Chapter 6: Electromagnetic Induction

In 1820 Oersted showed that an electric current produces a magnetic field. The natural question was the converse: does a magnetic field produce a current? After more than a decade of failed experiments, Michael Faraday (and independently Joseph Henry) discovered in 1831 that the answer is "yes — but only when the magnetic flux changes". This single observation — that changing magnetic fields drive currents — is the foundation of every electric generator on the planet, of transformers, induction motors, induction cooktops, MRI machines, RFID tags, microphones, electric guitars, and electromagnetic brakes. In this chapter we develop Faraday's law and explore six of its principal consequences.

Concept Map

• Magnetic flux: $\Phi_B = \int\vec B\cdot d\vec A$, scalar, units weber (Wb)
• Faraday's law: $\varepsilon = -\dfrac{d\Phi_B}{dt}$, Lenz: the minus sign encodes energy conservation
• Motional EMF: $\varepsilon = \int(\vec v\times\vec B)\cdot d\vec\ell$; straight rod: $\varepsilon = Bv\ell$; rotating rod: $\varepsilon = \tfrac12 B\omega\ell^2$
• Eddy currents: induced loops in solid conductors $\to$ heating / damping
• Self-inductance: $\Phi = LI$, $\varepsilon = -L\,dI/dt$, energy $U = \tfrac12 LI^2$; solenoid $L = \mu_0 n^2 A\ell$
• Mutual inductance: $\Phi_{12} = M\,I_2$, $\varepsilon_1 = -M\,dI_2/dt$; coaxial solenoids $M = \mu_0 n_1 n_2 A\ell$; reciprocity $M_{12} = M_{21}$
• AC generator: rotating coil in $\vec B$, $\varepsilon = NBA\omega\sin\omega t$

6.1 Magnetic Flux

Definition

The magnetic flux through a surface $S$ is

$\Phi_B = \int_S \vec B\cdot d\vec A,$

where $d\vec A$ is an oriented area element. For a uniform $\vec B$ and a flat surface of area $\vec A$,

$\Phi_B = \vec B\cdot\vec A = BA\cos\theta,$

with $\theta$ the angle between $\vec B$ and the surface normal. The SI unit is the weber, $1\,\text{Wb} = 1\,\text{T}\,\text{m}^2$. The CGS unit is the maxwell, $1\,\text{Wb} = 10^8\,\text{Mx}$.

Sign convention

Choosing $d\vec A$ fixes a positive sense of circulation around the boundary by the right-hand rule. Once chosen, you must keep it consistent throughout the problem.

Derivation: flux through a coil of $N$ turns

If each turn intercepts the same flux $\Phi$, the linked flux (or flux linkage) is

$\Phi_{\text{linked}} = N\Phi.$

In Faraday's law it is always $N\Phi$ that matters.

Worked Example

A circular loop of radius $5\,\text{cm}$ is placed in a uniform field $0.30\,\text{T}$. Find the flux when the loop's normal makes (a) $0^\circ$, (b) $60^\circ$, (c) $90^\circ$ with the field.

$A = \pi(0.05)^2 = 7.854\times 10^{-3}\,\text{m}^2,$

$\Phi(0^\circ) = BA = 2.36\times 10^{-3}\,\text{Wb},$

$\Phi(60^\circ) = BA\cos 60^\circ = 1.18\times 10^{-3}\,\text{Wb},$

$\Phi(90^\circ) = 0.$

Pitfalls

• Flux is a signed scalar; reversing the chosen normal flips its sign.
• For a non-planar surface or non-uniform field you must do an integral.
• Only the net flux through a closed loop matters for induction — you can deform the surface freely.

6.2 Faraday's Experiments

Three classic experiments summarised by Faraday establish the law:

1. Magnet–coil experiment: a bar magnet pushed into a coil connected to a galvanometer deflects the needle. Pulling out gives an opposite deflection. Keeping the magnet stationary gives no current. The current arises only during relative motion.
2. Two-coil experiment (transformer): a primary coil connected to a battery via a switch, and a secondary coil connected to a galvanometer. The galvanometer deflects momentarily when the switch is closed (current in primary builds up) and again, oppositely, when it is opened. With a steady current there is no deflection.
3. Current-changing-by-rheostat: with the primary carrying a steady current that is then varied by sliding a rheostat, the secondary shows a current proportional to the rate of change of primary current.

Conclusion

The induced EMF depends not on $\Phi$ itself but on $d\Phi/dt$.

Worked Example (conceptual)

In experiment 2, why does the galvanometer also deflect when the switch is opened? Because $I$ drops from a finite value to zero — that is a change in flux, of opposite sign — Lenz's law then makes the induced current oppose the decrease, i.e., flow in the same sense as the original primary current.

Pitfalls

• Relative motion is what matters; if you and the magnet move together, no EMF.
• A steady current in the primary gives a steady $\Phi$ in the secondary — no EMF.

6.3 Faraday's Law — and Lenz's Law

Statement

The induced EMF in a closed loop is equal to the negative rate of change of the magnetic flux linked with the loop:

$\boxed{\;\varepsilon = -\frac{d\Phi_B}{dt}\;}$

For a coil of $N$ turns,

$\varepsilon = -N\,\frac{d\Phi_B}{dt}.$

Lenz's law

The direction of the induced current is such that it opposes the change in flux that produced it. The negative sign in Faraday's law is the mathematical expression of Lenz's law. It is a statement of energy conservation: if the induced current aided the change, we would get more energy out than we put in.

Derivation: Faraday from work done

Consider a rod sliding on rails, the rod of length $\ell$ moving with velocity $v$ in a field $B$ perpendicular to the plane (Fig. classic). The flux through the circuit changes as $d\Phi/dt = B\ell v$. By energy conservation, the work done by the external force pulling the rod must equal the electrical energy dissipated. That gives $F_{\text{ext}}v = \varepsilon I$, and with $\varepsilon = B\ell v$ and the magnetic force on the rod $F_{\text{mag}} = BI\ell$ exactly cancelling $F_{\text{ext}}$ at steady state, the sign works out.

Worked Example

A coil of 100 turns has its area perpendicular to a magnetic field that varies as $B(t) = 0.2\,t^2\,\text{T}$ (with $t$ in s) and area $0.01\,\text{m}^2$. The induced EMF at $t = 5\,\text{s}$ is

$\varepsilon = -N A\frac{dB}{dt} = -100\times 0.01\times 0.4\,t = -2.0\,\text{V}\quad\text{at }t=5\,\text{s}.$

The negative sign means the induced current flows in the sense that opposes the increase in flux.

Lenz applied to magnet-coil

If a bar magnet's N-pole approaches a coil from the right, the flux through the coil (taking rightward normal) increases. The induced current must produce flux to the left (opposite the change), so by the right-hand rule, it flows in a sense that makes the right face of the coil a south pole — repelling the approaching N. The external agent must do work against this repulsion — that work becomes electrical energy.

Pitfalls

• The induced EMF opposes the change in flux, not the flux itself.
• If $\Phi$ is decreasing, the induced current flows so as to maintain the original flux.
• Always specify the sense of the normal before applying Lenz quantitatively.

6.4 Motional EMF

Definition

A motional EMF is one driven by the magnetic Lorentz force $\vec F = q\vec v\times\vec B$ on free charges in a moving conductor.

Derivation 1: Straight rod on rails

A rod of length $\ell$ moves with constant velocity $v$ perpendicular to a uniform field $B$ (with $B$ perpendicular to the plane of the rails). A free positive charge in the rod feels a force $qvB$ along the rod, separating positive and negative charges and producing an EMF.

In time $dt$ the rod sweeps an area $\ell v\,dt$, so

$d\Phi = B\ell v\,dt\implies \boxed{\;\varepsilon = B\ell v\;}$

If the rails are connected by a resistor $R$, the current is $I = B\ell v/R$ and the magnetic braking force is

$F = BI\ell = \frac{B^2\ell^2 v}{R},$

dissipating power $P = I^2 R = (B\ell v)^2/R = Fv$. The kinetic energy of the rod (or work of the external agent) is converted to heat in $R$. Energy conservation is built in.

Derivation 2: Rotating rod

Take a rod of length $\ell$ rotating with angular velocity $\omega$ about one end, in a uniform field $B$ perpendicular to the plane of rotation. The element at distance $r$ moves with velocity $\omega r$, contributing a motional EMF $dV = B(\omega r)\,dr$. Integrating from $0$ to $\ell$:

$\varepsilon = \int_0^{\ell} B\omega r\,dr = \tfrac12 B\omega\ell^2.$

$\boxed{\;\varepsilon_{\text{rot}} = \tfrac{1}{2}B\omega\ell^2\;}$

The far end is at higher potential (for positive charges pushed outward when $\vec\omega\parallel\vec B$).

Derivation 3: General formula

For any rigid conductor moving in a field,

$\varepsilon = \oint(\vec v\times\vec B)\cdot d\vec\ell.$

For a rigid loop translating in a uniform field, the integral is zero — the EMF arises only from a relative motion that changes the flux. This subtlety is what makes Faraday's law more general than Lorentz alone.

Worked Example

A copper rod of length 1 m falls freely in a horizontal field $B = 0.10\,\text{T}$ pointing E. At $t = 2\,\text{s}$ after release the rod has $v = gt = 19.6\,\text{m/s}$, so

$\varepsilon = B\ell v = 0.10\times 1\times 19.6 = 1.96\,\text{V}.$

Pitfalls

• The "$Bv\ell$" formula assumes $\vec v\perp\vec B\perp\vec\ell$.
• For a rotating rod don't forget the factor of $\tfrac12$; it comes from $\int r\,dr$.
• If the rod's circuit is open, charges accumulate at the ends until the electric field cancels the magnetic force — a static EMF builds up, but no current flows.

6.5 Energy Considerations — Induced Current and Power

Setup

Take the sliding-rod circuit of 6.4. Let an external agent pull the rod with constant velocity $v$ against the magnetic force.

Energy balance

• External work rate: $P_{\text{ext}} = F_{\text{ext}}\cdot v$.
• Electrical power dissipated in the resistor: $P_{\text{elec}} = I^2 R = (B\ell v)^2/R$.
• Magnetic force on the rod: $F_{\text{mag}} = BI\ell = B^2\ell^2 v/R$, opposing motion.

At steady $v$ (no acceleration), $F_{\text{ext}} = F_{\text{mag}}$, so

$P_{\text{ext}} = F_{\text{mag}}\,v = \frac{B^2\ell^2 v^2}{R} = P_{\text{elec}}.$

Every joule of work done by the external agent ends up as joule heating in $R$. Magnetic forces do no work on the charges directly; they merely redirect, and the work is done by the external agent against the magnetic braking.

Induced charge

If the flux through a circuit of resistance $R$ changes by $\Delta\Phi$,

$q_{\text{ind}} = \int I\,dt = \int \frac{|\varepsilon|}{R}\,dt = \int\frac{1}{R}\left|\frac{d\Phi}{dt}\right|dt = \frac{|\Delta\Phi|}{R}.$

So the induced charge depends only on the change in flux, not on how fast it happens.

Worked Example

A loop of resistance $4\,\Omega$ has the flux through it change from $0.6\,\text{Wb}$ to $0.2\,\text{Wb}$ in $0.1\,\text{s}$. Mean induced EMF, current, and charge:

$\bar\varepsilon = \frac{\Delta\Phi}{\Delta t} = \frac{0.4}{0.1} = 4\,\text{V},\quad I = 1\,\text{A},\quad q = \frac{\Delta\Phi}{R} = 0.1\,\text{C}.$

Pitfalls

• Power dissipated is $P = \varepsilon^2/R$ only if the loop is purely resistive.
• Induced charge $q = \Delta\Phi/R$ is independent of how fast; this is the principle of the ballistic galvanometer.

6.6 Eddy Currents

Definition

When a bulk (extended) conductor experiences a changing flux, induced currents circulate in closed loops within the conductor — these are eddy currents. They dissipate energy as heat and produce a retarding force on the motion that caused them.

Applications

1. Induction cooktops: an AC magnetic field induces eddy currents in the ferromagnetic base of a pan; the resistive heating warms the food.
2. Induction furnaces: a strong AC field melts metals via $I^2 R$ heating.
3. Magnetic damping in galvanometers: a metallic frame around the coil dissipates oscillation energy, bringing the needle quickly to rest without overshoot.
4. Magnetic braking in trains: an electromagnet near a moving rail induces eddies in the rail; the resulting drag brings the train to a stop without mechanical friction. The braking force is proportional to velocity, giving smooth slowing.
5. Speedometers: a rotating magnet induces eddies in a metal cup; the cup tends to rotate with the magnet, but is restrained by a spring — its angle measures the magnet's speed.
6. Electromagnetic levitation, metal detectors, mass spectrometry all use eddy currents.

Minimisation: lamination

Where eddies are undesirable (transformer cores, motor armatures), the conductor is split into thin laminations insulated from each other and oriented so that eddy paths are broken. This drastically reduces the cross-section available for eddy circulation and hence the energy loss. (The power loss in laminations of thickness $t$ scales as $t^2$.)

Worked Example

A copper plate is dropped between the poles of a strong electromagnet. Why does it fall slowly?

As the plate moves, the flux through different sections changes, inducing eddy currents. By Lenz's law these oppose the motion, producing a magnetic braking force. The kinetic energy of the falling plate is converted into heat in the plate.

Pitfalls

• Eddy losses are reduced but not eliminated by lamination; hysteresis loss is separate.
• A perfect conductor (superconductor) would exclude the field (Meissner) rather than letting eddies dissipate — a different regime.

6.7 Self-Inductance

Definition

When the current in a coil changes, the flux it produces (linked with itself) also changes, inducing an EMF in the same coil that opposes the change. This phenomenon is self-induction. The flux linked is proportional to the current:

$\Phi_{\text{linked}} = LI,$

where $L$ is the self-inductance (or inductance). The induced EMF is

$\boxed{\;\varepsilon = -L\,\frac{dI}{dt}\;}$

SI unit: henry, $1\,\text{H} = 1\,\text{Wb/A} = 1\,\text{V s/A}$.

Derivation: Self-inductance of a long solenoid

Take a solenoid of length $\ell$, area $A$, with $n$ turns per unit length (total $N = n\ell$). For current $I$,

$B = \mu_0 n I\quad(\text{inside, uniform}),$

$\Phi\text{ per turn} = BA = \mu_0 n I A,$

$\Phi_{\text{linked}} = N\Phi = (n\ell)(\mu_0 n I A) = \mu_0 n^2 A\ell\,I.$

Therefore

$\boxed{\;L = \mu_0 n^2 A\ell\;}$

If the core has relative permeability $\mu_r$, $L = \mu_0\mu_r n^2 A\ell$ — using a soft iron core boosts $L$ by a factor of $\sim 10^3$ or more.

Energy stored in an inductor

To establish a current $I$ in an inductor, the source must do work against the back-EMF. Power delivered $= \varepsilon\,I_{\text{inst}} = (L\,dI/dt)I$. Integrating:

$U = \int_0^I L i\,di = \tfrac12 LI^2.$

$\boxed{\;U_L = \tfrac{1}{2}LI^2\;}$

For a solenoid, this can be re-expressed as energy density of the magnetic field:

$u_B = \frac{U}{\text{Vol}} = \frac{\tfrac12 LI^2}{A\ell} = \tfrac12\mu_0 n^2 I^2 = \frac{B^2}{2\mu_0}.$

Worked Example

A solenoid 50 cm long, $4\,\text{cm}^2$ cross section, with 500 turns. Self-inductance:

$n = \frac{500}{0.5} = 1000\,\text{m}^{-1},$

$L = \mu_0 n^2 A\ell = 4\pi\times 10^{-7}\times 10^6\times 4\times 10^{-4}\times 0.5 = 2.51\times 10^{-4}\,\text{H} = 0.251\,\text{mH}.$

If $I = 2\,\text{A}$ is established, $U = \tfrac12 LI^2 = 5.0\times 10^{-4}\,\text{J}$.

Pitfalls

• $L$ depends only on geometry and core material, not on current (in the linear regime).
• Inductors oppose change in current — they pass DC freely once steady, but resist sudden change.
• An "ideal" inductor has no resistance; real ones do, with $L$–$R$ time constant $\tau = L/R$.

6.8 Mutual Inductance

Definition

For two coils 1 and 2 near each other, current $I_2$ in coil 2 produces flux $\Phi_{12}$ linked with coil 1:

$\Phi_{12} = M_{12}\,I_2,\qquad \varepsilon_1 = -M_{12}\,\frac{dI_2}{dt}.$

$M_{12}$ is the mutual inductance of coil 1 due to coil 2. By symmetry (reciprocity theorem),

$\boxed{\;M_{12} = M_{21} \equiv M\;}$

SI unit: henry.

Derivation: Mutual inductance of two coaxial solenoids

Consider an outer solenoid of $n_1$ turns/m, and an inner solenoid of $n_2$ turns/m, both of length $\ell$, the inner one of area $A$ (cross-section). Both are coaxial. (For simplicity assume the inner solenoid sits well inside the outer.)

When current $I_1$ flows in the outer, the field inside it (uniform) is $B_1 = \mu_0 n_1 I_1$. This field passes through every turn of the inner solenoid. Flux linked with the inner:

$\Phi_{21} = (n_2\ell)(B_1 A) = \mu_0 n_1 n_2 A\ell\,I_1,$

so

$\boxed{\;M = \mu_0 n_1 n_2 A\ell\;}$

By reciprocity, the same $M$ holds whether current flows in inner or outer.

Coupling coefficient

For any two inductors with self-inductances $L_1, L_2$ and mutual inductance $M$, define the coupling coefficient

$k = \frac{M}{\sqrt{L_1 L_2}},\quad 0\le k\le 1.$

$k=1$ for perfect coupling (no flux leakage, idealised transformer); $k \approx 0$ for widely separated coils.

Worked Example

A primary solenoid of length 1 m, area $20\,\text{cm}^2$ has $n_1 = 1000$ turns/m. Inside it, a secondary of $n_2 = 2000$ turns/m and the same area. Find $M$, and EMF in secondary when primary current changes at 50 A/s.

$M = 4\pi\times 10^{-7}\times 1000\times 2000\times 20\times 10^{-4}\times 1 = 5.03\times 10^{-3}\,\text{H},$

$\varepsilon_2 = M\,dI_1/dt = 5.03\times 10^{-3}\times 50 = 0.25\,\text{V}.$

Pitfalls

• $M$ depends on the relative geometry — orient one coil perpendicular to the other and $M$ can be made zero.
• Reciprocity $M_{12} = M_{21}$ is a theorem, not an approximation.
• $M$ can be positive or negative depending on chosen current senses; use the dot convention to track signs in circuit problems.

Inductors in series and parallel (mutual coupling)

For two inductors $L_1, L_2$ with mutual inductance $M$:

• Series, aiding fluxes: $L_{\text{eq}} = L_1 + L_2 + 2M$.
• Series, opposing fluxes: $L_{\text{eq}} = L_1 + L_2 - 2M$.
• Parallel, aiding: $L_{\text{eq}} = (L_1 L_2 - M^2)/(L_1 + L_2 - 2M)$.

6.9 AC Generator

Principle

The AC generator is an application of Faraday's law to a coil rotating in a magnetic field. The flux through the coil varies sinusoidally; the induced EMF is also sinusoidal.

Construction

• Armature: a coil of $N$ turns of area $A$ wound on a soft-iron core.
• Field magnet: produces a uniform $\vec B$ between its poles.
• Slip rings and brushes: provide a continuous electrical connection between the rotating coil and the external circuit.
• The armature is driven by an external mechanical agency (turbine, engine).

Derivation: EMF as a function of time

Let the coil rotate with constant angular velocity $\omega$ about an axis perpendicular to $\vec B$. At time $t$, the angle between the coil's normal $\hat n$ and $\vec B$ is $\theta = \omega t$ (taking $\theta = 0$ when $\hat n\parallel\vec B$).

$\Phi(t) = NBA\cos(\omega t).$

By Faraday's law,

$\varepsilon(t) = -\frac{d\Phi}{dt} = NBA\omega\sin(\omega t).$

$\boxed{\;\varepsilon(t) = \varepsilon_0\sin\omega t,\quad \varepsilon_0 = NBA\omega\;}$

The peak EMF $\varepsilon_0$ depends on the number of turns, area, field strength and angular velocity. The frequency $f = \omega/(2\pi)$ — typically 50 Hz in India and 60 Hz in the USA.

Physical interpretation

• When $\hat n\parallel\vec B$, $\Phi$ is maximum, but $d\Phi/dt = 0$: instantaneous EMF is zero.
• When $\hat n\perp\vec B$, $\Phi = 0$, but $d\Phi/dt$ is maximum: EMF is at peak.

Worked Example

A coil of $N=200$ turns, area $A = 0.05\,\text{m}^2$, rotates in a field $B = 0.40\,\text{T}$ at $f = 50\,\text{Hz}$. Peak EMF:

$\omega = 2\pi f = 314.16\,\text{rad/s},$

$\varepsilon_0 = NBA\omega = 200\times 0.40\times 0.05\times 314.16 = 1256.6\,\text{V}.$

DC vs AC generator

A DC generator differs only in the use of a split-ring commutator instead of slip rings — this reverses the connection to the external circuit every half-cycle, so the output is always of one sign (pulsating DC).

Pitfalls

• $\varepsilon_0 = NBA\omega$, not $NBA$ — always include $\omega$.
• For a coil with axis parallel to $\vec B$ at $t=0$, $\varepsilon$ is a $\sin$; with axis perpendicular at $t=0$, it is a $\cos$.
• An AC generator delivers an alternating EMF; whether the current is sinusoidal depends on the load (resistive, inductive, etc.) — see Chapter 7.

Solved Problems

Problem 1 — Flux through a tilted loop

A square loop of side 10 cm lies in a uniform field $B = 0.50\,\text{T}$ with its plane making $30^\circ$ with the field. Find the flux.

The plane makes $30^\circ$ with $\vec B$, so the normal makes $60^\circ$.

$\Phi = BA\cos 60^\circ = 0.50\times 0.01\times 0.5 = 2.5\times 10^{-3}\,\text{Wb}.$

Problem 2 — EMF in a coil with changing $B$

A coil of 50 turns and area $4.0\times 10^{-2}\,\text{m}^2$ is placed perpendicular to a field that varies as $B = 0.5\sin(100t)$ T. Find the maximum induced EMF.

$\varepsilon = -NA\frac{dB}{dt} = -NA(0.5)(100)\cos(100t).$

$\varepsilon_{\max} = 50\times 4.0\times 10^{-2}\times 50 = 100\,\text{V}.$

Problem 3 — Sliding rod problem

A rod of length $50\,\text{cm}$ and resistance $0.10\,\Omega$ slides on frictionless rails (negligible resistance) closing a circuit. $B = 0.20\,\text{T}$ perpendicular to the plane. The rod is pulled at $v = 5.0\,\text{m/s}$.

$\varepsilon = B\ell v = 0.20\times 0.50\times 5.0 = 0.50\,\text{V},\quad I = \frac{\varepsilon}{R} = 5.0\,\text{A},$

$F_{\text{ext}} = BI\ell = 0.20\times 5.0\times 0.50 = 0.50\,\text{N},$

$P_{\text{ext}} = F_{\text{ext}}v = 2.5\,\text{W} = P_{\text{diss}} = I^2 R = 2.5\,\text{W}.\,\checkmark$

Problem 4 — Rotating rod

A conducting rod of length 1.0 m rotates with $\omega = 400\,\text{rad/s}$ about one end in $B = 0.50\,\text{T}$ perpendicular to the plane. Find the EMF between the ends.

$\varepsilon = \tfrac12 B\omega\ell^2 = 0.5\times 0.5\times 400\times 1.0 = 100\,\text{V}.$

Problem 5 — Induced charge

A coil of $200$ turns and area $50\,\text{cm}^2$ is taken out of a field $B = 0.40\,\text{T}$ in $0.10\,\text{s}$. Resistance $= 4\,\Omega$. Find the induced charge.

$\Delta\Phi_{\text{linked}} = N\,B\,A = 200\times 0.40\times 50\times 10^{-4} = 0.040\,\text{Wb}.$

$q = \frac{\Delta\Phi_{\text{linked}}}{R} = \frac{0.040}{4} = 0.010\,\text{C} = 10\,\text{mC}.$

(Independent of the $0.10\,\text{s}$ — that's the point of the formula.)

Problem 6 — Inductance and energy

An ideal inductor $L = 0.10\,\text{H}$ carries $I = 10\,\text{A}$. Find stored energy. If the current is reduced to zero in $1\,\text{ms}$, find the back-EMF (assume linear ramp).

$U = \tfrac12 LI^2 = \tfrac12\times 0.10\times 100 = 5.0\,\text{J},$

$\varepsilon = L\frac{dI}{dt} = 0.10\times\frac{10}{10^{-3}} = 1000\,\text{V}.$

(That's why opening an inductive circuit produces sparks at the switch.)

Problem 7 — AC generator

The armature of an AC generator has 100 turns and area $200\,\text{cm}^2$. It rotates at 50 rev/s in a field of $0.10\,\text{T}$.

$\omega = 2\pi\times 50 = 314.16\,\text{rad/s},$

$\varepsilon_0 = NBA\omega = 100\times 0.10\times 0.02\times 314.16 = 62.83\,\text{V}.$

JEE/NEET Edge Cases

• Conductor in a uniform field translating without changing flux — no EMF. EMF requires changing linked flux. A square loop moving parallel to its plane in a uniform field has zero EMF (flux doesn't change).
• Conductor entering a magnetic field region — EMF exists only at the edges where flux is changing; once fully inside (uniform field), EMF is zero again.
• Two concentric coils, perpendicular axes — $M = 0$ (no flux linkage).
• Self-inductance of a toroid: $L = \mu_0 N^2 A/(2\pi r)$ where $r$ is the mean radius.
• Inductance with iron core — replace $\mu_0\to\mu_0\mu_r$ everywhere; iron core gives factor of $10^3$–$10^4$.
• LR-circuit time constants — building up: $I = I_0(1-e^{-t/\tau})$, decaying: $I = I_0 e^{-t/\tau}$, $\tau = L/R$.
• Energy from a battery into an LR circuit: half is dissipated in $R$ during the transient (when $L$ is being charged), and half is stored in $L$. After the source is disconnected and replaced by a wire (short), the stored energy is dissipated.
• Induced electric field: A changing $\vec B$ creates a non-conservative $\vec E$ field even in the absence of charges; $\oint\vec E\cdot d\vec\ell = -d\Phi_B/dt$ holds along any closed path.
• Trap: Lenz's law says the current opposes the change in flux; it does not say the field of the induced current opposes the external field.
• Trap: A bar magnet falling through a vertical conducting tube reaches a terminal velocity well below free-fall — eddy currents in the tube produce a velocity-proportional braking force.

Quick Recap

• Flux: $\Phi = \int\vec B\cdot d\vec A$; unit weber.
• Faraday: $\varepsilon = -d\Phi/dt$; Lenz: minus sign = energy conservation.
• Motional EMF: rod $\varepsilon = B\ell v$; rotating rod $\varepsilon = \tfrac12 B\omega\ell^2$.
• Induced charge $q = \Delta\Phi/R$ — independent of time.
• Self-inductance: $\varepsilon = -L\,dI/dt$, $U = \tfrac12 LI^2$, solenoid $L = \mu_0 n^2 A\ell$.
• Mutual inductance: $\varepsilon_1 = -M\,dI_2/dt$, $M_{12} = M_{21}$, coaxial solenoids $M = \mu_0 n_1 n_2 A\ell$.
• Eddy currents — used in damping/braking, minimised by lamination.
• AC generator: $\varepsilon = NBA\omega\sin\omega t$.

Formula Sheet