 V and internal resistance 0.2 Ω, are connected in series to a 3.4 Ω resistor. Find the current.
Equivalent Emf:
Equivalent Internal Resistance:
Total Resistance:
Current:
Answer: The current is 1.125 A.
6.13 Cells Connected in Parallel
Identical cells connected in parallel have the same equivalent emf as one cell, but their equivalent internal resistance is reduced.
For n identical cells,
and
This arrangement can be advantageous when supplying a relatively large current to a low-resistance load.
Question: Four identical 12 V batteries, each having internal resistance 0.8 Ω, are connected in parallel to a 3.8 Ω load. Find the current supplied to the load.
Equivalent Internal Resistance:
Total Resistance:
Current:
Answer: The load current is 3 A.
6.14 Potential Dividers
A potential divider uses resistors in series to obtain a desired fraction of a supply voltage.
For two resistors R1 and R2 in series across a supply V, the current is
The voltage across R2 is
Therefore,
Question: Two resistors of 2 kΩ and 4 kΩ are connected in series across an 18 V supply. Find the voltage across the 4 kΩ resistor.
Formula:
Solution:
Answer: The potential difference across the 4 kΩ resistor is 12 V.
The simple divider formula assumes that the output is not significantly loaded by another component. If a substantial load is connected across one resistor, the effective resistance changes and the circuit must be re-analysed.
6.15 Current Division
In a parallel network, total current divides among branches. The branch with lower resistance carries the larger current.
For two parallel resistors,
Because both branches have the same voltage,
Thus,
and
Question: A total current of 6 A enters two parallel resistors of 4 Ω and 8 Ω. Determine the current in each branch.
Current Through 4 Ω:
Current Through 8 Ω:
Check:
Answer: The 4 Ω branch carries 4 A, while the 8 Ω branch carries 2 A.
6.16 Wheatstone Bridge
The Wheatstone bridge is a circuit arrangement used for comparing resistances and determining an unknown resistance.
A basic bridge contains four resistors arranged in a diamond-shaped network and a detector connected between two intermediate junctions.
At balance, no current flows through the detector. If the four arms have resistances R1, R2, R3 and R4, the balance condition is
Equivalently,
Question: A balanced Wheatstone bridge has R1=100 Ω, R2=50 Ω, and R3=200 Ω. Find R4.
Balance Condition:
Substitution:
Answer: The unknown resistance is 100 Ω.
6.17 RC Circuits
A circuit containing both a resistor and capacitor is called an RC circuit. Unlike a purely resistive DC circuit, an RC circuit exhibits time-dependent behaviour while the capacitor charges or discharges.
When an initially uncharged capacitor is connected to a DC source through a resistor, current initially flows strongly. As charge accumulates on the capacitor, its voltage increases and the current gradually decreases.
When a charged capacitor is disconnected from its source and allowed to discharge through a resistor, both the capacitor charge and current decrease exponentially with time.
6.18 Charging of a Capacitor
Consider a series circuit containing a source of emf ε, a resistor R and a capacitor C.
Kirchhoff's loop rule gives
Since
the charging equation is
Its solution for an initially uncharged capacitor is
Since the final charge is
the charging equation may be written as
The capacitor voltage is
The charging current is
At t=0,
while after a very long time,
and
Question: A 10 μF capacitor is connected through a 2 MΩ resistor to a 12 V source. Find the maximum charge on the capacitor.
Formula:
Solution:
Answer: The maximum charge is 120 μC.
6.19 Discharging of a Capacitor
Suppose a capacitor initially carries charge Q0 and is then connected across a resistor with no external source.
The charge decreases according to
The capacitor voltage is
The magnitude of the discharge current is
where
Question: A capacitor initially charged to 100 V discharges through a resistor. If RC=5 s, find its voltage after 10 s.
Formula:
Solution:
Answer: The capacitor voltage is approximately 13.5 V.
6.20 Time Constant
The time constant of an RC circuit is the product of its resistance and capacitance.
The SI unit of RC is the second.
During charging, after one time constant,
Thus the capacitor has reached approximately 63.2% of its final charge.
During discharging,
Thus approximately 36.8% of the initial charge remains after one time constant.
Question: Find the time constant of a circuit containing a 4.7 kΩ resistor and a 100 μF capacitor.
Given:
Formula:
Solution:
Answer: The time constant is 0.47 s.
Question: A capacitor has final charge 50 μC when fully charged. Approximately how much charge does it contain after one time constant?
Formula:
Solution:
Answer: The capacitor contains approximately 31.6 μC.
6.21 Power and Energy in DC Networks
Kirchhoff's laws can also be checked through energy conservation. In a steady DC network, the total power supplied by sources equals the total power absorbed or dissipated by circuit elements, provided all energy transfers are included.
For a resistor,
For a source delivering current out of its positive terminal, the rate at which it supplies electrical energy is
Question: A 12 V ideal battery supplies a current of 2 A through two series resistors of 2 Ω and 4 Ω. Verify power conservation.
Power Supplied:
Power in 2 Ω Resistor:
Power in 4 Ω Resistor:
Total Power Dissipated:
Answer: The source supplies 24 W, exactly equal to the 24 W dissipated by the resistors.
6.22 Practical Circuit Analysis
Successful circuit analysis requires more than memorising formulas. A circuit should first be examined to determine its actual topology: which elements share the same current, which elements share the same voltage, and which nodes and independent loops are present.
For complex DC circuits, a reliable approach is to identify known quantities, assign current directions, simplify any obvious series or parallel groups, and then apply Kirchhoff's laws to the remaining network.
Question: A 20 V source with internal resistance 1 Ω supplies two external resistors of 6 Ω and 3 Ω connected in parallel. Determine the total current, terminal voltage and current through each external resistor.
Equivalent External Resistance:
Total Circuit Resistance:
Total Current:
Terminal Voltage:
Current Through 6 Ω:
Current Through 3 Ω:
Check:
The small difference from 6.67 A results from rounding.
Answer: Total current ≈ 6.67 A; terminal voltage ≈ 13.33 V; branch currents ≈ 2.22 A and 4.44 A.
Question: At a junction, current I1 enters and currents I2 and I3 leave. If I1=7 A and I2=2.5 A, find I3.
Junction Rule:
Solution:
Answer: The third current is 4.5 A.
Question: A 9 V source with internal resistance 0.5 Ω is connected to an unknown external resistor. The current is 1.5 A. Determine the external resistance.
Formula:
Rearrange:
Solution:
Answer: The external resistance is 5.5 Ω.
Kirchhoff's laws extend the principles of current, voltage and resistance to circuits that cannot be solved by simple series-parallel reduction. The junction rule embodies conservation of charge, while the loop rule embodies conservation of energy. These laws provide the foundation for systematic electrical network analysis and lead naturally to the study of magnetic fields produced by currents.
Chapter Summary
- A node is a connection point, a branch is a path between nodes, and a loop is a closed circuit path.
- Kirchhoff's junction rule states that total current entering a junction equals total current leaving it.
- The junction rule follows from conservation of electric charge.
- Kirchhoff's loop rule states that the algebraic sum of potential changes around a closed loop is zero.
- The loop rule follows from conservation of energy.
- Across a resistor, potential falls in the direction of conventional current and rises when traversed opposite to the current.
- Across a source, traversal from negative to positive terminal gives a potential rise equal to the emf.
- Multi-loop circuits are solved using simultaneous junction and loop equations.
- A negative calculated current means that the actual current direction is opposite to the direction initially assumed.
- Cells connected in series have their emfs and internal resistances added.
- Identical cells connected in parallel retain the same emf as one cell while reducing effective internal resistance.
- A potential divider provides a selected fraction of an input voltage.
- Current divides inversely with branch resistance in a parallel network.
- A balanced Wheatstone bridge satisfies R1/R2=R3/R4.
- An RC circuit exhibits time-dependent charging and discharging.
- The RC time constant is τ=RC.
- During charging, Q=Qmax(1-e-t/RC).
- During discharging, Q=Q0e-t/RC.
- Power conservation provides an important check on DC circuit calculations.
CHAPTER SEVEN
Magnetic Fields and Magnetic Forces
7.1 Introduction to Magnetism
Magnetism is closely connected with moving electric charges. A stationary electric charge produces an electric field, whereas a moving charge produces both electric and magnetic effects. Electric currents, which consist of moving charges, therefore generate magnetic fields.
Magnetic phenomena occur in permanent magnets, current-carrying conductors, electric motors, generators, transformers, loudspeakers, particle accelerators and many other devices. At the microscopic level, magnetic behaviour in matter is associated principally with the magnetic moments of electrons arising from their orbital motion and intrinsic spin.
The central quantity used to describe magnetic interactions is the magnetic field, represented by the vector B.
7.2 Magnetic Field
A magnetic field is a vector field in which a moving electric charge, a current-carrying conductor or a magnetic dipole can experience a magnetic force or torque.
The SI unit of magnetic field is the tesla (T). From the magnetic force relation,
one tesla may be expressed as
Because magnetic field is a vector quantity, both magnitude and direction must be specified.
7.3 Magnetic Field Lines
Magnetic field lines provide a graphical representation of a magnetic field. The tangent to a field line at any point gives the direction of B at that point, while closer spacing of the lines represents a stronger field.
Outside a bar magnet, magnetic field lines are conventionally drawn from the north pole toward the south pole. Within the magnet, they continue from south to north, forming continuous closed loops.
Unlike electric field lines, magnetic field lines do not begin or terminate on isolated magnetic charges. No isolated magnetic monopole has been experimentally established in ordinary classical electromagnetism.
Magnetic field lines never intersect. If two lines intersected, the magnetic field would have two different directions at the same point, which is impossible.
7.4 Magnetic Flux
Magnetic flux is a measure of the magnetic field passing through a specified surface.
For a uniform magnetic field passing through a flat surface,
where θ is the angle between the magnetic field and the normal to the surface.
More generally,
The SI unit of magnetic flux is the weber:
Question: A uniform magnetic field of 0.40 T passes through a rectangular surface of area 0.20 m2. The field makes an angle of 60° with the normal to the surface. Determine the magnetic flux.
Given:
Required: Magnetic flux ΦB.
Formula:
Solution:
Answer: The magnetic flux is 0.040 Wb.
7.5 Magnetic Force on a Moving Charge
A charged particle moving through a magnetic field experiences a force when its velocity has a component perpendicular to the field.
The vector expression is
The magnitude is
Here θ is the angle between the velocity and magnetic field vectors.
Three important cases follow immediately:
| Orientation | Angle | Magnetic Force |
|---|---|---|
| Velocity parallel to field | 0° | Zero |
| Velocity perpendicular to field | 90° | Maximum: F=|q|vB |
| Velocity antiparallel to field | 180° | Zero |
Direction of Magnetic Force:
For a positive charge, the direction of v×B is determined using the right-hand rule. Point the fingers of the right hand in the direction of the velocity and curl them toward the magnetic field; the thumb indicates the direction of force for a positive charge.
For a negative charge, the force is opposite to the direction obtained from the right-hand rule.
Question: A proton moves at 3.0×106 m/s perpendicular to a magnetic field of 0.20 T. Calculate the magnetic force. Take e=1.602×10-19 C.
Given:
Formula:
Solution:
Answer: The magnitude of the magnetic force is approximately 9.61×10-14 N.
7.6 Lorentz Force
A charged particle may simultaneously experience electric and magnetic fields. The total electromagnetic force is called the Lorentz force.
The electric part, qE, can act on a charge whether it is moving or stationary. The magnetic part requires motion and depends on the orientation of the velocity relative to the magnetic field.
The electric force can do work on a charged particle and change its speed. A purely magnetic force is always perpendicular to the instantaneous velocity and therefore does no work on the particle.
Consequently, a magnetic field alone can change the direction of a particle's velocity but not its kinetic energy or speed.
7.7 Motion of Charged Particles in Magnetic Fields
The path of a charged particle in a magnetic field depends on the orientation of its velocity.
If the velocity is parallel to the magnetic field, no magnetic force acts and the particle continues in a straight line at constant speed, provided no other force is present.
If the velocity is perpendicular to the magnetic field, the magnetic force is always perpendicular to the motion and acts as a centripetal force. The particle therefore follows a circular path.
If the velocity has both parallel and perpendicular components, the perpendicular component produces circular motion while the parallel component remains unchanged. The resulting trajectory is a helix.
7.8 Circular Motion of Charged Particles
For a charged particle moving perpendicular to a uniform magnetic field,
The centripetal force required for circular motion is
Equating the two forces,
Therefore,
The radius increases with mass and speed but decreases as the charge magnitude or magnetic field strength increases.
Question: A proton of mass 1.67×10-27 kg moves perpendicular to a 0.50 T magnetic field at 2.0×106 m/s. Find the radius of its circular path.
Given:
Formula:
Solution:
Answer: The radius is approximately 4.17 cm.
Period and Frequency:
The period of the circular motion is
Substituting r=mv/(|q|B),
The frequency is therefore
In the non-relativistic approximation, the period and frequency are independent of the particle's speed.
7.9 Velocity Selector
A velocity selector uses mutually perpendicular electric and magnetic fields to allow charged particles of a particular speed to pass through without deflection.
For the selected particle, the electric and magnetic forces have equal magnitudes and opposite directions:
Hence,
Question: A velocity selector has an electric field of 6.0×104 N/C and a perpendicular magnetic field of 0.20 T. Find the speed of particles that pass through undeflected.
Formula:
Solution:
Answer: The selected speed is 3.0×105 m/s.
7.10 Magnetic Force on a Current-Carrying Conductor
A current-carrying wire placed in a magnetic field experiences a force because the moving charge carriers within the wire experience magnetic forces.
For a straight conductor of vector length L,
The magnitude is
Here θ is the angle between the direction of conventional current and the magnetic field.
Question: A straight wire 0.50 m long carries a current of 4 A perpendicular to a uniform magnetic field of 0.30 T. Find the magnetic force.
Formula:
Solution:
Answer: The magnetic force is 0.60 N.
7.11 Force Between Parallel Current-Carrying Conductors
A current in a wire produces a magnetic field. Consequently, when two parallel wires carry currents, each wire lies in the magnetic field produced by the other and experiences a magnetic force.
For two long, straight parallel conductors separated by distance d, the force per unit length is
where
Parallel currents in the same direction attract, whereas currents in opposite directions repel.
Question: Two long parallel wires are 0.10 m apart and carry currents of 5 A and 8 A in the same direction. Find the force per metre between them.
Formula:
Solution:
Answer: The force per unit length is 8.0×10-5 N/m, and the wires attract.
7.12 Torque on a Current Loop
A current loop placed in a uniform magnetic field can experience a torque. This principle is fundamental to electric motors and moving-coil instruments.
For a rectangular coil containing N turns, each of area A, carrying current I,
where θ is the angle between the normal to the plane of the coil and the magnetic field.
The torque is maximum when θ=90° and zero when the magnetic dipole moment is parallel or antiparallel to the field.
Question: A coil of 20 turns has area 0.015 m2 and carries a current of 2 A. It is placed in a 0.40 T magnetic field with its normal perpendicular to the field. Determine the torque.
Formula:
Solution:
Answer: The torque is 0.24 N·m.
7.13 Magnetic Dipole Moment
The magnetic dipole moment of a current loop is a vector quantity that characterizes the magnetic strength and orientation of the loop.
For a coil of N turns,
and its magnitude is
The direction of μ is perpendicular to the plane of the loop and is determined by the right-hand rule.
The torque may therefore be written compactly as
7.14 Magnetic Potential Energy
A magnetic dipole in an external magnetic field possesses orientation-dependent potential energy:
For a uniform field,
The potential energy is minimum when the magnetic dipole moment is aligned with the field and maximum when it is oppositely directed.
7.15 Applications of Magnetic Forces
The principles developed in this chapter have numerous applications. Electric motors use torque on current-carrying coils to produce rotational motion. Charged-particle beams are controlled by magnetic fields in accelerators and analytical instruments. Velocity selectors separate particles according to speed, while related arrangements allow mass-to-charge ratios to be measured.
Magnetic forces also underlie the operation of loudspeakers, moving-coil meters and many electromechanical systems. The important point is that magnetic fields interact with moving charges and currents in predictable ways governed by vector relationships.
Chapter Summary
- A magnetic field is represented by the vector B and measured in tesla.
- Magnetic field lines form continuous closed curves and do not intersect.
- Magnetic flux is ΦB=∫B·dA; for a uniform field and flat surface, ΦB=BAcosθ.
- The magnetic force on a moving charge is F=qv×B.
- A magnetic field alone does no work on a charged particle.
- A charge moving perpendicular to a uniform magnetic field follows a circular path of radius r=mv/(|q|B).
- A velocity selector transmits undeflected particles satisfying v=E/B.
- The force on a straight current-carrying conductor is F=IL×B.
- Parallel currents in the same direction attract; opposite currents repel.
- A current loop in a magnetic field experiences torque τ=NIABsinθ.
- The magnetic dipole moment of a coil is μ=NIA.
- The potential energy of a magnetic dipole is U=-μBcosθ.
CHAPTER EIGHT
Sources of Magnetic Fields: Biot-Savart and Ampère's Laws
8.1 Magnetic Fields Produced by Electric Currents
The preceding chapter considered how magnetic fields exert forces on moving charges and currents. We now examine the complementary question: how are magnetic fields produced?
An electric current produces a magnetic field in the surrounding space. Since electric current is an organized flow of charge, this observation establishes the fundamental connection between moving electric charges and magnetism.
Two important laws are used to calculate magnetic fields generated by steady currents: the Biot-Savart law and Ampère's circuital law. The Biot-Savart law is generally applicable to steady-current distributions, whereas Ampère's law is especially convenient when the geometry possesses sufficient symmetry.
8.2 Biot-Savart Law
The Biot-Savart law gives the contribution to the magnetic field at a point due to an infinitesimal element of a steady current.
In vector form,
The magnitude is
where:
- I is the current;
- dℓ is the length of the current element;
- r is the distance from the current element to the observation point;
- θ is the angle between dℓ and the displacement toward the observation point;
- r̂ is a unit vector directed from the current element to the observation point.
The total field is obtained by integrating over the complete current distribution:
The direction of the field contribution follows from the cross product and can be found using the right-hand rule.
8.3 Magnetic Field of a Long Straight Conductor
For a very long straight wire carrying current I, integration of the Biot-Savart law gives the magnetic field at perpendicular distance r:
The field lines are concentric circles centred on the wire.
Direction of the Field:
The direction is obtained using the right-hand grip rule. Point the thumb of the right hand in the direction of conventional current; the curled fingers show the direction of the magnetic field around the wire.
Question: A long straight wire carries a current of 10 A. Determine the magnetic field at a point 5.0 cm from the wire.
Given:
Formula:
Solution:
Answer: The magnetic field is 4.0×10-5 T, or 40 μT.
Question: At what distance from a long straight wire carrying 15 A is the magnetic field 3.0×10-5 T?
Formula:
Required: Rearrange for r.
Solution:
Answer: The point is 0.10 m, or 10 cm, from the wire.
8.4 Magnetic Field of a Circular Current Loop
A circular loop carrying current produces a magnetic field whose direction along the axis of the loop follows the right-hand rule.
At the centre of a circular loop of radius R, the Biot-Savart law gives
For a closely wound coil containing N turns,
Question: A circular coil has 50 turns, radius 0.10 m, and carries a current of 2 A. Determine the magnetic field at its centre.
Formula:
Solution:
Answer: The magnetic field at the centre is approximately 6.28×10-4 T.
8.5 Magnetic Field on the Axis of a Circular Loop
The magnetic field at a point on the axis of a circular loop is also obtained from the Biot-Savart law.
For a loop of radius R, carrying current I, at an axial distance x from the centre,
For N closely wound turns,
At the centre, x=0, and the expression reduces to
8.6 Magnetic Field of a Current Arc
For a circular arc of radius R carrying current I and subtending an angle φ in radians at its centre, the magnetic field at the centre is
For a complete circle, φ=2π, giving
as expected.
Question: A semicircular wire of radius 0.20 m carries a current of 5 A. Find the magnetic field at the centre due to the semicircular arc alone.
Given:
Formula:
Solution:
Answer: The magnetic field at the centre is approximately 7.85 μT.
8.7 Superposition of Magnetic Fields
When several currents contribute to the magnetic field at the same point, the resultant field is the vector sum of the individual fields:
The directions must be considered carefully. Fields in the same direction add; fields in opposite directions subtract. For non-collinear fields, vector addition is required.
Question: Two long parallel wires are separated by 0.20 m and carry equal currents of 10 A in the same direction. Determine the net magnetic field at the midpoint between them.
The midpoint is 0.10 m from each wire.
Field Due to Either Wire:
By the right-hand rule, at the midpoint the two fields point in opposite directions because the currents are parallel and in the same direction.
Answer: The net magnetic field at the midpoint is zero.
8.8 Ampère's Circuital Law
For steady currents, the line integral of the magnetic field around a closed path equals the permeability of free space multiplied by the net current enclosed by that path.
Ampère's law is always valid for magnetostatic situations, but it is most useful computationally when symmetry allows the magnitude of B to be treated as constant along a suitable integration path.
Common high-symmetry applications include:
- a long straight conductor;
- a long solenoid;
- a toroid;
- highly symmetric current distributions.
8.9 Ampère's Law for a Long Straight Wire
Consider a long straight wire carrying current I. Choose a circular Amperian path of radius r centred on the wire.
By symmetry, B has the same magnitude everywhere on this circle and is tangent to the path. Therefore,
Since the circumference is 2πr,
Hence,
This agrees with the result obtained using the Biot-Savart law.
8.10 Magnetic Field Inside a Long Solenoid
A solenoid is a long cylindrical coil consisting of many closely spaced turns of conducting wire.
For an ideal long solenoid, the magnetic field inside is approximately uniform and parallel to the axis, while the external field is comparatively weak.
If the solenoid contains N turns over length L, the number of turns per unit length is
Ampère's law gives the field inside an ideal long solenoid as
Equivalently,
Question: A long solenoid contains 1200 turns uniformly distributed over a length of 0.60 m. It carries a current of 1.5 A. Find the magnetic field inside the solenoid, assuming an air core.
Given:
Turns per Unit Length:
Formula:
Solution:
Answer: The magnetic field is approximately 3.77×10-3 T, or 3.77 mT.
8.11 Magnetic Field of a Toroid
A toroid is essentially a solenoid bent into a closed circular form so that its windings form a ring.
For an ideal toroid containing N turns carrying current I, choose a circular Amperian path of radius r inside the windings.
Ampère's law gives
Therefore,
Unlike the ideal long solenoid, the magnetic field inside a toroid varies with radial distance r.
Question: An air-core toroid has 500 turns and carries a current of 2 A. Find the magnetic field at a radius of 0.10 m within the toroid.
Formula:
Solution:
Answer: The magnetic field is 2.0 mT.
8.12 Magnetic Dipoles and Current Loops
A small current loop behaves as a magnetic dipole. Its magnetic dipole moment is
This provides a useful bridge between current-generated magnetic fields and the behaviour of magnetic materials. At distances much larger than the dimensions of the loop, the field pattern resembles that of a small bar magnet.
The magnetic dipole model is also useful in describing microscopic magnetic effects in matter, although atomic magnetism ultimately requires quantum mechanics for a complete description.
8.13 Magnetic Properties of Matter
Materials respond differently when placed in external magnetic fields. Their behaviour depends on atomic and molecular magnetic moments and on how these moments respond collectively to an applied field.
At introductory level, magnetic materials are commonly classified as:
- diamagnetic;
- paramagnetic;
- ferromagnetic.
These categories describe substantially different responses to an applied magnetic field.
8.14 Diamagnetic, Paramagnetic and Ferromagnetic Materials
Diamagnetic Materials:
Diamagnetic materials develop an induced magnetic response opposite to the applied field and are therefore weakly repelled from regions of stronger magnetic field.
Examples include bismuth, copper, silver and water.
Paramagnetic Materials:
Paramagnetic substances possess microscopic magnetic moments that tend to align partially with an applied magnetic field. Thermal agitation prevents complete alignment under ordinary conditions, so the attraction is usually weak.
Examples include aluminium and oxygen.
Ferromagnetic Materials:
Ferromagnetic materials exhibit strong collective alignment of magnetic moments. They can become strongly magnetized and may retain significant magnetization after the external field is removed.
Iron, cobalt and nickel are familiar examples.
| Property | Diamagnetic | Paramagnetic | Ferromagnetic |
|---|---|---|---|
| Response to external field | Weak opposition | Weak alignment | Strong alignment |
| Typical interaction | Weak repulsion | Weak attraction | Strong attraction |
| Persistence after field removal | No significant permanent magnetization | No significant permanent magnetization under ordinary conditions | May retain magnetization |
| Examples | Copper, bismuth, water | Aluminium, oxygen | Iron, cobalt, nickel |
8.15 Magnetic Energy Density
Energy can be stored in a magnetic field. In vacuum, the magnetic energy density is
This quantity gives the magnetic energy stored per unit volume of space.
Question: Determine the magnetic energy density in vacuum where the magnetic field is 0.20 T.
Formula:
Solution:
Answer: The magnetic energy density is approximately 1.59×104 J/m3.
8.16 Applications of Current-Produced Magnetic Fields
Current-generated magnetic fields form the operating basis of electromagnets, relays, motors, magnetic actuators, inductors, transformers and numerous measurement instruments.
A solenoid provides a useful means of producing an approximately uniform magnetic field within a cylindrical region. Introducing a suitable magnetic core can greatly modify and often strengthen the field. Toroidal windings are valuable where magnetic flux is intended to remain largely confined to a closed magnetic path.
The Biot-Savart law and Ampère's law are complementary rather than competing laws. The Biot-Savart law is particularly useful when the detailed contribution from current elements must be integrated, whereas Ampère's law provides a much shorter route for highly symmetric current distributions.
Question: A long straight wire carries 20 A. What current must another long straight wire carry to produce the same magnetic field at twice the perpendicular distance?
Principle:
For the fields to be equal,
Given
Solution:
Answer: The second wire must carry 40 A.
Question: A long solenoid is required to produce a magnetic field of 5.0×10-3 T. If it has 2500 turns per metre, determine the required current.
Given:
Formula:
Required:
Solution:
Answer: The required current is approximately 1.59 A.
Chapter Summary
- Steady electric currents produce magnetic fields.
- The Biot-Savart law relates an infinitesimal current element to its contribution to the magnetic field.
- The field of a long straight wire is B=μ0I/(2πr).
- The magnetic field at the centre of a circular loop is B=μ0I/(2R).
- For N turns, the centre field becomes B=μ0NI/(2R).
- The field of a circular arc subtending angle φ is B=μ0Iφ/(4πR).
- Magnetic fields obey the principle of vector superposition.
- Ampère's circuital law for steady currents is ∮B·dℓ=μ0Ienc.
- For an ideal long solenoid, B=μ0nI.
- For an ideal toroid, B=μ0NI/(2πr) within the windings.
- A current loop behaves as a magnetic dipole with magnetic moment μ=NIA.
- Materials may be classified according to magnetic response as diamagnetic, paramagnetic or ferromagnetic.
- The magnetic energy density in vacuum is uB=B2/(2μ0).
- Biot-Savart and Ampère's laws provide complementary methods for determining magnetic fields produced by electric currents.
CHAPTER NINE
Electromagnetic Induction and Faraday-Lenz Laws
9.1 Introduction to Electromagnetic Induction
A magnetic field can exert forces on moving charges, but the relationship between electricity and magnetism goes further. A changing magnetic environment can produce an electric effect even when no battery or electrostatic source is present. This phenomenon is known as electromagnetic induction.
Electromagnetic induction is the production of an electromotive force in a conductor or circuit when the magnetic flux linked with the conductor or circuit changes.
The induced electromotive force may result from a changing magnetic field, motion of a conductor through a magnetic field, rotation of a coil, or any process that changes the magnetic flux linkage.
Electromagnetic induction is the operating principle behind electric generators, transformers, induction cookers, many sensors, wireless charging systems and numerous electrical machines.
9.2 Magnetic Flux and Flux Linkage
Magnetic flux through a surface was introduced earlier as
For a uniform magnetic field passing through a flat surface,
where θ is the angle between the magnetic field and the normal to the surface.
For a coil containing N turns, each turn links magnetic flux ΦB. The total flux linkage is
Question: A coil of 200 turns has area 4.0×10-3 m2 and is placed perpendicular to a uniform magnetic field of 0.50 T. Find the magnetic flux through one turn and the total flux linkage.
Given:
The field is parallel to the area vector, so θ=0°.
Flux Through One Turn:
Flux Linkage:
Answer: Flux per turn = 2.0×10-3 Wb; total flux linkage = 0.40 Wb-turn.
9.3 Faraday's Law of Electromagnetic Induction
The magnitude of the induced electromotive force in a circuit is equal to the rate of change of magnetic flux linkage through the circuit.
For a coil of N turns,
For an average change over a finite time interval,
The negative sign represents Lenz's law and determines the direction of the induced emf.
Question: The magnetic flux through each turn of a 150-turn coil decreases uniformly from 8.0×10-4 Wb to 2.0×10-4 Wb in 0.030 s. Find the magnitude of the induced emf.
Given:
Formula:
Solution:
Answer: The induced emf has magnitude 3.0 V.
Ways of Changing Magnetic Flux:
Since
magnetic flux can change by changing:
- the magnetic field strength B;
- the area A enclosed by the circuit;
- the angle θ between the field and the area vector;
- the position of the coil in a non-uniform magnetic field.
9.4 Lenz's Law
The induced current flows in such a direction that the magnetic field produced by the induced current opposes the change in magnetic flux that produced it.
The word change is essential. The induced field does not necessarily oppose the original magnetic field. It opposes the increase or decrease in magnetic flux.
For example, if magnetic flux directed into a loop increases, the induced current produces a field directed out of the loop. If the inward flux decreases, the induced current produces an inward field in an attempt to maintain the original flux.
Lenz's law is a direct consequence of conservation of energy. If the induced current reinforced the flux change that produced it, energy could increase spontaneously without external work.
9.5 Direction of Induced Current
The direction of induced current can be found systematically:
- determine the direction of the original magnetic flux;
- determine whether that flux is increasing or decreasing;
- choose the direction of the induced magnetic field required to oppose the change;
- use the right-hand grip rule to determine the corresponding current direction.
Question: A magnetic field directed into the page through a conducting loop is increasing. What is the direction of the induced current?
Reasoning: The inward magnetic flux is increasing. The induced field must therefore point out of the page in order to oppose the increase.
Using the right-hand rule, a magnetic field out of the page is produced by a counterclockwise current.
Answer: The induced current flows counterclockwise.
9.6 Motional Electromotive Force
An emf may be induced when a conductor moves through a magnetic field. The charge carriers within the moving conductor experience the magnetic force
Positive and negative charges are driven toward opposite ends of the conductor, producing charge separation and therefore a potential difference.
Motional emf is an emf produced when a conductor moves through a magnetic field in such a way that free charges experience a magnetic force along the conductor.
For a straight rod of length ℓ moving with speed v perpendicular both to its length and to a uniform magnetic field B,
More generally, if the relevant angle is θ,
Question: A metal rod of length 0.80 m moves at 5.0 m/s perpendicular to a uniform magnetic field of 0.60 T. Find the induced emf.
Formula:
Solution:
Answer: The induced emf is 2.4 V.
9.7 Moving Conducting Rod in a Magnetic Field
Consider a conducting rod sliding along conducting rails in a uniform magnetic field perpendicular to the plane of the rails. If the rails form a closed circuit with total resistance R, the induced current is
Using ε=Bℓv,
The induced current in the rod experiences a magnetic force
Substituting for I,
The force acts opposite to the motion, in agreement with Lenz's law.
Question: A conducting rod of length 0.50 m moves at 4.0 m/s across rails in a uniform magnetic field of 0.80 T. The total circuit resistance is 2.0 Ω. Find the induced emf and current.
Emf:
Current:
Answer: Induced emf = 1.6 V; induced current = 0.80 A.
9.8 Induced Electric Fields
Faraday's law has a deeper interpretation than the production of current in wires. A changing magnetic field produces an electric field in the surrounding space.
The electric field produced by changing magnetic flux is generally non-conservative and forms closed loops. Faraday's law may therefore be written as
This equation shows that a changing magnetic field can create an electric field even where no conducting wire is present.
This idea becomes one of the fundamental components of Maxwell's equations.
9.9 Rotating Coil and Alternating Emf
Consider a coil of N turns, each of area A, rotating with constant angular speed ω in a uniform magnetic field B.
If the angle between the field and the area vector is
then the flux through one turn is
The total flux linkage is
From Faraday's law,
Therefore,
The maximum emf is
Thus,
This is a sinusoidally alternating emf.
Question: A coil of 100 turns, each of area 0.020 m2, rotates at angular speed 50 rad/s in a uniform magnetic field of 0.30 T. Find the maximum induced emf.
Formula:
Solution:
Answer: The maximum induced emf is 30 V.
9.10 Electric Generator Principle
An electric generator is a device that converts mechanical energy into electrical energy through electromagnetic induction.
In a simple generator, a coil rotates in a magnetic field. Because the magnetic flux through the coil changes continuously, an emf is induced.
A basic alternating-current generator produces an emf of the form
Mechanical work must be supplied to keep the coil rotating because the induced current produces magnetic effects that oppose the rotation. This is an application of Lenz's law and energy conservation.
9.11 Eddy Currents
Eddy currents are circulating currents induced within bulk conducting materials when the magnetic flux through different regions of the conductor changes.
Eddy currents are not confined to a single wire path. They circulate through the conducting material itself.
Like all induced currents, their direction is determined by Lenz's law. They produce magnetic fields that oppose the change responsible for their creation.
9.12 Applications of Eddy Currents
Eddy currents may be useful or undesirable depending on the application.
Induction Heating:
Alternating magnetic fields induce strong eddy currents in conducting materials. The electrical resistance of the material converts the induced electrical energy into thermal energy.
Electromagnetic Damping:
Eddy currents can oppose the motion of conducting components and reduce unwanted oscillations in measuring instruments and mechanical systems.
Metal Detection:
Changing magnetic fields induce currents in nearby conductive objects. The magnetic fields produced by these currents can be detected electronically.
Energy Losses:
In transformer and machine cores, eddy currents can produce unwanted heating. These losses can be reduced by constructing cores from thin insulated laminations rather than solid blocks.
9.13 Electromagnetic Braking
Electromagnetic braking is a practical application of Lenz's law. When a conducting disc or plate moves through a magnetic field, eddy currents are induced. Their magnetic interaction with the applied field produces a force opposing the motion.
Because there is no requirement for mechanical contact between rubbing surfaces, electromagnetic braking can reduce mechanical wear.
However, the lost mechanical energy is converted primarily into thermal energy in the conductor.
9.14 Energy Conservation in Electromagnetic Induction
When a conductor moves through a magnetic field and an induced current flows, the induced magnetic force opposes the motion. External work must therefore be done to maintain constant speed.
For a rod of length ℓ moving with speed v through magnetic field B in a circuit of resistance R,
and
The magnetic force opposing the motion is
Therefore,
The mechanical power required to maintain the motion is
Hence,
Electrical power dissipated is
Substituting I=Bℓv/R,
Thus,
The mechanical energy supplied is converted into electrical and ultimately thermal energy.
Question: A rod of length 0.40 m moves at 6.0 m/s through a magnetic field of 0.50 T. The circuit resistance is 3.0 Ω. Determine the induced current and electrical power dissipated.
Emf:
Current:
Power:
Answer: Induced current = 0.40 A; electrical power dissipated = 0.48 W.
9.15 Practical Applications of Electromagnetic Induction
Electromagnetic induction is central to modern electrical technology. Generators convert mechanical energy to electrical energy. Transformers use changing magnetic flux to transfer electrical energy between circuits. Induction cookers produce heating directly in conducting cookware. Dynamic microphones and magnetic pickups convert mechanical vibrations into electrical signals.
Inductive sensors detect position, speed, metal objects and changes in magnetic environments. Wireless power transfer systems use changing magnetic fields to transfer energy between nearby coils without direct electrical contact.
Faraday's law tells us how rapidly changing magnetic flux produces emf, while Lenz's law determines the direction of the induced effect. Together, they provide a complete introductory description of electromagnetic induction and lead directly to inductance and transformers.
Chapter Summary
- Electromagnetic induction is the production of emf when magnetic flux linkage changes.
- Magnetic flux for a uniform field through a flat surface is ΦB=BAcosθ.
- Flux linkage for an N-turn coil is NΦB.
- Faraday's law is ε=-N dΦB/dt.
- Lenz's law states that induced current opposes the change in magnetic flux that produced it.
- Changing field strength, area, orientation or position can change magnetic flux.
- A conductor moving through a magnetic field can develop motional emf.
- For a rod moving appropriately through a uniform field, ε=Bℓv.
- A changing magnetic field produces a circulating electric field described by ∮E·dℓ=-dΦB/dt.
- A rotating coil in a uniform magnetic field produces sinusoidal emf ε=NBAωsinωt.
- The maximum generator emf is ε0=NBAω.
- Eddy currents are circulating currents induced in bulk conductors.
- Eddy currents are useful in induction heating and electromagnetic braking but can also produce undesirable energy losses.
- Lenz's law is consistent with conservation of energy.
CHAPTER TEN
Inductance, Magnetic Energy and Transformers
10.1 Introduction to Inductance
When current flows through a circuit, it produces a magnetic field. If the current changes, the magnetic field changes, and therefore the magnetic flux linked with the circuit also changes. According to Faraday's law, this changing flux produces an induced emf.
This effect gives rise to the important property known as inductance. Inductance describes how strongly a circuit responds electromagnetically to changes in current.
Inductance is the property of an electrical circuit by which a change in current produces an induced electromotive force that opposes the change in current.
10.2 Self-Induction
Self-induction is the production of an induced emf in a circuit due to a change in the current flowing through that same circuit.
Suppose current through a coil increases. The increasing current strengthens the coil's magnetic field and increases the magnetic flux through its turns. The resulting induced emf acts in a direction that opposes the increase in current.
If the current decreases, the induced emf acts in a direction that tends to maintain the current.
For this reason, an inductor tends to resist sudden changes in current.
10.3 Self-Inductance
For a coil in which the magnetic flux linkage is proportional to current,
where L is the self-inductance.
Therefore,
The SI unit of inductance is the henry (H).
From the induced-emf relation,
Question: A coil of 250 turns carries a current of 2 A. The magnetic flux through each turn is 4×10-4 Wb. Determine the inductance.
Formula:
Solution:
Answer: The inductance is 0.050 H, or 50 mH.
10.4 Inductance of a Solenoid
For a long air-core solenoid of length ℓ, cross-sectional area A, and N turns, the magnetic field inside is approximately
The magnetic flux through one turn is
Therefore,
The flux linkage is
Since NΦB=LI,
If a magnetic material is present and behaves approximately linearly, μ0 is replaced by the permeability μ of the material.
Question: An air-core solenoid has 1000 turns, length 0.50 m and cross-sectional area 2.0×10-4 m2. Find its inductance.
Formula:
Solution:
Answer: The inductance is approximately 5.03×10-4 H, or 0.503 mH.
10.5 Induced Emf in an Inductor
From Faraday's law and the definition of inductance,
The induced emf is proportional to the rate of change of current, not simply to the current itself.
A rapidly changing current therefore produces a larger self-induced emf than a slowly changing current.
Question: The current through a 0.20 H inductor decreases uniformly from 5 A to 1 A in 0.10 s. Find the magnitude of the induced emf.
Formula:
Solution:
Answer: The induced emf has magnitude 8 V.
10.6 RL Circuits
An RL circuit contains resistance R and inductance L. When such a circuit is connected to a DC source, current does not immediately reach its final steady value because the inductor produces a self-induced emf opposing the increase.
For a series RL circuit connected to a source of emf ε, Kirchhoff's loop rule gives
Equivalently,
This differential equation describes the growth of current.
10.7 Growth of Current in an RL Circuit
If the current is initially zero and the switch is closed at t=0, the current rises according to
where
and
Immediately after the switch is closed, the current is zero because the inductor initially opposes the sudden current change.
After a long time, dI/dt approaches zero, so the ideal inductor behaves like a conducting wire in steady DC operation and the current approaches
Question: A 12 V battery is connected in series with a 6 Ω resistor and an ideal inductor. Find the final steady current.
Formula:
Solution:
Answer: The final steady current is 2 A.
10.8 Decay of Current in an RL Circuit
When the source is removed from an RL circuit while a conducting path remains, the inductor attempts to maintain the existing current.
The current decays exponentially according to
where
Question: An RL circuit initially carries 4 A. Its time constant is 2.0 s. Find the current after 4.0 s after the source is removed.
Formula:
Solution:
Answer: The current is approximately 0.54 A.
10.9 RL Time Constant
The time constant of an RL circuit is the ratio of inductance to resistance.
Its SI unit is seconds.
During current growth, after one time constant,
During decay, after one time constant,
Question: A 0.60 H inductor is connected in series with a 3.0 Ω resistor. Find the RL time constant.
Formula:
Solution:
Answer: The time constant is 0.20 s.
10.10 Energy Stored in an Inductor
An inductor stores energy in its magnetic field while current flows through it.
The energy stored in an inductor is
The energy is zero when the current is zero and increases with the square of the current.
Question: A 0.50 H inductor carries a current of 4.0 A. Find the magnetic energy stored.
Formula:
Solution:
Answer: The stored energy is 4.0 J.
10.11 Energy Density in a Magnetic Field
The energy stored in an inductor may be regarded as energy stored throughout its magnetic field.
For a magnetic field in vacuum, the energy density is
This expression was introduced earlier and now receives a direct interpretation through the energy stored in an inductor.
For a linear magnetic medium of permeability μ, a corresponding expression is
10.12 Mutual Induction
A changing current in one circuit can induce an emf in a nearby second circuit because the changing current produces a changing magnetic field and therefore changing magnetic flux through the second circuit.
Mutual induction is the production of an emf in one circuit due to a changing current in a neighbouring circuit.
If changing current I1 in circuit 1 produces flux linkage in circuit 2, the induced emf in circuit 2 is
where M is the mutual inductance.
10.13 Mutual Inductance
Mutual inductance measures how effectively a changing current in one circuit produces magnetic flux linkage and induced emf in another circuit.
If current I1 produces flux Φ21 through each of N2 turns of a second coil, then
The SI unit of mutual inductance is also the henry.
Question: Two coils have mutual inductance 0.25 H. The current in the first coil changes uniformly at 8 A/s. Find the magnitude of the emf induced in the second coil.
Formula:
Solution:
Answer: The induced emf is 2 V.
10.14 The Transformer Principle
A transformer is an electrical device that transfers alternating electrical energy between circuits by mutual induction, usually while changing voltage and current levels.
A basic transformer consists of two coils wound around a common magnetic core:
- the primary coil, connected to the input supply;
- the secondary coil, connected to the output circuit.
Alternating current in the primary produces a changing magnetic flux in the core. This changing flux links the secondary coil and induces an alternating emf according to Faraday's law.
A steady direct current produces a constant magnetic field after the transient has ended. A constant magnetic flux does not continuously induce emf in the secondary. Transformers therefore operate on changing current, normally alternating current.
10.15 Ideal Transformer Equations
For an ideal transformer, assume that all magnetic flux produced by the primary links the secondary and that there are no energy losses.
The rms or instantaneous voltage ratio follows the turns ratio:
Equivalently,
Here:
- Vp = primary voltage;
- Vs = secondary voltage;
- Np = primary turns;
- Ns = secondary turns.
Question: A transformer has 500 primary turns and 2000 secondary turns. If the primary voltage is 120 V, find the secondary voltage for ideal operation.
Formula:
Solution:
Answer: The secondary voltage is 480 V.
10.16 Step-Up Transformers
A step-up transformer increases the voltage from primary to secondary.
For a step-up transformer,
and therefore
In an ideal transformer, increasing voltage is accompanied by a corresponding decrease in current.
10.17 Step-Down Transformers
A step-down transformer decreases the voltage from primary to secondary.
For a step-down transformer,
and therefore
Question: A transformer has 1200 turns on its primary and 100 turns on its secondary. It is connected to a 240 V AC source. Find the secondary voltage.
Formula:
Solution:
Answer: The secondary voltage is 20 V.
10.18 Current and Power Relations in Transformers
For an ideal transformer, input power equals output power:
Therefore,
Combining this with the turns ratio gives
Thus voltage and current ratios are reciprocal in an ideal transformer.
Question: An ideal transformer steps 240 V down to 24 V. If the secondary current is 5 A, find the primary current.
Power Relation:
Solution:
Answer: The primary current is 0.50 A.
10.19 Transformer Efficiency and Energy Losses
Real transformers are not perfectly efficient. Their efficiency is
Since AC electrical power depends on the nature of the load, a simple resistive-load expression is
Important transformer losses include:
Copper Losses:
The windings possess electrical resistance and dissipate energy at a rate approximately
Eddy-Current Losses:
Changing magnetic flux induces currents in the iron core. Laminating the core into insulated sheets reduces the available paths for large circulating currents.
Hysteresis Losses:
Repeated magnetization and demagnetization of magnetic core material dissipates energy. Appropriate magnetic materials are chosen to reduce this loss.
Flux Leakage:
Not all magnetic flux produced by the primary necessarily links the secondary. Good core design improves magnetic coupling between the coils.
Question: A transformer receives 500 W at its primary and delivers 450 W to its load. Find its efficiency.
Formula:
Solution:
Answer: The transformer efficiency is 90%.
Transformers in Power Transmission:
Electrical power transmitted over long distances may be written approximately as
For a fixed transmitted power, increasing voltage allows current to be reduced:
Power lost as heat in transmission lines is
Reducing current therefore greatly reduces line losses. This is why transformers are used to step voltage up for long-distance transmission and step it down again for distribution and use.
Question: A power line transmits 100 kW. Compare the current when the transmission voltage is 1.0 kV with the current when the voltage is raised to 10 kV, assuming unity power factor for this introductory calculation.
At 1.0 kV:
At 10 kV:
The current decreases by a factor of ten. Since line heating varies as I2, the same line resistance would then produce one-hundredth of the resistive power loss.
Answer: Current falls from 100 A to 10 A.
10.20 Applications of Inductors and Transformers
Inductors are used whenever circuits must oppose rapid current changes, store magnetic energy or interact with alternating signals. They occur in filters, switching power supplies, electromagnetic actuators, radio-frequency circuits and energy-storage systems.
Transformers are essential in power transmission, electronic power supplies, electrical isolation systems, impedance matching and signal coupling. Their operation is one of the clearest practical demonstrations of Faraday's law and mutual induction.
Inductance also forms the magnetic counterpart to capacitance. A capacitor stores energy in an electric field, while an inductor stores energy in a magnetic field. This parallel becomes especially important in the study of electromagnetic oscillations and alternating-current circuits.
Self-induction describes how a circuit responds to changes in its own current, while mutual induction describes electromagnetic coupling between different circuits. The transformer is the most important practical application of mutual induction, and the energy-storage property of inductors provides the foundation for electromagnetic oscillations.
Chapter Summary
- Inductance describes the opposition of a circuit to changes in current through induced emf.
- Self-induction occurs when a changing current induces emf in the same circuit.
- For a linear inductor, flux linkage and current are related by NΦ=LI.
- The SI unit of inductance is the henry.
- The inductance of an ideal long air-core solenoid is L=μ0N2A/ℓ.
- The self-induced emf is ε=-L dI/dt.
- An RL circuit does not reach steady current instantaneously.
- During RL current growth, I=Imax(1-e-t/τ).
- During RL current decay, I=I0e-t/τ.
- The RL time constant is τ=L/R.
- Energy stored in an inductor is U=(1/2)LI2.
- Magnetic energy density in vacuum is uB=B2/(2μ0).
- Mutual induction occurs when changing current in one circuit induces emf in another.
- The mutually induced emf is ε=-M dI/dt.
- A transformer transfers alternating electrical energy between circuits by mutual induction.
- For an ideal transformer, Vs/Vp=Ns/Np.
- For an ideal transformer, VpIp=VsIs.
- A step-up transformer increases voltage, while a step-down transformer decreases voltage.
- Real transformers experience copper, eddy-current, hysteresis and flux-leakage losses.
- High-voltage power transmission reduces current and therefore reduces I2R transmission losses.
CHAPTER ELEVEN
Maxwell's Equations and Electromagnetic Waves
11.1 From Electricity and Magnetism to Electromagnetism
The study of electricity and magnetism begins with phenomena that may initially appear separate. Electric charges produce electric fields, electric currents produce magnetic fields, magnetic fields exert forces on moving charges, and changing magnetic fields produce electric fields. When these ideas are considered together, they form a unified physical theory known as electromagnetism.
The mathematical framework that brings the principal classical laws of electricity and magnetism together is expressed through Maxwell's equations. These equations describe how electric charges and currents produce electromagnetic fields and how changing electric and magnetic fields are related to one another.
One of the most important consequences of Maxwell's theory is that changing electric and magnetic fields can sustain each other and propagate through space as waves. These are electromagnetic waves. Visible light is one example, but radio waves, microwaves, infrared radiation, ultraviolet radiation, X-rays and gamma rays belong to the same physical family.
11.2 Maxwell's Equations: An Overview
Maxwell's equations are the fundamental equations of classical electromagnetism. In integral form, they relate electric and magnetic fields to electric charge, electric current and changing electromagnetic flux.
In vacuum, their integral forms may be written as follows:
| Law | Integral Form | Central Meaning |
|---|---|---|
| Gauss's law for electricity | ∮E·dA = Qenc/ε0 | Electric charge acts as a source or sink of electric field. |
| Gauss's law for magnetism | ∮B·dA = 0 | The net magnetic flux through any closed surface is zero. |
| Faraday's law | ∮E·dℓ = -dΦB/dt | A changing magnetic flux produces a circulating electric field. |
| Ampère-Maxwell law | ∮B·dℓ = μ0Ienc + μ0ε0dΦE/dt | Electric current and changing electric flux produce magnetic fields. |
These four relations contain much of the classical theory of electric and magnetic fields. The purpose at this level is not to develop their full mathematical treatment, but to understand their physical content and use their simpler consequences.
11.3 Gauss's Law for Electricity
Gauss's law relates the electric flux through a closed surface to the net electric charge enclosed by that surface:
The quantity on the left is the total electric flux through a closed surface, commonly called a Gaussian surface. The quantity Qenc is the algebraic sum of the charges enclosed by the surface.
The law shows that electric charge is a source of electric field. Positive charges produce outward electric flux, whereas negative charges produce inward electric flux.
Question: A closed surface encloses a net charge of 4.0 μC. Determine the total electric flux through the surface.
Given:
Required: Total electric flux.
Formula:
Solution:
Answer: The total electric flux is approximately 4.52×105 N·m2/C.
11.4 Gauss's Law for Magnetism
The magnetic counterpart of Gauss's law states that
The net magnetic flux through every closed surface is therefore zero.
This result expresses a fundamental difference between electric and magnetic fields. Electric field lines may originate on positive charges and terminate on negative charges. Magnetic field lines, by contrast, form continuous closed loops.
If a closed surface surrounds part of a magnet, every magnetic field line entering the surface eventually leaves it. The inward and outward contributions therefore give zero net magnetic flux.
Gauss's law for magnetism is consistent with the experimental fact that isolated magnetic monopoles have not been established in classical electromagnetism. Cutting a bar magnet produces two smaller magnets, each with north and south poles, rather than isolated north and south poles.
11.5 Faraday's Law in Integral Form
Faraday's law states that a changing magnetic flux produces a circulating electric field:
The left side represents the circulation of the electric field around a closed path. The right side describes the rate at which magnetic flux through the surface bounded by that path changes.
This formulation extends the concept of electromagnetic induction beyond conducting wires. A changing magnetic field creates an electric field in space even if no conductor is present.
The negative sign expresses Lenz's law: the induced effect is directed so as to oppose the change in magnetic flux responsible for it.
11.6 Ampère-Maxwell Law
Ampère's circuital law for steady currents was introduced as
For situations involving changing electric fields, this relation requires an additional term. The more general Ampère-Maxwell law is
The first term represents the magnetic field associated with ordinary conduction current. The second term represents the magnetic effect of a changing electric flux.
Maxwell's addition of the changing-electric-flux term is crucial because it establishes a symmetry between changing electric and magnetic fields: a changing magnetic field produces an electric field, and a changing electric field produces a magnetic field.
11.7 Displacement Current
The quantity
is called the displacement current.
Displacement current is the quantity associated with a changing electric flux that contributes to the magnetic field in the Ampère-Maxwell law in the same mathematical manner as conduction current.
It is important to understand that displacement current does not necessarily represent physical charges crossing the region in question. A familiar example occurs between the plates of a charging capacitor.
Conduction current flows through the connecting wires, but charges do not cross the dielectric gap between the capacitor plates. Nevertheless, the electric field between the plates changes as the capacitor charges. This changing electric field contributes to the magnetic field through the displacement-current term.
Question: The electric flux through a surface changes uniformly by 2.0×108 N·m2/C in 4.0×10-3 s. Determine the magnitude of the corresponding displacement current.
Given:
Formula:
Solution:
Answer: The displacement current is approximately 0.443 A.
11.8 Physical Interpretation of Maxwell's Equations
Maxwell's equations may be understood conceptually through four statements.
| Equation | Physical Interpretation |
|---|---|
| Gauss's law for electricity | Electric charges produce electric fields. |
| Gauss's law for magnetism | Magnetic field lines form continuous loops; there is no net magnetic flux through a closed surface. |
| Faraday's law | A changing magnetic field produces a circulating electric field. |
| Ampère-Maxwell law | Electric currents and changing electric fields produce magnetic fields. |
The last two relations are especially important for electromagnetic waves. They show that time-varying electric and magnetic fields are dynamically coupled.
11.9 Production of Electromagnetic Waves
A stationary charge produces an electrostatic field. A charge moving steadily produces electric and magnetic fields, but electromagnetic radiation is associated with accelerating charges.
Suppose charges oscillate back and forth in a transmitting antenna. Their changing motion produces time-varying electric and magnetic fields. A changing electric field contributes to a magnetic field through the Ampère-Maxwell law, while a changing magnetic field produces an electric field through Faraday's law.
The coupled disturbance can detach from the source region and propagate through space. This propagating combination of electric and magnetic fields is an electromagnetic wave.
An electromagnetic wave is a self-propagating disturbance consisting of time-varying electric and magnetic fields that carry energy and momentum through space.
11.10 Properties of Electromagnetic Waves
Electromagnetic waves have several important properties:
- they can propagate through vacuum and therefore do not require a material medium;
- in vacuum, they travel at the speed of light;
- they are transverse waves;
- their electric and magnetic fields are mutually perpendicular;
- both fields are perpendicular to the direction of propagation;
- they carry energy and momentum;
- they can undergo reflection, refraction, interference, diffraction and polarization;
- they are produced by accelerated electric charges.
For a plane electromagnetic wave travelling in a particular direction, the electric field E, magnetic field B and propagation direction are mutually perpendicular.
11.11 Speed of Electromagnetic Waves
Maxwell's equations predict that electromagnetic waves travel through vacuum with speed
Using
and
we obtain approximately
This is the measured speed of light in vacuum. The agreement revealed that light itself is an electromagnetic wave.
Question: Verify approximately the speed predicted for an electromagnetic wave in vacuum using μ0=4π×10-7 H/m and ε0=8.854×10-12 F/m.
Formula:
Solution:
Answer: The predicted speed is approximately 3.00×108 m/s.
11.12 Frequency, Wavelength and Wave Speed
Like other periodic waves, electromagnetic waves satisfy
where v is wave speed, f is frequency and λ is wavelength.
In vacuum, v=c, so
Therefore,
and
Frequency and wavelength are inversely related when wave speed is fixed. Increasing frequency therefore decreases wavelength.
Question: A radio wave has frequency 100 MHz. Find its wavelength in vacuum.
Given:
Formula:
Solution:
Answer: The wavelength is 3.00 m.
Question: An electromagnetic wave in vacuum has wavelength 600 nm. Determine its frequency.
Given:
Formula:
Solution:
Answer: The frequency is 5.00×1014 Hz.
11.13 Electric and Magnetic Fields in an Electromagnetic Wave
For a plane electromagnetic wave travelling through vacuum, the electric and magnetic fields oscillate in phase. Their magnitudes are related by
Therefore,
Although the numerical values of E and B have different units, their amplitudes are directly related through the speed of light.
Question: The maximum electric field of an electromagnetic wave in vacuum is 120 V/m. Determine the maximum magnetic field.
Given:
Formula:
Solution:
Answer: The maximum magnetic field is 4.00×10-7 T.
11.14 Energy Transport in Electromagnetic Waves
Electromagnetic waves transport energy. The flow of electromagnetic energy is described by the Poynting vector,
The direction of S gives the direction in which electromagnetic energy propagates.
For perpendicular electric and magnetic fields, the instantaneous magnitude is
For a sinusoidal plane electromagnetic wave in vacuum, the average intensity may be expressed as
or equivalently,
Here E0 and B0 are the field amplitudes.
11.15 Electromagnetic Spectrum
The electromagnetic spectrum is the complete range of electromagnetic radiation arranged according to frequency or wavelength.
| Region | General Characteristics | Representative Applications |
|---|---|---|
| Radio waves | Lowest-frequency broad region; relatively long wavelengths | Broadcasting, communication, navigation |
| Microwaves | Shorter wavelengths than conventional radio bands | Radar, satellite communication, microwave heating |
| Infrared | Associated strongly with thermal radiation from ordinary-temperature objects | Thermal imaging, remote controls, sensing |
| Visible light | Range detectable by the human eye | Vision, illumination, optical systems |
| Ultraviolet | Higher frequency than visible light | Fluorescence, sterilization, analytical applications |
| X-rays | High-frequency, strongly penetrating ionizing radiation | Medical imaging, material inspection |
| Gamma rays | Very high-frequency electromagnetic radiation | Nuclear and astrophysical processes, radiotherapy and scientific applications |
The boundaries between neighbouring regions of the electromagnetic spectrum are conventional rather than perfectly sharp. All regions are governed by the same electromagnetic principles.
As frequency increases, wavelength decreases. The photon energy associated with electromagnetic radiation also increases with frequency according to E=hf, although a detailed treatment of photon physics belongs to quantum physics.
11.16 Applications of Electromagnetic Waves
Electromagnetic waves are central to communication, medicine, scientific measurement, navigation, remote sensing and modern information technology.
Radio-frequency electromagnetic waves carry broadcasting and communication signals. Microwaves are used in radar, satellite links and wireless systems. Infrared radiation is useful for thermal imaging and sensing. Visible light carries visual information and is used extensively in optical communication and instrumentation.
Higher-frequency ultraviolet radiation has applications in sterilization and fluorescence. X-rays are widely used in medical and industrial imaging because of their penetrating ability, while gamma radiation has important applications in medicine, nuclear science and astrophysics.
Electrostatics describes fields produced by stationary charges. Magnetostatics describes fields associated with steady currents. Faraday's law introduces electric fields produced by changing magnetic fields, while the Ampère-Maxwell law shows that changing electric fields also produce magnetic fields. Maxwell's equations unite these ideas and predict electromagnetic waves.
Chapter Summary
- Maxwell's equations provide the fundamental classical description of electric and magnetic fields.
- Gauss's law for electricity is ∮E·dA=Qenc/ε0.
- Gauss's law for magnetism is ∮B·dA=0.
- Faraday's law in integral form is ∮E·dℓ=-dΦB/dt.
- The Ampère-Maxwell law includes both conduction current and changing electric flux.
- Displacement current is associated with changing electric flux and is given by Id=ε0dΦE/dt.
- Accelerating electric charges can produce electromagnetic radiation.
- Electromagnetic waves consist of mutually perpendicular oscillating electric and magnetic fields.
- Electromagnetic waves do not require a material medium.
- The speed of electromagnetic waves in vacuum is c=1/√(μ0ε0)≈3.00×108 m/s.
- Frequency and wavelength are related by c=fλ in vacuum.
- For a plane electromagnetic wave in vacuum, E=cB.
- Electromagnetic waves carry energy and momentum.
- The Poynting vector describes electromagnetic energy flow.
- Radio waves, microwaves, infrared, visible light, ultraviolet, X-rays and gamma rays are all electromagnetic radiation.
CHAPTER TWELVE
Alternating Current Circuits
12.1 Introduction to Alternating Current
Direct current flows in one direction, although its magnitude need not always be perfectly constant. In contrast, an alternating current changes direction periodically and usually changes magnitude continuously with time.
Alternating current, abbreviated AC, is an electric current whose instantaneous magnitude and direction vary periodically with time.
The most important practical form of alternating current is sinusoidal. It is naturally produced when a coil rotates uniformly in a magnetic field, as discussed in electromagnetic induction.
A sinusoidal current may be written as
where I0 is the peak current, ω is the angular frequency and φ is the phase constant.
12.2 Sinusoidal Alternating Voltage and Current
A sinusoidal alternating voltage may be represented by
where V0 is the maximum or peak voltage.
When the phase constant is zero, the equations reduce to
and
During one complete cycle, a sinusoidal current passes through positive and negative values. Its average value over a complete cycle is zero. This does not mean that AC cannot transfer energy. The heating and power effects are described using root-mean-square values.
12.3 Period, Frequency, Phase and Angular Frequency
The period, T, is the time required for one complete cycle. The frequency, f, is the number of cycles completed per second.
Frequency is measured in hertz.
Angular frequency is related to frequency by
Therefore,
Phase specifies the position of a sinusoid within its cycle relative to a chosen reference. Two sinusoidal quantities of the same frequency may reach corresponding values at different times and are then said to have a phase difference.
Question: An AC supply has frequency 50 Hz. Determine its period and angular frequency.
Given:
Period:
Angular Frequency:
Answer: The period is 0.020 s and the angular frequency is approximately 314 rad/s.
12.4 Peak and Root-Mean-Square Values
Because the instantaneous value of alternating current changes continuously, a useful equivalent value is required for power calculations. This is the root-mean-square, or rms, value.
The rms value of an alternating current is the value of direct current that would produce the same average heating effect in a given resistor.
For a sinusoidal current,
Similarly,
Hence,
and
Question: A sinusoidal AC voltage has a peak value of 325 V. Find its rms value.
Formula:
Solution:
Answer: The rms voltage is approximately 230 V.
Question: A sinusoidal alternating current has an rms value of 5.0 A. Determine its peak value.
Formula:
Solution:
Answer: The peak current is approximately 7.07 A.
12.5 Alternating Current Through a Resistor
Consider a pure resistor connected to a sinusoidal AC source. Ohm's law applies instantaneously:
If
then
Since
we obtain
Therefore, voltage and current in a pure resistor are in phase. They reach zero, positive maximum and negative maximum at the same times.
Using rms quantities,
Question: A 40 Ω resistor is connected to a 120 V rms AC supply. Find the rms current.
Formula:
Solution:
Answer: The rms current is 3.0 A.
12.6 Alternating Current Through an Inductor
For an ideal inductor, the voltage is related to the rate of change of current:
If the current is sinusoidal,
then
Therefore,
Since
the voltage leads the current by 90°. Equivalently, the current lags the voltage by 90°.
In a purely inductive AC circuit, voltage leads current by 90°, or current lags voltage by 90°.
12.7 Inductive Reactance
An inductor opposes alternating current through a quantity known as inductive reactance.
Since ω=2πf,
The SI unit of reactance is the ohm.
For a pure inductor,
Inductive reactance increases as frequency increases. An inductor therefore offers greater opposition to higher-frequency alternating current.
Question: A 0.20 H ideal inductor is connected to a 50 Hz, 100 V rms supply. Determine the inductive reactance and rms current.
Inductive Reactance:
Current:
Answer: XL≈62.8 Ω and Irms≈1.59 A.
12.8 Alternating Current Through a Capacitor
For a capacitor,
Since current is the rate of flow of charge,
If
then
and
Thus the current leads the voltage by 90°. Equivalently, the voltage lags the current by 90°.
In a purely capacitive AC circuit, current leads voltage by 90°.
12.9 Capacitive Reactance
The opposition offered by a capacitor to alternating current is described by capacitive reactance:
Since ω=2πf,
For a pure capacitor,
Capacitive reactance decreases as frequency increases. A capacitor therefore offers greater opposition to low-frequency AC and less opposition to high-frequency AC.
Question: A 100 μF capacitor is connected to a 50 Hz, 120 V rms AC supply. Find its capacitive reactance and rms current.
Given:
Capacitive Reactance:
Current:
Answer: XC≈31.8 Ω and Irms≈3.77 A.
12.10 Comparison of R, L and C in AC Circuits
| Element | Opposition | Frequency Dependence | Phase Relationship |
|---|---|---|---|
| Resistor | R | Ideal resistance taken as independent of frequency | Voltage and current are in phase |
| Inductor | XL=ωL | Increases with frequency | Current lags voltage by 90° |
| Capacitor | XC=1/(ωC) | Decreases with frequency | Current leads voltage by 90° |
These distinctions are essential when analysing circuits containing combinations of resistors, inductors and capacitors.
12.11 Series RLC Circuit
A series RLC circuit contains a resistor, inductor and capacitor connected in series to an AC source. Because the components are in series, the same instantaneous current flows through all three components.
However, the voltage drops across the components are not generally in phase:
- VR is in phase with current;
- VL leads current by 90°;
- VC lags current by 90°.
Consequently, the voltages cannot generally be added as ordinary scalar magnitudes. Their phase relationships must be taken into account.
12.12 Impedance
Impedance is the total effective opposition presented by an AC circuit to alternating current, including both resistance and reactance.
For a series RLC circuit, the magnitude of the impedance is
The rms current is therefore
The impedance is measured in ohms.
Question: A series RLC circuit has R=30 Ω, XL=50 Ω and XC=10 Ω. If connected to a 100 V rms supply, find the impedance and rms current.
Net Reactance:
Impedance:
Current:
Answer: The impedance is 50 Ω and the rms current is 2.0 A.
12.13 Phase Angle and Phasor Relationships
The phase angle φ between the source voltage and current in a series RLC circuit satisfies
The cosine of the phase angle is
Three cases should be distinguished.
| Condition | Circuit Behaviour | Phase Relationship |
|---|---|---|
| XL>XC | Net inductive | Current lags source voltage |
| XC>XL | Net capacitive | Current leads source voltage |
| XL=XC | Resonance | Current and source voltage are in phase |
Question: For the circuit in the previous example, determine the phase angle.
Given:
Formula:
Solution:
Since XL>XC, the circuit is inductive.
Answer: The phase angle is approximately 53.1°, with the current lagging the source voltage.
12.14 Resonance in a Series RLC Circuit
Resonance in a series RLC circuit occurs when the inductive reactance equals the capacitive reactance.
Therefore,
Multiplying by ωC,
Hence the resonant angular frequency is
Since ω=2πf, the resonant frequency is
At resonance,
and therefore the current reaches its maximum value for a fixed source voltage:
The source voltage and current are also in phase at resonance.
Question: A series RLC circuit contains an inductor of 0.20 H and a capacitor of 20 μF. Determine the resonant frequency.
Given:
Formula:
Solution:
Answer: The resonant frequency is approximately 79.6 Hz.
12.15 Power in Alternating Current Circuits
In a purely resistive circuit, voltage and current are in phase, and the average power is
In a general AC circuit where voltage and current have phase difference φ, the average power is
For a series RLC circuit, average power is dissipated in the resistor:
An ideal inductor and ideal capacitor do not consume net energy over a complete cycle. They temporarily store energy and return it to the circuit.
Question: An AC circuit operates at 240 V rms and draws 5.0 A rms. If the phase angle is 30°, calculate the average power.
Formula:
Solution:
Answer: The average power is approximately 1.04 kW.
12.16 Power Factor
The power factor of an AC circuit is the cosine of the phase angle between the source voltage and current.
For a series RLC circuit,
A power factor of unity means that voltage and current are in phase. For a given real power and supply voltage, a lower power factor requires a larger current.
Question: A series AC circuit has resistance 40 Ω and impedance 50 Ω. Find its power factor.
Formula:
Solution:
Answer: The power factor is 0.80.
12.17 Electromagnetic Oscillations in LC Circuits
An ideal LC circuit contains a capacitor of capacitance C and an inductor of inductance L. Such a circuit can sustain electromagnetic oscillations in which energy is transferred repeatedly between the electric field of the capacitor and the magnetic field of the inductor.
Suppose a capacitor is initially charged and then connected to an ideal inductor. The capacitor begins to discharge, producing current through the inductor. As the current increases, magnetic energy accumulates in the inductor.
When the capacitor becomes momentarily uncharged, the current reaches its maximum value and the energy is entirely magnetic. The inductor then maintains the current, charging the capacitor with opposite polarity. The process repeats.
The natural angular frequency of an ideal LC oscillator is
The corresponding frequency is
This is the same natural frequency that appears as the resonant frequency of an ideal series RLC circuit.
12.18 Energy Exchange in an LC Circuit
The electric energy stored in a capacitor is
The magnetic energy stored in an inductor is
In an ideal LC circuit with no resistance, the total energy remains constant:
Thus,
At maximum capacitor charge, the current is zero and all the energy is electric. At maximum current, the capacitor is momentarily uncharged and all the energy is magnetic.
Question: A 20 μF capacitor is initially charged to 100 V and connected to an ideal inductor. Determine the total energy in the resulting LC oscillator.
Given:
Formula:
Solution:
Answer: The total electromagnetic energy is 0.10 J.
12.19 Applications of AC and Resonant Circuits
Alternating-current circuits are fundamental to electrical power systems because AC voltage levels can be changed efficiently using transformers. This allows electrical energy to be transmitted at high voltage and comparatively low current, reducing resistive losses in transmission lines.
Resonant circuits are important in frequency selection. A circuit can be designed so that its response becomes particularly strong near a selected frequency. This principle is used in radio receivers, transmitters, filters, communication equipment and many electronic systems.
Inductors and capacitors also play important roles in power supplies, signal processing, filtering and energy storage. Their contrasting dependence on frequency makes them especially useful for controlling alternating signals.
12.20 Course Integration
The topics developed throughout this course form a connected description of electricity and magnetism. Electric charge produces electric fields, and Coulomb's law describes the force between stationary point charges. Gauss's law provides a powerful field method for symmetric charge distributions, while electric potential and capacitance introduce the concepts of electrical energy and energy storage.
When charges move collectively, they form electric current. Current, resistance, electromotive force and circuit laws provide the foundation for DC circuit analysis. Moving charges also generate magnetic fields and experience magnetic forces. The Biot-Savart and Ampère laws relate currents to the magnetic fields they produce.
When magnetic flux changes, Faraday's law predicts an induced electric field or emf, while Lenz's law determines the direction of the induced effect. Self-inductance describes the response of a circuit to changes in its own current, and mutual inductance explains electromagnetic coupling between circuits and the operation of transformers.
Maxwell's equations complete the classical unification by showing that electric and magnetic fields are dynamically connected. Changing magnetic fields produce electric fields, while changing electric fields contribute to magnetic fields. Electromagnetic waves emerge naturally from this coupling and propagate through vacuum at the speed of light.
Alternating-current circuits provide a major practical application of these principles. Resistors dissipate electrical energy, while capacitors and inductors alternately store and return energy in electric and magnetic fields. In RLC circuits, their combined behaviour leads to impedance, phase differences and resonance.
General Physics II reveals electricity and magnetism not as separate collections of laws but as different aspects of a unified electromagnetic theory. The progression from electric charge and fields to circuits, magnetism, induction, Maxwell's equations, electromagnetic waves and AC circuits provides the conceptual foundation required for more advanced study in physics, electrical and electronic engineering, telecommunications, computing hardware, instrumentation and related physical sciences.
Chapter Summary
- Alternating current changes magnitude and direction periodically with time.
- A sinusoidal current may be written as i=I0sin(ωt+φ).
- Frequency and period are related by f=1/T.
- Angular frequency is ω=2πf.
- For sinusoidal AC, Irms=I0/√2 and Vrms=V0/√2.
- In a pure resistor, voltage and current are in phase.
- For an ideal inductor, current lags voltage by 90°.
- Inductive reactance is XL=ωL=2πfL.
- For an ideal capacitor, current leads voltage by 90°.
- Capacitive reactance is XC=1/(ωC)=1/(2πfC).
- The impedance of a series RLC circuit is Z=√[R2+(XL-XC)2].
- The phase angle satisfies tanφ=(XL-XC)/R.
- Series resonance occurs when XL=XC.
- The resonant frequency is f0=1/[2π√(LC)].
- At series resonance, impedance is minimum and equals R, while current is maximum.
- Average AC power is P=VrmsIrmscosφ.
- The power factor is cosφ.
- An ideal LC circuit undergoes electromagnetic oscillations.
- In an ideal LC circuit, energy alternates between the capacitor's electric field and the inductor's magnetic field.
- Maxwell's equations unify the electric, magnetic and electromagnetic phenomena developed throughout the course.