Chapter 6 · Electromagnetic Induction

6.3 Self and Mutual Inductance and the AC Generator

A changing current in one coil induces an emf in a neighbour ($\varepsilon_2=-M\,dI_1/dt$) and in itself ($\varepsilon=-L\,dI/dt$); an inductor stores energy $\tfrac12LI^2$; and a coil turning in a field is an AC generator, $\varepsilon=NBA\omega\sin\omega t$.

A phone lying on a wireless charging pad has no electrical contact with it. Inside the pad a flat coil carries an alternating current; a second flat coil in the phone sits a few millimetres above it, and the changing flux from the first coil induces an emf in the second. How well the pad works depends on a single number that describes the pair of coils — their mutual inductance. Switch off the current to a large coil (the ignition coil of a petrol engine, or the electromagnet of a scrapyard crane) and the coil fights the change: a spark can jump at the switch, because the coil induces a large emf in itself — its self-inductance.

This section defines both kinds of inductance, calculates them for long solenoids, finds the energy an inductor stores in its magnetic field, and ends with the most important application of the whole chapter: the AC generator, a coil turning in a magnetic field, which produces nearly all of the world's electricity.

Mutual Inductance

Place two coils near each other. A current $I_1$ in coil 1 makes a magnetic field, part of which threads coil 2. Since the field of coil 1 is proportional to $I_1$ (Biot–Savart law, Chapter 4), so is the flux linkage of coil 2: $N_2\Phi_2\propto I_1$. The constant of proportionality depends only on the coils — their sizes, shapes, numbers of turns, separation, orientation and the material around them.

📘 Definition 6.3 — Mutual inductance and the henry
The mutual inductance $M$ of two coils is the flux linkage of one coil per unit current in the other: $$N_2\Phi_2=MI_1 .$$ When $I_1$ changes, the emf induced in coil 2 is therefore $$\varepsilon_2=-M\frac{dI_1}{dt}.$$ The SI unit of inductance is the henry (H): $1\ \text{H}=1\ \text{Wb/A}=1\ \text{V s/A}=1\ \Omega\,\text{s}$. Two coils have a mutual inductance of one henry if a current changing at one ampere per second in one induces an emf of one volt in the other. Practical values range from microhenries to a few henries.

Two long coaxial solenoids. Let an inner solenoid S1 of radius $r_1$, length $l$ and $n_1=N_1/l$ turns per metre be wound tightly inside an outer solenoid S2 of the same length with $n_2=N_2/l$ turns per metre. A current $I_1$ in S1 makes a field $B_1=\mu_0n_1I_1$ inside S1 (Chapter 4) and practically none outside it. So the flux through each turn of S2 is $B_1\pi r_1^2$ — only the area of the inner solenoid counts — and $$N_2\Phi_2=(n_2l)(\mu_0n_1I_1)(\pi r_1^2)\quad\Rightarrow\quad M_{21}=\mu_0n_1n_2\pi r_1^2l .$$ Now run a current $I_2$ in the outer solenoid instead. Its field $\mu_0n_2I_2$ fills S2, but only the part inside S1 threads S1: $N_1\Phi_1=(n_1l)(\mu_0n_2I_2)(\pi r_1^2)$, so $M_{12}=\mu_0n_1n_2\pi r_1^2l$ — the same. This is true for any pair of coils (it can be proved in general).

📐 Result 6.8 — Mutual inductance of two long coaxial solenoids

For two long coaxial solenoids of common length $l$, the inner of radius $r_1$, with $n_1$ and $n_2$ turns per metre (end effects neglected): $$M=\mu_0n_1n_2\pi r_1^2\,l=\frac{\mu_0N_1N_2\pi r_1^2}{l}.$$ For any two coils $M_{12}=M_{21}$ (reciprocity): a given rate of change of current in either coil induces the same emf in the other. Filling the solenoids with a material of relative permeability $\mu_r$ (Chapter 5) multiplies $M$ by $\mu_r$. $M$ falls when the coils are moved apart and is nearly zero when their axes are perpendicular.

✏️ Example 1 — A pair of coaxial coils
A long solenoid of length $0.50$ m and radius $2.0$ cm has $500$ turns. A second coil of $100$ turns is wound tightly over its middle. Find (a) the mutual inductance, and (b) the emf induced in the second coil when the current in the solenoid rises steadily from $0$ to $2.0$ A in $40$ ms. (c) What emf would be induced in the solenoid if, instead, the current in the second coil changed at the same rate?
Setup The outer coil has the same cross-section as the solenoid (it is wound tightly over it), so the flux through each of its turns is $B_1\pi r^2$ with $B_1=\mu_0\dfrac{N_1}{l}I_1$.
(a) M $M=\dfrac{\mu_0N_1N_2\pi r^2}{l}$ $=\dfrac{(4\pi\times10^{-7})(500)(100)\pi(0.020)^2}{0.50}$ $\approx1.6\times10^{-4}$ H $=0.16$ mH.
(b) emf $\dfrac{dI_1}{dt}=\dfrac{2.0}{0.040}=50$ A/s. $|\varepsilon_2|=M\dfrac{dI_1}{dt}$ $=(1.58\times10^{-4})(50)$ $\approx7.9\times10^{-3}$ V $=7.9$ mV, in the sense that opposes the growing flux.
(c) Reciprocity The same, $7.9$ mV, since $M_{12}=M_{21}$.
Check Interactive 1 starts with these two coils and this current ramp; during the ramp the secondary emf reads $7.9$ mV.

Self-Inductance

A coil's own field also threads the coil itself. If the current in the coil changes, its own flux linkage changes, and Faraday's law says an emf is induced in the coil — by Lenz's law, in the sense that opposes the change of current. Because it acts against the battery when the current is rising, it is often called a back emf.

📘 Definition 6.4 — Self-inductance
The self-inductance $L$ of a coil is its flux linkage per unit current in it: $$N\Phi_B=LI .$$ When the current changes, the self-induced (back) emf is $$\varepsilon=-L\frac{dI}{dt}.$$ $L$ is measured in henries and depends on the geometry of the coil and on the core material. A coil used for its self-inductance is called an inductor. The induced emf opposes any change of current — it slows the rise when a circuit is switched on and tries to keep the current going when it is switched off.

Self-inductance plays the part of inertia in an electric circuit. Mass resists a change of velocity ($F=m\,dv/dt$); inductance resists a change of current ($\varepsilon=L\,dI/dt$). A steady current through an ideal inductor meets no opposition at all; it is only changes that are resisted.

📐 Result 6.9 — Self-inductance of a long solenoid

A long solenoid of length $l$, cross-sectional area $A$, $N$ turns ($n=N/l$ per metre), with air (vacuum) inside, has field $B=\mu_0nI$ and flux linkage $N\Phi_B=(nl)(\mu_0nI)A$, so $$L=\mu_0n^2Al=\frac{\mu_0N^2A}{l}.$$ With a core of relative permeability $\mu_r$ filling it: $L=\mu_r\mu_0n^2Al$. For a fixed length, $L\propto N^2$: doubling the turns doubles both the field and the number of turns it threads.

✏️ Example 2 — The inductance of a solenoid, and a spark at the switch
A solenoid $40$ cm long, of radius $2.0$ cm, has $1000$ turns. (a) Find its self-inductance with air inside. (b) Find it when a soft-iron core of relative permeability $400$ fills it. (c) A current of $2.0$ A in the coil is switched off and falls to zero in $1.0$ ms. Find the average back emf in each case, and explain why a spark may jump across the switch.
(a) Air $A=\pi(0.020)^2\approx1.26\times10^{-3}\ \text{m}^2$. $L=\dfrac{\mu_0N^2A}{l}$ $=\dfrac{(4\pi\times10^{-7})(1000)^2(1.26\times10^{-3})}{0.40}$ $\approx3.9\times10^{-3}$ H $=3.9$ mH.
(b) Iron core $L=400\times3.95\times10^{-3}\approx1.6$ H.
(c) Back emf $|\varepsilon|=L\dfrac{\Delta I}{\Delta t}$: air: $(3.95\times10^{-3})\dfrac{2.0}{1.0\times10^{-3}}\approx7.9$ V; iron core: $(1.58)(2.0\times10^3)\approx3.2\times10^3$ V. The emf acts to keep the current flowing; thousands of volts across the opening contacts of the switch can ionise the air between them, and the current jumps across as a spark. That is how the ignition coil of a petrol engine makes the spark for its spark plugs.

Energy Stored in an Inductor

To build up a current in an inductor, the source must push charge against the back emf. When the current is $i$ and rising at the rate $di/dt$, the back emf is $L\,di/dt$, and the source does work against it at the rate $$P=\varepsilon\,i=Li\frac{di}{dt}.$$ In a short time $dt$ the work done is $dW=Li\,di$. Adding these up as $i$ grows from $0$ to $I$ (uses integration, or the area of the triangle under the graph of $Li$ against $i$): $$W=\int_0^ILi\,di=\tfrac12LI^2 .$$ This work is not lost as heat (an ideal inductor has no resistance): it is stored in the magnetic field, and is given back when the current falls — it is the energy that makes the spark in Example 2.

📐 Result 6.10 — Energy stored in an inductor

An inductor of inductance $L$ carrying a current $I$ stores energy $$U=\tfrac12LI^2$$ in its magnetic field. For a long solenoid, $U=\tfrac12(\mu_0n^2Al)I^2=\dfrac{B^2}{2\mu_0}(Al)$ with $B=\mu_0nI$, so the magnetic field stores energy per unit volume $$u_B=\frac{B^2}{2\mu_0}$$ (in vacuum or air) — the magnetic counterpart of $\tfrac12\varepsilon_0E^2$ for an electric field (Section 2.4). Compare: $\tfrac12LI^2$ for an inductor, $\tfrac12CV^2$ for a capacitor, $\tfrac12mv^2$ for a moving mass.

✏️ Example 3 — Energy in an inductor and in its field
(a) Find the energy stored in a $50$ mH inductor carrying $4.0$ A. (b) What current would make it store $1.0$ J? (c) For the air-cored solenoid of Example 2 carrying $2.0$ A, find the field inside, the energy density, and the energy stored, and check the answer with $\tfrac12LI^2$.
(a) Energy $U=\tfrac12LI^2=\tfrac12(0.050)(4.0)^2=0.40$ J.
(b) Current $I=\sqrt{\dfrac{2U}{L}}=\sqrt{\dfrac{2.0}{0.050}}=\sqrt{40}\approx6.3$ A.
(c) Field energy $B=\mu_0\dfrac NlI$ $=(4\pi\times10^{-7})(2500)(2.0)$ $\approx6.3\times10^{-3}$ T. $u_B=\dfrac{B^2}{2\mu_0}$ $=\dfrac{(6.28\times10^{-3})^2}{2(4\pi\times10^{-7})}$ $\approx15.7\ \text{J/m}^3$. Volume $Al=(1.26\times10^{-3})(0.40)\approx5.0\times10^{-4}\ \text{m}^3$, so $U\approx7.9\times10^{-3}$ J.
Check $\tfrac12LI^2=\tfrac12(3.95\times10^{-3})(2.0)^2\approx7.9\times10^{-3}$ J ✓ — the energy really is in the field inside the solenoid.
✏️ Example 4 — Back emf from a current graph
The current in a $10$ mH inductor rises steadily from $0$ to $2.0$ A in $4.0$ ms, stays at $2.0$ A for $5.0$ ms, then falls steadily to zero in $2.0$ ms. Find the self-induced emf in each interval, sketch it, and find the greatest energy stored.
Rising $\dfrac{dI}{dt}=\dfrac{2.0}{4.0\times10^{-3}}=500$ A/s; $\varepsilon=-L\dfrac{dI}{dt}$ $=-(0.010)(500)$ $=-5.0$ V (opposing the rise).
Steady $dI/dt=0$, so $\varepsilon=0$ although $2.0$ A flows.
Falling $\dfrac{dI}{dt}=-1000$ A/s; $\varepsilon=+10$ V (now acting to keep the current going). The graph is a step of $-5.0$ V, zero, then a taller step of $+10$ V: the faster fall gives the bigger emf.
Energy Greatest at $2.0$ A: $U=\tfrac12(0.010)(2.0)^2=0.020$ J.
Interactive 1 — Mutual and Self-Induction in a Coil Pair

A long primary solenoid ($500$ turns, $0.50$ m long, radius $2.0$ cm) carries a current $I_1$ that you choose; a short secondary coil of $N_2$ turns is wound round it. Slide the secondary off the end to weaken the coupling (simple model: the field outside the solenoid is ignored), or fill the solenoid with an iron core (relative permeability $100$, illustrative). The three graphs share one time axis (shown in slow motion): $I_1$, the back emf of the primary $-L_1\,dI_1/dt$, and the emf induced in the secondary $-M\,dI_1/dt$. Both emfs appear only while $I_1$ is changing.

100100%
$L_1$; $M$—
primary current $I_1$—
back emf; secondary emf—
energy $\tfrac12L_1I_1^2$—

The AC Generator

An AC generator (alternator) turns mechanical energy into electrical energy using the third way of changing flux from Section 6.1 — changing the angle. Its parts are:

A turbine — driven by falling water, steam from burning coal or from a nuclear reactor, wind, or a diesel engine — turns the axle at a steady angular speed $\omega$. Let $\theta=\omega t$ be the angle between the normal to the coil and $\mathbf B$ at time $t$. The flux linkage is $N\Phi_B=NBA\cos\omega t$, and Faraday's law gives $$\varepsilon=-\frac{d}{dt}\left(NBA\cos\omega t\right)=NBA\omega\sin\omega t .$$ The emf is alternating: it varies sinusoidally, changing sign twice in every revolution. It is greatest when $\sin\omega t=\pm1$ — when the plane of the coil is parallel to $\mathbf B$, the flux is zero but changing fastest (the sides of the coil then cut straight across the field lines). It is zero when the plane of the coil is perpendicular to $\mathbf B$, where the flux is greatest but momentarily not changing.

📐 Result 6.11 — EMF of an AC generator

A coil of $N$ turns, each of area $A$, rotating with constant angular speed $\omega$ about an axis perpendicular to a uniform field $B$, with $\theta=\omega t$ the angle between its normal and $\mathbf B$, has an emf $$\varepsilon=\varepsilon_0\sin\omega t,\qquad \varepsilon_0=NBA\omega ,$$ of frequency $\nu=\dfrac{\omega}{2\pi}$ (one cycle per revolution). The peak emf $\varepsilon_0$ is proportional to $N$, $B$, $A$ and $\omega$. The mains supply in India has $\nu=50$ Hz.

In the large generators of power stations, it is usually the electromagnet that rotates (the rotor) while the heavy coils are fixed in the casing (the stator); the physics is the same, since only the relative motion matters, and the high-current output then needs no sliding contacts. With $N$, $B$ and $A$ fixed, the frequency and the peak emf are both set by the speed of rotation — which is why grid generators must be held at exactly the right speed. Alternating current itself, and how it behaves in circuits, is the subject of Chapter 7.

✏️ Example 5 — A small generator
A rectangular coil of $50$ turns, each of area $0.040\ \text{m}^2$, rotates at $50$ revolutions per second about an axis perpendicular to a uniform field of $0.30$ T. (a) Find the peak emf and the frequency. (b) Find the flux linkage and the emf at the instant when the plane of the coil makes $30^\circ$ with the field. (c) In what positions is the emf zero?
(a) Peak $\omega=2\pi(50)\approx314$ rad/s. $\varepsilon_0=NBA\omega$ $=(50)(0.30)(0.040)(314)$ $\approx1.9\times10^2$ V (188.5 V); $\nu=50$ Hz.
(b) Plane at 30° The normal then makes $\theta=60^\circ$ with $\mathbf B$. $N\Phi_B=NBA\cos60^\circ$ $=(0.60)(0.50)$ $=0.30$ Wb, and $\varepsilon=\varepsilon_0\sin60^\circ$ $\approx188.5\times0.866$ $\approx1.6\times10^2$ V.
(c) Zero emf When $\sin\theta=0$: the plane of the coil perpendicular to $\mathbf B$ ($\theta=0^\circ$ or $180^\circ$), twice per revolution — exactly when the flux linkage is at its largest, $\pm NBA=\pm0.60$ Wb.
Check Interactive 2 starts with this generator; at $\theta=60^\circ$ it reads $N\Phi_B=0.30$ Wb and $\varepsilon\approx163$ V.
✏️ Example 6 — How fast must it turn?
A generator coil has $200$ turns, each of area $0.025\ \text{m}^2$, in a field of $0.20$ T. At what angular speed, and how many revolutions per second, must it turn to give a peak emf of $311$ V (the peak value of a $220$ V mains supply)?
Rearrange $\omega=\dfrac{\varepsilon_0}{NBA}$ $=\dfrac{311}{(200)(0.20)(0.025)}$ $=\dfrac{311}{1.0}$ $=311$ rad/s.
Frequency $\nu=\dfrac{\omega}{2\pi}=\dfrac{311}{6.283}\approx49.5$ revolutions per second — very nearly the $50$ Hz of the mains, about $3000$ rpm.
Interactive 2 — The AC Generator: Flux and EMF Together

A coil turns between the poles of a magnet (drawn in slow motion; the graphs use the real time). The purple arrow is the normal to the coil, at angle $\theta=\omega t$ to $\mathbf B$. Compare the two graphs: the emf $\varepsilon=-d(N\Phi_B)/dt$ peaks where the flux graph crosses zero, and is zero where the flux is largest — a quarter-cycle apart. Change $N$, $B$, the area and the frequency and watch $\varepsilon_0=NBA\omega$.

500.3040050
angle $\theta=\omega t$—
flux linkage $N\Phi_B$—
emf $\varepsilon$—
peak $\varepsilon_0=NBA\omega$—
⚠️ Common mistakes with inductance and generators

1. Using the outer radius for $M$. For coaxial solenoids the flux is confined to the inner one, so $M$ uses $\pi r_1^2$ of the inner solenoid.

2. Thinking an inductor opposes current. It opposes changes of current. A steady current flows through an ideal inductor unopposed, with zero emf across it.

3. $L\propto N$. For a solenoid of fixed length $L\propto N^2$: twice the turns, four times the inductance.

4. Getting the generator's phase backwards. The emf is greatest when the flux is zero (coil plane parallel to $\mathbf B$) and zero when the flux is greatest (coil plane perpendicular to $\mathbf B$).

5. Rev/s versus rad/s. $\varepsilon_0=NBA\omega$ needs $\omega$ in rad/s: $\omega=2\pi\nu$, and $\nu=\text{rpm}/60$.

🚀 Beyond the board syllabus (JEE Advanced) — current growth and decay in an LR circuit

In JEE Advanced 2026 only (it lists LR circuits with d.c. sources); not in the CBSE, JEE Main or NEET 2026 syllabi, which stop at self- and mutual inductance.

Connect a battery of emf $\varepsilon$ to an inductor $L$ in series with a resistance $R$. Kirchhoff's loop rule (Chapter 3) with the back emf gives $\varepsilon-L\dfrac{di}{dt}-iR=0$. The current cannot jump; it grows as (uses integration) $$i=\frac{\varepsilon}{R}\left(1-e^{-t/\tau}\right),\qquad \tau=\frac LR ,$$ reaching $63\%$ of its final value $\varepsilon/R$ after one time constant $\tau$. If the battery is then removed and the coil left connected across $R$, the current decays as $i=I_0e^{-t/\tau}$, the stored energy $\tfrac12LI_0^2$ turning into heat in $R$. For example, $L=0.20$ H, $R=10\ \Omega$, $\varepsilon=12$ V: $\tau=20$ ms, the final current is $1.2$ A, after $20$ ms it is $1.2(1-e^{-1})\approx0.76$ A, and the energy finally stored is $\tfrac12(0.20)(1.2)^2\approx0.14$ J.

Coupling coefficient. For two coils of self-inductances $L_1$ and $L_2$, $M=k\sqrt{L_1L_2}$ with $0\le k\le1$; $k=1$ when every field line of each coil threads the other (two coils wound together on one iron core), and $k\to0$ for coils far apart or at right angles. Both results are tested in JEE Main and JEE Advanced.

Exercises

I
Mutual inductance. $\mu_0=4\pi\times10^{-7}\ \text{T m/A}$.
  • 1 Define mutual inductance and its SI unit. On what factors does the mutual inductance of two coils depend?
  • 2 Two coils have a mutual inductance of $1.5$ H. The current in one rises steadily from $0$ to $20$ A in $0.50$ s. Find the emf induced in the other.
  • 3 An emf of $25$ mV is induced in coil B when the current in coil A changes at $5.0$ A/s. Find the mutual inductance, and the flux linkage of coil B when the current in A is a steady $2.0$ A.
  • 4 An inner solenoid of $1200$ turns, radius $1.5$ cm and length $0.60$ m sits coaxially inside an outer solenoid of $300$ turns, radius $2.5$ cm and the same length. Find their mutual inductance. Does the radius of the outer solenoid matter?
  • 5 For the two solenoids of part 4, show that the flux linkage of the outer solenoid per unit current in the inner one equals the flux linkage of the inner solenoid per unit current in the outer one.
  • 6 How does the mutual inductance of two coils change if (a) a soft-iron rod is inserted through both, (b) the coils are moved further apart, (c) one coil is turned so that the axes are perpendicular?
1 $M$ is the flux linkage of one coil per unit current in the other, $N_2\Phi_2=MI_1$, so that $\varepsilon_2=-M\,dI_1/dt$. Unit: henry ($1\ \text{H}=1\ \text{Wb/A}=1\ \text{V s/A}$). It depends on the numbers of turns, sizes and shapes of the coils, their separation and relative orientation, and the permeability of the material around/inside them.
2 $|\varepsilon|=M\dfrac{dI}{dt}=1.5\times\dfrac{20}{0.50}=60$ V.
3 $M=\dfrac{25\times10^{-3}}{5.0}$ $=5.0\times10^{-3}$ H $=5.0$ mH; $N_B\Phi_B=MI_A$ $=(5.0\times10^{-3})(2.0)$ $=1.0\times10^{-2}$ Wb.
4 $M=\dfrac{\mu_0N_1N_2\pi r_1^2}{l}$ $=\dfrac{(4\pi\times10^{-7})(1200)(300)\pi(0.015)^2}{0.60}$ $\approx5.3\times10^{-4}$ H. The outer radius does not matter: the inner solenoid's field (and the flux it shares) is confined to its own cross-section.
5 Current $I_1$ inside: $B_1=\mu_0n_1I_1$ over area $\pi r_1^2$; linkage of the outer coil $=N_2\mu_0n_1I_1\pi r_1^2$, giving $\mu_0n_1n_2\pi r_1^2l$ per unit current. Current $I_2$ outside: $B_2=\mu_0n_2I_2$ fills the outer coil, but the inner coil encloses only $\pi r_1^2$ of it; linkage $=N_1\mu_0n_2I_2\pi r_1^2$, giving $\mu_0n_1n_2\pi r_1^2l$ again. So $M_{12}=M_{21}$.
6 (a) Increases (by roughly the relative permeability of the iron); (b) decreases, since less of one coil's flux threads the other; (c) falls to almost zero — the field of one coil then runs nearly parallel to the plane of the other.
II
Self-inductance and back emf.
  • 1 Define self-inductance. Why is the self-induced emf called a "back emf"? In what sense is inductance like mass?
  • 2 Find the self-inductance of an air-cored solenoid $0.30$ m long with $600$ turns and cross-sectional area $4.0\ \text{cm}^2$.
  • 3 How does the inductance of a long solenoid change if (a) the number of turns is doubled with the length unchanged, (b) the length is doubled with the number of turns unchanged?
  • 4 The current through a $0.20$ H inductor falls steadily from $5.0$ A to $1.0$ A in $0.040$ s. Find the induced emf. Does it act with or against the current?
  • 5 A back emf of $12$ V appears across a coil when its current changes at $80$ A/s. Find its self-inductance.
  • 6 A steady current of $1.5$ A in a coil produces a flux linkage of $6.0$ mWb. Find the self-inductance.
  • 7 The current in a $0.10$ H inductor is $i=2.0\sin(100\pi t)$ A. Find the greatest self-induced emf, and the current at the instants when the emf is greatest.
1 $L$ is a coil's flux linkage per unit current in itself, $N\Phi_B=LI$, so $\varepsilon=-L\,dI/dt$. By Lenz's law the emf opposes the change of current — when a source drives the current up, the emf acts against the source ("back"). Like mass, which resists changes of velocity, inductance resists changes of current; neither resists steady motion.
2 $L=\dfrac{\mu_0N^2A}{l}$ $=\dfrac{(4\pi\times10^{-7})(600)^2(4.0\times10^{-4})}{0.30}$ $\approx6.0\times10^{-4}$ H $=0.60$ mH.
3 $L=\mu_0N^2A/l$: (a) $\times4$; (b) $\times\tfrac12$.
4 $|\varepsilon|=L\left|\dfrac{dI}{dt}\right|$ $=0.20\times\dfrac{4.0}{0.040}$ $=20$ V. The current is falling, so the emf acts in the direction of the current, trying to keep it up.
5 $L=\dfrac{12}{80}=0.15$ H.
6 $L=\dfrac{N\Phi_B}{I}=\dfrac{6.0\times10^{-3}}{1.5}=4.0\times10^{-3}$ H $=4.0$ mH.
7 $\varepsilon=-L\dfrac{di}{dt}$ $=-(0.10)(2.0)(100\pi)\cos(100\pi t)$; the greatest value is $20\pi\approx63$ V, at the instants when $\cos(100\pi t)=\pm1$, i.e. when $i=0$ — the current is then changing fastest.
III
Energy stored in an inductor and in a magnetic field.
  • 1 Show that the work done in establishing a current $I$ in an inductor $L$ is $\tfrac12LI^2$.
  • 2 Find the energy stored in a $0.40$ H inductor carrying $3.0$ A.
  • 3 An inductor stores $2.5$ J when it carries $5.0$ A. Find its inductance.
  • 4 By what factor does the stored energy change if the current is doubled? By what factor must the current change to double the energy?
  • 5 The field inside a long air-cored solenoid is $0.50$ T. Find the energy density, and the energy stored in the $2.0\times10^{-4}\ \text{m}^3$ of space inside it.
  • 6 Modelling. An MRI scanner has a field of about $3.0$ T in a region of roughly $1.0\ \text{m}^3$. Estimate the energy stored in the field and compare it with the kinetic energy of a $1000$ kg car at $30$ m/s.
1 While the current is $i$ and rising, the source works against the back emf $L\,di/dt$ at the rate $P=Li\,di/dt$. In time $dt$: $dW=Li\,di$. Summing from $0$ to $I$: $W=\int_0^ILi\,di=\tfrac12LI^2$ (the area of the triangle under the graph of $Li$ against $i$). With no resistance, none of it becomes heat; it is stored in the field.
2 $U=\tfrac12(0.40)(3.0)^2=1.8$ J.
3 $L=\dfrac{2U}{I^2}=\dfrac{5.0}{25}=0.20$ H.
4 $U\propto I^2$: doubling $I$ gives $4\times$ the energy. To double $U$, multiply $I$ by $\sqrt2\approx1.41$.
5 $u_B=\dfrac{B^2}{2\mu_0}=\dfrac{0.25}{2(4\pi\times10^{-7})}$ $\approx9.9\times10^4\ \text{J/m}^3$; $U\approx(9.95\times10^4)(2.0\times10^{-4})\approx20$ J.
6 $u_B=\dfrac{(3.0)^2}{2(4\pi\times10^{-7})}\approx3.6\times10^6\ \text{J/m}^3$, so about $3.6$ MJ. The car: $\tfrac12(1000)(30)^2=4.5\times10^5$ J — the magnet stores about eight times as much. That is why switching such a magnet off is done slowly and carefully.
IV
The AC generator.
  • 1 Describe the main parts of an AC generator and derive the expression $\varepsilon=NBA\omega\sin\omega t$ for its emf.
  • 2 A coil of $100$ turns, each of area $0.050\ \text{m}^2$, rotates at $1500$ rpm in a field of $0.40$ T, about an axis perpendicular to the field. Find the peak emf and the frequency.
  • 3 For the generator of part 2, find the emf at the instant when the normal to the coil makes $30^\circ$ with the field.
  • 4 The speed of rotation of a generator is doubled. What happens to the peak emf and to the frequency?
  • 5 A generator of peak emf $200$ V is connected to a $100\ \Omega$ resistor (generator resistance negligible). Find the peak current and the peak power delivered.
  • 6 Modelling. A bicycle dynamo has a coil of $500$ turns, each of area $2.0\ \text{cm}^2$, in an effective field of $0.10$ T, and the tyre turns its spindle at $50$ revolutions per second. Find the peak emf. What happens to the lamp's brightness as the cyclist slows down?
Check: AC generator
$\omega$; frequency—
peak emf $NBA\omega$—
emf at $\theta$—
1 Armature coil ($N$ turns, area $A$) on an axle turned by a turbine; field magnets giving $B$ perpendicular to the axle; slip rings on the axle joined to the coil ends; carbon brushes pressing on the rings to take the current out. With $\theta=\omega t$ between the normal and $\mathbf B$: $N\Phi_B=NBA\cos\omega t$, so $\varepsilon=-\dfrac{d(N\Phi_B)}{dt}=NBA\omega\sin\omega t$.
2 $\omega=\dfrac{2\pi\times1500}{60}\approx157$ rad/s; $\varepsilon_0=(100)(0.40)(0.050)(157)$ $\approx3.1\times10^2$ V (314 V); $\nu=25$ Hz.
3 $\varepsilon=\varepsilon_0\sin30^\circ\approx157$ V.
4 Both double: $\varepsilon_0\propto\omega$ and $\nu=\omega/2\pi$.
5 $I_0=\dfrac{200}{100}=2.0$ A; peak power $\varepsilon_0I_0=400$ W.
6 $\omega=2\pi(50)\approx314$ rad/s; $\varepsilon_0=(500)(0.10)(2.0\times10^{-4})(314)$ $\approx3.1$ V. As the cyclist slows, $\omega$ falls, so both the emf and the frequency fall and the lamp dims (and at very low speed it flickers visibly).

Original problems in the style of the CBSE board paper, JEE Main, JEE Advanced and NEET. Try each one against the clock (about 2 minutes for an MCQ, 5 minutes for a long answer) before opening the solution.

🎯 Exam corner — inductance and the AC generator BoardJEE MainJEE AdvancedNEET

E1. Board · 3 marks Define self-inductance. Derive an expression for the self-inductance of a long air-cored solenoid of length $l$, cross-sectional area $A$ and $N$ turns, and evaluate it for $l=0.50$ m, $A=5.0\ \text{cm}^2$, $N=1000$.

Definition The flux linkage per unit current, $L=N\Phi_B/I$; equivalently the back emf per unit rate of change of current.
Derivation $B=\mu_0\dfrac NlI$; flux per turn $BA$; linkage $N\Phi_B=\dfrac{\mu_0N^2A}{l}I$; so $L=\dfrac{\mu_0N^2A}{l}$ (end effects neglected).
Number $L=\dfrac{(4\pi\times10^{-7})(10^6)(5.0\times10^{-4})}{0.50}$ $\approx1.3\times10^{-3}$ H $=1.3$ mH. Exam tip: convert cm² to m² ($\times10^{-4}$) before substituting.

E2. Board · 2 marks State the principle of an AC generator. Show that the emf of a coil rotating uniformly in a magnetic field varies sinusoidally with time, and state the positions of the coil for which the emf is maximum and zero.

Principle Electromagnetic induction: a coil rotating in a magnetic field has a continuously changing flux linkage, so an emf is induced in it.
Derivation $N\Phi_B=NBA\cos\omega t\Rightarrow\varepsilon=NBA\omega\sin\omega t$.
Positions Maximum ($\varepsilon_0=NBA\omega$) when the plane of the coil is parallel to $\mathbf B$; zero when it is perpendicular to $\mathbf B$. Exam tip: say "plane of the coil" or "normal to the coil" explicitly — examiners penalise "the coil is parallel".

E3. JEE Main (Numerical value.) The current in a $20$ mH coil increases from $2.0$ A to $5.0$ A. Find the increase in the energy stored, in mJ.

Formula $\Delta U=\tfrac12L\left(I_2^2-I_1^2\right)$ — not $\tfrac12L(\Delta I)^2$.
Answer $\Delta U=\tfrac12(0.020)(25-4.0)=0.21$ J $=\mathbf{210}$ mJ. Exam tip: $\tfrac12L(3.0)^2=90$ mJ is the trap answer.

E4. NEET Two coils have a mutual inductance of $5.0$ mH. The current in the first is $i_1=4.0\sin(100\pi t)$ A. The greatest emf induced in the second coil is (A) $2\pi$ V (B) $2\pi$ mV (C) $20$ V (D) $0.020$ V.

Concept $\varepsilon_2=-M\dfrac{di_1}{dt}$ $=-M(4.0)(100\pi)\cos(100\pi t)$.
Answer (A) $\varepsilon_\text{max}=(5.0\times10^{-3})(400\pi)=2\pi\approx6.3$ V. Exam tip: (D) is $Mi_\text{max}$ — a flux linkage in Wb, not an emf.

E5. JEE Advanced (One or more correct.) A coil of $100$ turns, each of area $0.020\ \text{m}^2$, rotates at $100$ rad/s about an axis perpendicular to a uniform field of $0.50$ T. (A) The peak emf is $100$ V. (B) The emf is zero at the instants when the plane of the coil is perpendicular to the field. (C) When the plane of the coil is parallel to the field, the flux linkage is $1.0$ Wb. (D) The frequency of the emf is about $15.9$ Hz.

(A) $NBA\omega$ $=(100)(0.50)(0.020)(100)$ $=100$ V ✓.
(B), (C) Plane perpendicular to $\mathbf B$: flux linkage greatest, $NBA=1.0$ Wb, and momentarily not changing — emf zero (B) ✓. Plane parallel to $\mathbf B$: flux linkage zero (the $1.0$ Wb belongs to the perpendicular position) — (C) ✗.
(D) $\nu=\dfrac{100}{2\pi}\approx15.9$ Hz ✓.
Answer (A), (B), (D). Exam tip: flux and emf are a quarter-cycle out of step — when one is at its largest the other is zero.

E6. Board · Case study In a laboratory demonstration two flat coils are placed face to face, a short distance apart. The current in the first coil is made to rise steadily at $200$ A/s, and a meter shows a steady emf of $5.0$ mV across the second coil. (a) Find the mutual inductance of the pair, and the emf if the current rose at $500$ A/s. (2 marks) (b) What happens to the emf if the second coil is moved further away or tilted? (1 mark) (c) Why does a steady current in the first coil produce no reading at all? (1 mark)

(a) $M=\dfrac{|\varepsilon_2|}{dI_1/dt}=\dfrac{5.0\times10^{-3}}{200}=2.5\times10^{-5}$ H $=25\ \mu\text{H}$. At $500$ A/s: $\varepsilon_2=(2.5\times10^{-5})(500)=12.5$ mV.
(b) Less of the first coil's flux threads the second, so $M$ — and the emf — falls.
(c) A steady current gives a constant flux linkage; $dI_1/dt=0$, so $\varepsilon_2=0$. This is why a wireless charger and a transformer must use alternating current. Exam tip: in case studies, quote the formula $\varepsilon_2=-M\,dI_1/dt$ in every part — it carries the marks.
🔬 Research–Teaching NexusResearch-tutored · optional, about 40 minutes

Research question. How quickly does the mutual inductance of a wireless-charging coil pair fall as the gap between them grows — and does a simple model predict it?

Why it matters. Wireless charging is used not only for phones but for electric toothbrushes, medical implants (whose batteries are recharged through the skin) and, on a much larger scale, electric buses that charge from coils in the road at bus stops — trials have run in cities in Korea, Europe and the USA. Engineers must know how much coupling is lost when the coils are misaligned or further apart.

What to do.

  1. Illustrative measurements for two identical flat coils of radius $a=20$ mm (each with self-inductance $L=24\ \mu\text{H}$), at gap $d$: $d=2$ mm: $M=18.0\ \mu\text{H}$; $5$ mm: $16.6\ \mu\text{H}$; $10$ mm: $13.5\ \mu\text{H}$; $15$ mm: $9.2\ \mu\text{H}$; $20$ mm: $6.7\ \mu\text{H}$. Plot $M$ against $d$.
  2. The on-axis field of a circular loop falls off as $\left(\dfrac{a^2}{a^2+d^2}\right)^{3/2}$ (Chapter 4). Test the model $M(d)=M_0\left(\dfrac{a^2}{a^2+d^2}\right)^{3/2}$ with $M_0=18.5\ \mu\text{H}$: compute the model value at each gap and the percentage difference.
  3. The transmitter current is $1.5$ A in amplitude at $125$ kHz, so its greatest rate of change is $2\pi\nu I_0\approx1.18\times10^6$ A/s. Find the peak emf in the receiver at each gap.
  4. The phone needs a peak emf of at least $10$ V. Up to what gap does charging work?

Tools: Spreadsheet or calculator; graph paper.

Hand in: Your table (gap, measured $M$, model $M$, % difference, peak emf); the graph with both data and model; the largest working gap; and a short comment on where the model fails and why.

Going further. The fraction $k=M/\sqrt{L_1L_2}$ measures how well the coils are coupled. Compute it at each gap. Why do charging pads use a sheet of ferrite (a magnetic ceramic) behind each coil, and why do some phones contain magnets that pull them into alignment?

Instructor note: model values $18.2,\ 16.9,\ 13.2,\ 9.5,\ 6.5\ \mu\text{H}$ — all within about $3\%$ of the data (real coils are flat spirals, not single loops, and the turns are spread out). Peak emfs at $1.18\times10^6$ A/s: $21.2,\ 19.6,\ 15.9,\ 10.8,\ 7.9$ V, so charging works up to a gap of about $15$ mm. Coupling $k=M/L$: $0.75,\ 0.69,\ 0.56,\ 0.38,\ 0.28$. Ferrite guides the flux of each coil towards the other (raising $M$ and $k$) and shields the phone's electronics.

🧪 Self-check quiz

0/5 correct

Q1. The henry is equivalent to

Q2. The number of turns of a long solenoid is doubled, its length and area staying the same. Its self-inductance becomes

Q3. The energy stored in an inductor of inductance $L$ carrying a current $I$ is

Q4. In an AC generator, the emf is zero at the instant when

Q5. The self-induced emf in a coil always opposes

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