13.3 Magnetic Fields, Ampere's Law, Faraday's Law, Lenz's Law & Transformers

Magnetic Fields, Ampere's Law, Faraday's Law, Lenz's Law & Transformers

In the Intermediate Physics Part II syllabus and the UET ECAT engineering entrance examination, electromagnetism and electromagnetic induction represent the core of classical field physics. Candidates are rigorously tested on magnetic deflections of charged particles, galvanometer modification derivations, Amperian loop evaluations, Lenz's law directions, motional emf equations, and transformer turns ratio power calculations without calculator support. Mastering these derivations and vector relationships is essential for securing a competitive engineering admission score.

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1. Magnetic Field and the Lorentz Magnetic Force

The Magnetic Induction Field ($\mathbf{B}$)

A moving electric charge or a steady electric current generates a magnetic field in the surrounding space. The magnetic field is characterized by the vector magnetic flux density $\mathbf{B}$ (also termed magnetic induction). The SI unit of $\mathbf{B}$ is the tesla (T):

$1\text{ T} = 1\text{ N}/(\text{A}\cdot\text{m}) = 1\text{ Wb/m}^2 = 10^4\text{ gauss (G)}$

The Lorentz Magnetic Force Equation

When a particle bearing electric charge $q$ moves with velocity $\mathbf{v}$ through an external magnetic field $\mathbf{B}$, it experiences a velocity-dependent Lorentz magnetic force $\mathbf{F}_B$:

$\mathbf{F}_B = q(\mathbf{v} \times \mathbf{B})$

The magnitude of the magnetic force is:

$F_B = |q| v B \sin\theta$

where $\theta$ is the angle between the velocity vector $\mathbf{v}$ and the magnetic field vector $\mathbf{B}$:

• Direction: Determined by the right-hand cross-product rule. For a positive charge, curl the fingers of the right hand from $\mathbf{v}$ toward $\mathbf{B}$; the extended thumb indicates the force direction. For a negative charge (e.g., an electron), the force direction is strictly inverted ($180^\circ$).
• Zero Force Conditions: The force is zero if the particle is stationary ($v = 0$), or if the particle moves parallel ($\theta = 0^\circ$) or antiparallel ($\theta = 180^\circ$) to the magnetic field lines.
• Maximum Force Condition: The force reaches maximum magnitude $F_{\max} = q v B$ when the particle moves perpendicularly through the magnetic field ($\theta = 90^\circ$).

The Zero-Work Theorem of Magnetic Forces

Because the cross product $(\mathbf{v} \times \mathbf{B})$ is mutually perpendicular to both $\mathbf{v}$ and $\mathbf{B}$, the magnetic force is always strictly orthogonal to the instantaneous velocity vector of the charged particle:

$\mathbf{F}_B \cdot \mathbf{v} = q(\mathbf{v} \times \mathbf{B}) \cdot \mathbf{v} = 0$

The instantaneous mechanical power delivered by the magnetic field is:

$P = \frac{dW}{dt} = \mathbf{F}_B \cdot \mathbf{v} = 0 \implies W = \int \mathbf{F}_B \cdot d\mathbf{r} = 0$

Fundamental Theorem: A static magnetic field does zero work on a moving charged particle. A magnetic force can change the direction of motion (acting purely as a centripetal force), but it can never alter the speed or kinetic energy of the particle.

Trajectory of a Charged Particle in Uniform $\mathbf{B}$

1. Perpendicular Entry ($\theta = 90^\circ$): The magnetic force provides the required centripetal acceleration, deflecting the particle into a stable planar circular orbit of radius $r$:

$q v B = \frac{m v^2}{r} \implies r = \frac{m v}{q B} = \frac{p}{q B} = \frac{\sqrt{2m K}}{q B}$

The period $T$ of one complete circular revolution and the cyclotron frequency $f_c$ are:

$T = \frac{2\pi r}{v} = \frac{2\pi m}{q B}, \quad f_c = \frac{1}{T} = \frac{q B}{2\pi m}$

Remarkably, both period and cyclotron frequency are completely independent of the particle's speed and orbital radius.

1. Oblique Entry ($0 < \theta < 90^\circ$): The velocity resolves into parallel component $v_\parallel = v\cos\theta$ (unaffected by $\mathbf{B}$) and perpendicular component $v_\perp = v\sin\theta$ (which undergoes circular motion). The composite trajectory is a helix along the direction of $\mathbf{B}$ with pitch $p_{\text{pitch}} = v_\parallel T = (v\cos\theta)\left(\frac{2\pi m}{q B}\right)$.

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2. Magnetic Force on Current Conductors & Galvanometer Mechanics

Force on a Current-Carrying Conductor (Laplace's Force)

A current-carrying wire of length $L$ placed in an external magnetic field $\mathbf{B}$ experiences a total magnetic force equal to the vector sum of Lorentz forces acting on its conduction electrons:

$\mathbf{F} = I(\mathbf{L} \times \mathbf{B}) \implies F = I L B \sin\alpha$

where $\mathbf{L}$ is a vector of magnitude $L$ oriented along the direction of conventional current, and $\alpha$ is the angle between the conductor and $\mathbf{B}$.

Torque on a Planar Current Loop

Consider a rectangular coil of $N$ turns, length $a$, and width $b$ (area $A = ab$) carrying current $I$ suspended in a uniform magnetic field $\mathbf{B}$. The forces on opposite sides cancel translationally, but create a net couple torque:

$\tau = N I A B \cos\alpha$

where $\alpha$ is the angle between the plane of the coil and the magnetic field lines. In vector notation:

$\boldsymbol{\tau} = \mathbf{M} \times \mathbf{B} \implies \tau = M B \sin\theta$

where $\mathbf{M} = N I \mathbf{A}$ is the magnetic dipole moment of the coil (in $\text{A}\cdot\text{m}^2$), and $\theta$ is the angle between the normal vector of the coil plane and $\mathbf{B}$ (note $\theta = 90^\circ - \alpha$).

The Moving-Coil Galvanometer

A moving-coil galvanometer detects and measures minute electrical currents. It consists of a rectangular coil suspended between the concave magnetic pole pieces of a permanent magnet with a soft iron cylinder mounted at its core:

1. Radial Magnetic Field: The concave pole pieces and soft iron core produce a radial magnetic field, ensuring the plane of the coil is always parallel to the field lines ($\alpha = 0^\circ$) regardless of deflection angle $\theta$. Thus, deflecting torque is always maximum: $\tau_{\text{def}} = N I A B$.
2. Restoring Torque: As the coil deflects through angle $\theta$, a suspension strip (phosphor-bronze) twists, generating an opposing restoring torque: $\tau_{\text{rest}} = C \theta$, where $C$ is the torsional couple per unit twist.
3. Equilibrium Condition:

$\tau_{\text{def}} = \tau_{\text{rest}} \implies N I A B = C \theta \implies I = \left(\frac{C}{N A B}\right) \theta$

Because $C, N, A, B$ are constants, the angular deflection is directly proportional to current:

$\theta \propto I$

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3. Conversion of Galvanometer to Ammeter and Voltmeter

A bare galvanometer cannot handle large currents (its full-scale deflection current $I_g$ is typically only $1$–$10\text{ mA}$, with coil resistance $R_g \sim 50$–$100\,\Omega$). It must be modified for practical measurements.

Conversion to an Ammeter

An ammeter measures electrical current in a circuit branch and must be connected in series. To measure large currents $I$ without burning the coil or altering circuit current:

1. A very small resistance called a shunt resistor ($R_s$) is connected in parallel across the galvanometer coil.
2. At full-scale deflection, current $I$ divides at the node: current $I_g$ passes through the galvanometer, and the bypass current $(I - I_g)$ flows through the shunt.
3. Because both elements are in parallel, potential drops across them are equal:

$V_g = V_s \implies I_g R_g = (I - I_g) R_s$

$R_s = \frac{I_g R_g}{I - I_g}$

• The equivalent resistance of the completed ammeter is $R_A = \frac{R_g R_s}{R_g + R_s} \approx R_s$.
• An ideal ammeter has zero internal resistance ($R_A = 0$) so that inserting it into a circuit produces zero additional voltage drop.

Conversion to a Voltmeter

A voltmeter measures the potential difference between two circuit nodes and must be connected in parallel. To measure a high voltage range $V$ without drawing significant current:

1. A very large resistance called a multiplier resistor ($R_h$) is connected in series with the galvanometer coil.
2. When maximum voltage $V$ is applied across the combination, the current flowing through both is limited to $I_g$:

$V = I_g(R_g + R_h) \implies \frac{V}{I_g} = R_g + R_h$

$R_h = \frac{V}{I_g} - R_g$

• The equivalent resistance of the completed voltmeter is $R_V = R_g + R_h \approx R_h$.
• An ideal voltmeter has infinite internal resistance ($R_V = \infty$) so that it draws zero current from the branch under measurement.

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4. Ampere's Circuital Law & Solenoid Fields

Statement of Ampere's Law

Ampere's Circuital Law is the magnetic analog of Gauss's Law in electrostatics. It states that the line integral of magnetic flux density $\mathbf{B}$ around any arbitrary closed loop (termed an Amperian loop) equals the permeability of free space $\mu_0$ multiplied by the net electric current $I_{\text{enc}}$ enclosed by the loop:

$\oint_C \mathbf{B} \cdot d\mathbf{L} = \mu_0 I_{\text{enc}}$

where $\mu_0 = 4\pi \times 10^{-7}\text{ T}\cdot\text{m/A} \approx 1.257 \times 10^{-6}\text{ T}\cdot\text{m/A}$ is the permeability of free space.

Applications of Ampere's Circuital Law

1. Long Straight Current-Carrying Conductor

Construct a circular Amperian loop of radius $r$ centered on the wire. By circular symmetry, $\mathbf{B}$ is constant in magnitude and tangential everywhere along the loop ($d\mathbf{L} \parallel \mathbf{B}$):

$\oint \mathbf{B} \cdot d\mathbf{L} = B(2\pi r) = \mu_0 I \implies B = \frac{\mu_0 I}{2\pi r}$

2. Ideal Solenoid

A solenoid is a helical coil of wire wound closely into a cylinder of length $L$ and total turns $N$. When current $I$ flows:

• Outside an infinitely long ideal solenoid, field lines spread out into infinite space, so $B_{\text{out}} \approx 0$.
• Inside the core, field lines are intensely concentrated, uniform, and parallel to the central axis.

Constructing a rectangular Amperian loop of length $l$ with one side inside the solenoid and one side outside:

$\oint \mathbf{B} \cdot d\mathbf{L} = B l + 0 + 0 + 0 = B l$

The enclosed current is $I_{\text{enc}} = (n l) I$, where $n = \frac{N}{L}$ is the number of turns per unit length:

$B l = \mu_0 (n l I) \implies B = \mu_0 n I = \mu_0 \frac{N}{L} I$

ECAT Edge Note: At the extreme ends of a semi-infinite solenoid, the magnetic field drops to exactly half its central value: $B_{\text{end}} = \frac{1}{2} \mu_0 n I$.

3. Toroid (Endless Solenoid)

A toroid is formed by bending a solenoid into a closed circular ring of mean radius $r$. Constructing a circular Amperian loop of radius $r$ within the core enclosing all $N$ turns:

$B(2\pi r) = \mu_0 N I \implies B = \frac{\mu_0 N I}{2\pi r} = \mu_0 n I$

The field is strictly confined within the toroidal ring; it is identically zero outside the inner and outer radii.

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5. Faraday's Law, Lenz's Law & Motional EMF

Magnetic Flux ($\Phi_B$)

Magnetic flux is the measure of the total number of magnetic field lines passing perpendicularly through an area $A$:

$\Phi_B = \mathbf{B} \cdot \mathbf{A} = B A \cos\theta$

SI unit of magnetic flux: weber (Wb) ($1\text{ Wb} = 1\text{ T}\cdot\text{m}^2 = 1\text{ V}\cdot\text{s}$). Flux is maximum when field lines are normal to surface ($\theta = 0^\circ$) and zero when parallel ($\theta = 90^\circ$).

Faraday's Law of Electromagnetic Induction

In 1831, Michael Faraday established that whenever the magnetic flux linking an electrical circuit changes, an electromotive force (emf) is induced in the circuit. The magnitude of the induced emf $\mathcal{E}$ is directly proportional to the time rate of change of magnetic flux:

$\mathcal{E} = -N \frac{\Delta \Phi_B}{\Delta t}$

where $N$ is the number of turns in the coil.

Lenz's Law and Conservation of Energy

In 1834, Heinrich Lenz formulated the physical principle that determines the direction of the induced current: the direction of an induced current is always such that its own magnetic field opposes the change in magnetic flux that produced it.

• The negative sign in Faraday's equation is the mathematical embodiment of Lenz's Law.
• Energy Conservation Proof: If a north magnetic pole is pushed toward a conducting loop, Lenz's law dictates that the induced current establishes a north pole on the near face of the loop, producing a repulsive magnetic force. An external agent must perform positive mechanical work against this repulsion to move the magnet. This mechanical work is converted into the electrical energy of the induced current. If the opposite occurred (induced south pole attracting the magnet), the magnet would self-accelerate without external energy input, creating energy from nothing and violating the First Law of Thermodynamics.

Motional Electromotive Force

When a straight metallic rod of length $L$ moves with velocity $\mathbf{v}$ through a uniform perpendicular magnetic field $\mathbf{B}$, free electrons in the rod experience Lorentz force $\mathbf{F}_m = -e(\mathbf{v} \times \mathbf{B})$, accumulating at one end and creating an opposing electrostatic field $\mathbf{E}$. Equilibrium is reached when electric and magnetic forces balance: $e E = e v B \implies E = v B$.

The potential difference developed between the rod terminals is the motional emf:

$\mathcal{E} = -v B L \sin\theta$

where $\theta$ is the angle between velocity $\mathbf{v}$ and field $\mathbf{B}$. If the rod slides along stationary conductive rails forming a closed circuit of resistance $R$:

• Induced current: $I = \frac{|\mathcal{E}|}{R} = \frac{v B L}{R}$.
• Opposing magnetic braking force: $F = I L B = \frac{v B^2 L^2}{R}$.
• Mechanical power required to maintain velocity: $P = F v = \frac{v^2 B^2 L^2}{R} = I^2 R$ (matching electrical Joule heating exactly).

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6. Inductance & Energy Stored in Inductors

Mutual Inductance ($M$)

When changing current $\frac{\Delta I_p}{\Delta t}$ in a primary coil changes the magnetic flux $\Phi_s$ threading an adjacent secondary coil, an emf $\mathcal{E}_s$ is induced in the secondary:

$\mathcal{E}_s = -M \frac{\Delta I_p}{\Delta t}$

where $M$ is the mutual inductance between the coils, measured in henries (H) ($1\text{ H} = 1\text{ V}\cdot\text{s/A}$). For two coaxial solenoids: $M = \mu_0 n_1 n_2 A L$.

Self-Inductance ($L$)

A changing current $\frac{\Delta I}{\Delta t}$ in a single coil produces a changing flux through its own turns, inducing a self-opposing back-emf:

$\mathcal{E}_L = -L \frac{\Delta I}{\Delta t}$

where $L$ is the self-inductance of the coil. For a solenoid of length $l$, area $A$, and $N$ turns:

$L = \mu_0 n^2 A l = \frac{\mu_0 N^2 A}{l}$

Magnetic Potential Energy Stored in an Inductor

To establish a steady current $I$ in an inductor, work must be performed by the voltage source against the induced back-emf. The instantaneous power is $P = \mathcal{E}_L i = L i \frac{di}{dt}$. The total work done is:

$W = \int_0^I P \, dt = \int_0^I L i \, di = \frac{1}{2} L I^2$

This energy is stored within the magnetic field of the inductor: $U_B = \frac{1}{2} L I^2$.

Dividing by core volume ($V = A l$) yields the volumetric magnetic energy density $u_B$:

$u_B = \frac{U_B}{A l} = \frac{B^2}{2\mu_0}$

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7. Alternating Current Generators & Practical Transformers

The AC Generator

A generator converts mechanical energy into alternating electrical energy. When a coil of $N$ turns and area $A$ rotates with uniform angular velocity $\omega$ in a uniform magnetic field $\mathbf{B}$, the instantaneous flux linkage is $\Phi_B(t) = B A \cos(\omega t)$.

By Faraday's Law, the induced alternating electromotive force is:

$\mathcal{E}(t) = -N \frac{d\Phi_B}{dt} = -N \frac{d}{dt}[B A \cos(\omega t)] = N B A \omega \sin(\omega t)$

$\mathcal{E}(t) = \mathcal{E}_0 \sin(\omega t)$

where $\mathcal{E}_0 = N B A \omega$ is the peak (maximum) output voltage.

Transformers: Working Principle & Turns Ratio

A transformer is a static electrical machine that transfers alternating electrical energy between two or more circuits via electromagnetic mutual induction, stepping voltage up or down at fixed frequency.

It consists of two isolated coils—the primary coil ($N_p$ turns) and the secondary coil ($N_s$ turns)—wound upon a common laminated ferromagnetic core. When an AC voltage $V_p$ is applied to the primary, it induces an alternating magnetic flux in the core that threads the secondary.

By Faraday's law, assuming zero leakage flux:

$V_p = -N_p \frac{\Delta \Phi}{\Delta t}, \quad V_s = -N_s \frac{\Delta \Phi}{\Delta t}$

Dividing the two equations gives the fundamental Transformer Turns Ratio:

$\frac{V_s}{V_p} = \frac{N_s}{N_p} = k$

• Step-Up Transformer ($N_s > N_p$): $V_s > V_p$ and $I_s < I_p$. Used at power generating stations to step voltage up to hundreds of kilovolts ($132$–$500\text{ kV}$ in Pakistan) to minimize $I^2 R$ transmission line losses.
• Step-Down Transformer ($N_s < N_p$): $V_s < V_p$ and $I_s > I_p$. Used at grid substations and household distribution poles to step voltage down to $220\text{ V}$.

Ideal Transformer vs. Real Energy Losses

In an ideal transformer, efficiency is $100\%$ with zero power dissipation: $P_{\text{in}} = P_{\text{out}} \implies V_p I_p = V_s I_s$:

$\frac{I_p}{I_s} = \frac{V_s}{V_p} = \frac{N_s}{N_p}$

In real engineering transformers, efficiency is $\eta = \frac{P_{\text{out}}}{P_{\text{in}}} \times 100\%$ ($90$–$98\%$) due to four primary physical loss mechanisms:

1. Eddy Current Losses: Changing core flux induces circulating eddy currents in the bulk iron core, dissipating energy as Joule heat ($I^2 R$). Minimized by building the core out of thin, insulated laminations of high-resistivity silicon steel.
2. Hysteresis Losses: Repeated magnetization and demagnetization of core domains each AC cycle dissipates energy proportional to the area of the material's hysteresis loop. Minimized by utilizing soft iron or silicon alloy with an extremely narrow hysteresis loop.
3. Copper Losses ($I^2 R$ Losses): Resistive heat generated in primary and secondary copper windings. Minimized by utilizing thick, low-resistance copper conductors for the high-current winding.
4. Flux Leakage: Not all magnetic flux generated by primary links with secondary. Minimized by shell-type core geometry or winding secondary directly over primary.

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8. Electromagnetism and Induction Summary Table

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9. DC Motor and Back EMF

A DC motor is a current-carrying coil in a magnetic field. The torque on the coil is $\tau = NIAB\cos\alpha$, where $\alpha$ is the angle between the plane of the coil and the field. A split-ring commutator reverses the current every half turn so that the torque keeps acting in the same rotational sense.

As the coil turns, it cuts flux and acts as a generator, producing a back emf $\varepsilon_b$ that opposes the supply voltage (Lenz's law). With armature resistance $R$:

$V = \varepsilon_b + IR \implies I = \frac{V - \varepsilon_b}{R}$

• At start-up the coil is at rest, $\varepsilon_b = 0$, so the current is very large; a starting resistance limits it.
• As speed rises, $\varepsilon_b$ rises and the current falls.
• The mechanical power developed is $P_{\text{mech}} = \varepsilon_b I$.

Worked example: A 220 V motor has armature resistance $2\,\Omega$ and draws 5 A at full speed. Back emf $\varepsilon_b = 220 - 5(2) = 210\text{ V}$; mechanical power $= 210 \times 5 = 1050\text{ W}$; efficiency (ignoring other losses) $= \frac{210}{220} \approx 95\%$. At the instant of starting, $I = \frac{220}{2} = 110\text{ A}$.

A DC generator is the same machine run in reverse: mechanical rotation produces an emf, and a commutator turns the alternating emf of the coil into a unidirectional (pulsating) output.