Chapter 6 · Electromagnetic Induction

6.1 Magnetic Flux, Faraday's Law and Lenz's Law

The magnetic flux through a coil is $\Phi_B=\mathbf B\cdot\mathbf A=BA\cos\theta$. Whenever it changes, an emf $\varepsilon=-N\,d\Phi_B/dt$ appears in the coil, driving a current whose own field opposes the change (Lenz's law).

Tap your bank card on a shop's card reader and, without a battery in the card, its chip wakes up and answers. The reader's coil carries a current that changes direction more than ten million times a second; the changing magnetic field it makes threads a flat coil of wire hidden inside the card, and a voltage appears in that coil — enough to power the chip. An electric guitar works the same way in reverse: each steel string vibrates above a small magnet wrapped in thousands of turns of fine wire, the vibrating string changes the magnetic field through the coil, and the tiny voltage that results is amplified into sound.

In both cases nothing touches the coil and no battery is connected to it. The voltage — an induced emf — comes from a changing magnetic field alone. This section describes the experiments in which the effect was discovered, defines the quantity that has to change (the magnetic flux), states Faraday's law, which gives the size of the emf, and Lenz's law, which gives its direction, and shows that Lenz's law is the law of conservation of energy in disguise.

The Experiments of Faraday and Henry

By 1820 it was known that an electric current makes a magnetic field (Chapter 4). Many scientists then looked for the reverse effect — could a magnetic field make a current? A strong magnet held still beside a coil gave nothing, which puzzled everyone. In 1831 Michael Faraday in England, and Joseph Henry in America at about the same time, found what was missing: change. Three simple experiments show it.

Experiment 1 — a magnet and a coil. Connect a coil to a sensitive centre-zero galvanometer. Push the N pole of a bar magnet towards the coil and the needle deflects; hold the magnet still, even inside the coil, and the needle returns to zero; pull the magnet away and the needle deflects the other way. Using the S pole reverses both deflections. Moving the coil towards a fixed magnet has exactly the same effect as moving the magnet: only the relative motion matters. The faster the motion, the larger the deflection.

Experiment 2 — two coils, one moving. Replace the magnet by a second coil carrying a steady current from a battery. Moving this current-carrying coil towards or away from the first coil again deflects the galvanometer, in opposite directions for approach and retreat.

Experiment 3 — two coils, both at rest. Keep both coils still, side by side, and put a switch (a tapping key) in the battery circuit. At the instant the switch is closed the galvanometer kicks one way and then returns to zero, although a steady current keeps flowing in the first coil; at the instant it is opened the galvanometer kicks the other way. Winding both coils on one soft-iron rod makes the kicks much larger.

What do the three experiments have in common? In each of them the magnetic field passing through the second coil changes: because a magnet or a coil moves, or because a current is switched on or off. A steady field, however strong, induces nothing. The quantity that measures "how much field passes through a coil" is the magnetic flux.

Magnetic Flux

We defined electric flux in Section 1.3 as $\Phi_E=\mathbf E\cdot\mathbf A$, the "amount of field" crossing a surface. Magnetic flux is defined in exactly the same way, with $\mathbf B$ in place of $\mathbf E$. Think of it as counting the magnetic field lines that thread a loop: more lines when the field is stronger, when the loop is larger, and when the loop faces the field squarely; none at all when the field lines run parallel to the plane of the loop.

📘 Definition 6.1 — Magnetic flux and the weber
The magnetic flux through a plane surface of area $A$ in a uniform magnetic field $\mathbf B$ is $$\Phi_B=\mathbf B\cdot\mathbf A=BA\cos\theta ,$$ where $\mathbf A=A\hat{\mathbf n}$ is the area vector (normal to the surface) and $\theta$ is the angle between $\mathbf B$ and the normal to the surface. For a curved surface or a field that varies from place to place, the surface is split into small pieces $d\mathbf A$ and the contributions are added: $\Phi_B=\sum\mathbf B\cdot d\mathbf A$.
Flux is a scalar; it is positive or negative according to the direction chosen for $\hat{\mathbf n}$. Its SI unit is the weber (Wb): $1\ \text{Wb}=1\ \text{T m}^2$. For a coil of $N$ identical turns the total flux through all the turns, $N\Phi_B$, is called the flux linkage of the coil.

The angle needs care. When the field is perpendicular to the plane of a coil, it is parallel to the normal: $\theta=0$ and the flux is greatest, $\Phi_B=BA$. When the field lies in the plane of the coil, $\theta=90^\circ$ and $\Phi_B=0$. If a question gives the angle between the field and the plane, the angle you need is $90^\circ$ minus that.

✏️ Example 1 — Flux through a tilted coil
A square coil of side $10$ cm has $50$ turns and lies in a uniform magnetic field of $0.40$ T. Find the flux through one turn and the flux linkage of the coil when (a) the normal to the coil makes an angle of $60^\circ$ with the field, (b) the plane of the coil makes an angle of $60^\circ$ with the field. (c) In what position is the flux zero?
Area $A=(0.10\ \text{m})^2=1.0\times10^{-2}\ \text{m}^2$, so $BA=(0.40)(1.0\times10^{-2})=4.0\times10^{-3}\ \text{Wb}$.
(a) Normal at 60° $\Phi_B=BA\cos60^\circ$ $=4.0\times10^{-3}\times0.50$ $=2.0\times10^{-3}$ Wb per turn; flux linkage $N\Phi_B=50\times2.0\times10^{-3}$ $=0.10$ Wb.
(b) Plane at 60° Then the normal is at $90^\circ-60^\circ=30^\circ$ to $\mathbf B$: $\Phi_B=4.0\times10^{-3}\cos30^\circ\approx3.5\times10^{-3}$ Wb per turn, and $N\Phi_B\approx0.17$ Wb.
(c) Zero flux When the plane of the coil is parallel to $\mathbf B$ ($\theta=90^\circ$): no field lines thread the coil.

Faraday's Law of Induction

Faraday measured the induced emf carefully and found that its size depends on just one thing: how fast the flux through the circuit changes. It does not matter whether the flux changes because the field changes, because the area of the loop changes, or because the loop turns — only the rate of change of $\Phi_B$ counts.

📐 Law 6.1 — Faraday's law of electromagnetic induction

Whenever the magnetic flux through a circuit changes, an emf is induced in the circuit. The induced emf equals the negative of the rate of change of the flux: $$\varepsilon=-\frac{d\Phi_B}{dt}.$$ For a coil of $N$ turns, each threaded by the same flux $\Phi_B$, the emfs of the turns add: $$\varepsilon=-N\frac{d\Phi_B}{dt}=-\frac{d(N\Phi_B)}{dt}.$$ When the flux changes by $\Delta\Phi_B$ in a time $\Delta t$, the average emf is $\varepsilon_\text{av}=-N\dfrac{\Delta\Phi_B}{\Delta t}$. The minus sign gives the direction of the emf (Lenz's law, below). Since $\Phi_B=BA\cos\theta$, the flux — and hence an emf — can be changed by changing $B$, $A$ or $\theta$.

Check the units: $1\ \text{Wb/s}=1\ \text{T m}^2\text{/s}$. With $1\ \text{T}=1\ \text{N/(A m)}$ (from $F=BIl$), $1\ \text{T m}^2\text{/s}=1\ \text{N m/(A s)}=1\ \text{J/C}=1$ V. So $1\ \text{Wb}=1\ \text{V s}$: a flux changing at one weber per second induces one volt.

The emf drives a current if the circuit is closed. If the coil and whatever is connected to it have total resistance $R$, the induced current is $I=\varepsilon/R$. An interesting result follows for the total charge that flows while the flux changes.

📐 Result 6.2 — Induced current and induced charge

In a closed circuit of total resistance $R$, the induced current is $I=\dfrac{|\varepsilon|}{R}=\dfrac{N}{R}\left|\dfrac{d\Phi_B}{dt}\right|$. The charge that flows while the flux changes by $\Delta\Phi_B$ is $$Q=I\,\Delta t=\frac{N\,|\Delta\Phi_B|}{R}.$$ $Q$ depends only on the total change of flux, not on how quickly the change happens: a fast change gives a large current for a short time, a slow change a small current for a long time, and the same charge. (For a varying rate, add up $I\,dt$ over small intervals — the same result follows.)

Result 6.2 is the basis of the search coil, a small coil of known area connected to a meter that records the charge passing through it. Snatching the coil out of a field, or flipping it over, lets you measure the field without knowing how fast your hand moved (Example 4).

✏️ Example 2 — Average emf when a field changes
A coil of $200$ turns, each of area $5.0\times10^{-3}\ \text{m}^2$, lies with its plane perpendicular to a uniform magnetic field. The field falls steadily from $0.60$ T to $0.10$ T in $0.25$ s. The coil and the meter joined to it have a total resistance of $8.0\ \Omega$. Find (a) the induced emf, (b) the induced current, (c) the charge that flows. (d) How would the answers change if the field fell in $0.50$ s instead?
(a) emf $\theta=0$, so $\Delta\Phi_B=A\,\Delta B$ $=(5.0\times10^{-3})(0.10-0.60)$ $=-2.5\times10^{-3}$ Wb per turn. $\varepsilon=-N\dfrac{\Delta\Phi_B}{\Delta t}$ $=-\dfrac{200\times(-2.5\times10^{-3})}{0.25}$ $=+2.0$ V.
(b) Current $I=\dfrac{\varepsilon}{R}=\dfrac{2.0}{8.0}=0.25$ A.
(c) Charge $Q=I\,\Delta t=0.25\times0.25=0.0625$ C. Check with Result 6.2: $Q=\dfrac{N|\Delta\Phi_B|}{R}=\dfrac{200\times2.5\times10^{-3}}{8.0}=0.0625$ C ✓.
(d) Slower The emf and current halve ($1.0$ V, $0.125$ A) but flow for twice as long; the charge is still $0.0625$ C.
Check Interactive 1 (first setting, $N=200$) runs this change between $t=2.0$ s and $t=2.25$ s: the emf trace jumps to $+2.0$ V there.
✏️ Example 3 — Instantaneous emf from a formula for the flux
The magnetic flux through a single loop of wire varies with time as $\Phi_B=(5t^2-4t+1)$ mWb, with $t$ in seconds. The loop has resistance $2.0\ \Omega$. Find (a) the emf as a function of time, (b) the emf and the current at $t=2.0$ s, (c) the instant at which the emf is zero, and the flux at that instant.
(a) Differentiate $\varepsilon=-\dfrac{d\Phi_B}{dt}=-(10t-4)$ mV.
(b) At 2.0 s $\varepsilon=-(20-4)=-16$ mV, so the emf has magnitude $16$ mV (the minus sign says it acts in the sense that opposes the growing flux). $I=\dfrac{16\times10^{-3}}{2.0}=8.0\times10^{-3}$ A $=8.0$ mA.
(c) Zero emf $10t-4=0$ at $t=0.40$ s. Then $\Phi_B=5(0.16)-4(0.40)+1=0.20$ mWb: the flux is at its minimum, so it is momentarily not changing. The flux is not zero there — it is the rate that is zero.
✏️ Example 4 — Measuring a field with a search coil
A search coil of $100$ turns, each of area $2.0\ \text{cm}^2$, is placed between the poles of an electromagnet with its plane perpendicular to the field. It is joined to a meter that records the total charge; the resistance of the whole circuit is $40\ \Omega$. When the coil is flipped through $180^\circ$, a charge of $15\ \mu\text{C}$ passes through the meter. Find the field.
Flux change Before: $\Phi_B=+BA$. After a half-turn the same field enters the coil from the other face: $\Phi_B=-BA$. So $|\Delta\Phi_B|=2BA$ per turn — twice as much as pulling the coil out of the field.
Charge $Q=\dfrac{N(2BA)}{R}$, so $B=\dfrac{QR}{2NA}$ $=\dfrac{(15\times10^{-6})(40)}{2\times100\times2.0\times10^{-4}}$.
Answer $B=\dfrac{6.0\times10^{-4}}{4.0\times10^{-2}}=1.5\times10^{-2}$ T $=15$ mT. The speed of the flip does not enter.
Interactive 1 — The EMF Is Minus the Slope of the Flux

A coil of $N$ turns, each of area $5.0\times10^{-3}\ \text{m}^2$, sits in a field pointing into the screen (⊗). Choose what changes — the field $B$, the area $A$ or the angle $\theta$ — and press Play. The top graph is the flux linkage $N\Phi_B$; the dashed line touching it is its tangent, whose slope is $d(N\Phi_B)/dt$. The bottom graph is the emf, which is always minus that slope. The arrows on the coil show the induced current (positive emf = clockwise as you look at the screen).

200
time $t$—
flux linkage $N\Phi_B$—
slope $d(N\Phi_B)/dt$—
emf $\varepsilon$—

Lenz's Law and Conservation of Energy

Faraday's law gives the size of the induced emf; the minus sign in it carries the direction. In 1834 Heinrich Lenz stated the direction rule in words that are much easier to use than the sign.

📐 Law 6.3 — Lenz's law

The induced emf has such a direction that the current it drives (in a closed circuit) produces a magnetic field that opposes the change in magnetic flux that produced it.

If the flux through a loop is increasing, the induced current's field points against the external field inside the loop; if the flux is decreasing, the induced field points along the external field, trying to keep the flux up. The law opposes the change, not the field itself.

🧭 Problem-solving strategy — finding the direction of an induced current
  1. Find the direction of the external field $\mathbf B$ through the loop.
  2. Decide whether the flux through the loop is increasing or decreasing (a magnet approaching or receding, a current switched on or off, the area growing or shrinking, the loop turning).
  3. Choose the direction of the induced field inside the loop: opposite to $\mathbf B$ if the flux is increasing, the same as $\mathbf B$ if it is decreasing.
  4. Use the right-hand rule for a current loop (curl the fingers of the right hand with the current; the thumb points along the field inside the loop) to get the current. As a check, think of the loop as a small magnet: the face where the current looks anticlockwise is its N pole, and it repels an approaching pole of the same name or attracts a receding one.

Why it must be so — energy. Push the N pole of a magnet towards a coil. By Lenz's law, the near face of the coil becomes an N pole, and it repels the magnet: you have to push. The work you do is the energy that appears as heat ($I^2R$) in the coil. Pull the magnet away, and the near face becomes an S pole that attracts the magnet back — again you do work. Now imagine the opposite rule. The approaching N pole would create an S pole facing it, which would pull the magnet in faster; the faster motion would make a bigger current and a stronger pull, and so on — the magnet would accelerate and the coil heat up with no work done by anyone. That would create energy from nothing. Lenz's law is therefore a consequence of the law of conservation of energy: the induced current always opposes the motion or change that causes it, so mechanical work must be done to keep the current going.

If the circuit is open (a coil with a gap, or a ring that is cut), the emf still appears across the gap but no current flows, so there is no opposing force and no energy is dissipated.

✏️ Example 5 — A loop round a solenoid
A long solenoid has $2000$ turns per metre and radius $3.0$ cm. A single circular loop of wire, of radius $5.0$ cm and resistance $0.010\ \Omega$, is placed round the middle of the solenoid, coaxial with it. The current in the solenoid is increasing at $50$ A/s. Find (a) the rate of change of the field inside the solenoid, (b) the emf and the current induced in the loop, and (c) the direction of the induced current. (Inside a long solenoid $B=\mu_0nI$, Chapter 4; outside it the field is negligible.)
(a) dB/dt $\dfrac{dB}{dt}=\mu_0n\dfrac{dI}{dt}$ $=(4\pi\times10^{-7})(2000)(50)$ $\approx0.126$ T/s.
(b) emf The field exists only inside the solenoid, so the flux through the loop is $B\times\pi r_s^2$ with $r_s=3.0$ cm, the solenoid's radius — not the loop's. $\varepsilon=\pi r_s^2\dfrac{dB}{dt}=\pi(0.030)^2(0.126)\approx3.6\times10^{-4}$ V $=0.36$ mV. $I=\dfrac{\varepsilon}{R}=\dfrac{3.55\times10^{-4}}{0.010}\approx36$ mA.
(c) Direction The flux through the loop is increasing along the solenoid's field. The induced current must make a field opposite to it inside the loop, so it circulates in the opposite sense to the current in the solenoid's windings.
Note A loop of any size or shape around the solenoid has the same emf. The charges in the loop are pushed although the field at the wire is almost zero — see the box at the end of this section.
✏️ Example 6 — A magnet falling through a ring
A copper ring is held horizontal. A bar magnet, held vertically above it with its N pole downwards, is dropped so that it falls through the ring. Looking down from above, find the direction of the current in the ring (a) as the magnet approaches the ring, and (b) after it has passed through and is falling away below it. (c) Compare the acceleration of the magnet with $g$ in each stage. (d) What changes if the ring has a narrow cut in it?
(a) Approaching Field lines leave the N pole, so the magnet's field through the ring points downwards and the downward flux is increasing. The induced field must point upwards inside the ring; by the right-hand rule the current is anticlockwise seen from above. The top face of the ring is an N pole, repelling the falling N pole.
(b) Leaving Below the ring, the magnet's field through the ring still points downwards (towards its S pole, now the top end), but the flux is decreasing. The induced field points downwards to keep it up: the current is clockwise seen from above. The bottom face of the ring is now an N pole, attracting the S pole at the top of the magnet — pulling it back.
(c) Acceleration In both stages the ring pushes or pulls upwards on the magnet, so its acceleration is less than $g$. The lost kinetic energy appears as heat in the ring.
(d) Cut ring An emf is still induced, but no current can flow, so there is no magnetic force: the magnet falls with acceleration $g$.
⚠️ Common mistakes with flux and induction

1. Measuring $\theta$ from the plane. In $\Phi_B=BA\cos\theta$, $\theta$ is measured from the normal. A field perpendicular to the plane gives the maximum flux.

2. Confusing flux with its rate of change. A large flux induces nothing if it is steady; a zero flux can be changing fast (Example 3, and a rotating coil in Section 6.3).

3. Forgetting $N$. Each turn contributes its own emf; the flux linkage is $N\Phi_B$.

4. "The induced field opposes the field." It opposes the change. When the flux is decreasing, the induced field is in the same direction as the external field.

5. Thinking a faster change drives more charge. It drives a larger current, for a shorter time; the charge $Q=N|\Delta\Phi_B|/R$ is the same.

Interactive 2 — Faraday's Magnet-and-Coil Lab

Repeat Experiment 1. A bar magnet moves along the axis of a coil joined to a centre-zero galvanometer (circuit resistance $20\ \Omega$). Choose the motion, its speed, the number of turns and which pole leads. The graphs show the flux linkage $N\Phi_B$ (top) and the induced emf (bottom) over one cycle; the label beside the coil shows the pole that Lenz's law gives its near face. Watch the charge readout: for a given push it does not depend on the speed. (Magnet strength and coil radius $4.5$ cm are illustrative.)

0.60400
magnet centre to coil—
flux linkage $N\Phi_B$—
emf; current—
charge since start of cycle—
🚀 Beyond the board syllabus (JEE Main · NEET · JEE Advanced) — the induced electric field

Not required by the CBSE syllabus. Faraday's law is listed by JEE Main, NEET and JEE Advanced; none names the induced electric field, which is Faraday's law written in terms of the field and appears mainly in JEE problems.

In Example 5 the loop sits where the magnetic field is almost zero, yet its electrons are pushed round it. What pushes them? A changing magnetic field creates an electric field in the space around it, whose lines form closed loops. Faraday's law can be written in terms of this field: $$\oint\mathbf E\cdot d\mathbf l=-\frac{d\Phi_B}{dt}.$$ The emf is the work done per unit charge by this field round the loop. Unlike the electrostatic field of Chapters 1–2, the induced field is non-conservative: its line integral round a closed path is not zero, so it cannot be described by a potential.

By symmetry, inside a long solenoid of radius $R$ with $dB/dt$ uniform, the lines of $\mathbf E$ are circles about the axis. Taking a circle of radius $r$ (uses the symmetric line integral): $E\cdot2\pi r=\pi r^2\dfrac{dB}{dt}$ inside, so $E=\dfrac r2\dfrac{dB}{dt}$ for $r\le R$, and $E=\dfrac{R^2}{2r}\dfrac{dB}{dt}$ for $r\ge R$. For Example 5 at $r=5.0$ cm: $E=\dfrac{(0.030)^2}{2(0.050)}(0.126)\approx1.1\times10^{-3}$ V/m, and $E\cdot2\pi r\approx3.6\times10^{-4}$ V — the emf found before ✓. This field accelerates electrons in a betatron.

Exercises

I
Magnetic flux.
  • 1 Define magnetic flux. Is it a scalar or a vector? Give its SI unit in terms of the tesla.
  • 2 A circular coil of radius $5.0$ cm has $40$ turns and lies with its plane perpendicular to a uniform field of $0.20$ T. Find the flux through one turn and the flux linkage of the coil.
  • 3 The coil of part 2 is turned so that its normal makes $60^\circ$ with the field. Find the flux through one turn. What is it when the plane of the coil is parallel to the field?
  • 4 Show that $1\ \text{Wb}=1\ \text{T m}^2=1\ \text{V s}$.
  • 5 A square loop of side $20$ cm lies in the $xy$-plane in a uniform field $\mathbf B=(0.30\,\hat{\mathbf i}+0.40\,\hat{\mathbf k})$ T. Find the flux through the loop, taking its normal along $+\hat{\mathbf k}$.
  • 6 A hemispherical bowl of radius $10$ cm has its axis along a uniform field of $0.50$ T. Find the flux through its curved surface. (Hint: the net magnetic flux through any closed surface is zero.)
1 $\Phi_B=\mathbf B\cdot\mathbf A=BA\cos\theta$ (uniform field, plane surface; $\theta$ between $\mathbf B$ and the normal), a measure of the field threading the surface. It is a scalar (a dot product), positive or negative with the choice of normal. Unit: weber, $1\ \text{Wb}=1\ \text{T m}^2$.
2 $\Phi_B=BA=0.20\times\pi(0.050)^2\approx1.6\times10^{-3}$ Wb; $N\Phi_B=40\times1.57\times10^{-3}\approx6.3\times10^{-2}$ Wb.
3 $\Phi_B=1.57\times10^{-3}\cos60^\circ\approx7.9\times10^{-4}$ Wb. Plane parallel to the field: $\theta=90^\circ$, $\Phi_B=0$.
4 From $\Phi_B=BA$: $1\ \text{Wb}=1\ \text{T m}^2$. From $\varepsilon=d\Phi_B/dt$: $1\ \text{V}=1\ \text{Wb/s}$, so $1\ \text{Wb}=1\ \text{V s}$. (Directly: $1\ \text{T}=1\ \text{N A}^{-1}\text{m}^{-1}$, so $1\ \text{T m}^2=1\ \text{N m/A}=1\ \text{J s/C}=1\ \text{V s}$.)
5 $\mathbf A=(0.20)^2\hat{\mathbf k}=0.040\,\hat{\mathbf k}\ \text{m}^2$. $\Phi_B=\mathbf B\cdot\mathbf A=0.40\times0.040=1.6\times10^{-2}$ Wb. The $x$-component of $\mathbf B$ runs along the plane and contributes nothing.
6 The bowl plus the flat circle across its rim make a closed surface, through which the net flux is zero. So the flux through the curved surface equals that through the flat circle: $B\pi r^2=0.50\times\pi(0.10)^2\approx1.6\times10^{-2}$ Wb.
II
Faraday's law: induced emf, current and charge.
  • 1 State Faraday's law of electromagnetic induction. List three ways of changing the flux through a coil.
  • 2 A coil of $500$ turns, each of area $4.0\times10^{-3}\ \text{m}^2$, is perpendicular to a field that grows steadily from $0$ to $0.30$ T in $0.15$ s. Find the induced emf.
  • 3 The flux through a loop is $\Phi_B=(2t^3-3t)$ mWb ($t$ in s). Find the emf at $t=2.0$ s.
  • 4 The flux linkage of a coil of resistance $5.0\ \Omega$ falls from $0.050$ Wb to $0.010$ Wb. How much charge flows through the coil? Does the answer depend on how quickly the flux falls?
  • 5 The flux through a loop rises steadily from $0$ to $6.0$ mWb between $t=0$ and $2.0$ s, stays constant until $4.0$ s, then falls steadily to zero at $5.0$ s. Find the emf in each interval and sketch the emf–time graph.
  • 6 A coil of $100$ turns and radius $3.0$ cm is perpendicular to a field $B=B_0\sin\omega t$ with $B_0=0.050$ T and frequency $50$ Hz. Find the greatest emf induced.
  • 7 A single loop of area $0.10\ \text{m}^2$ and resistance $0.50\ \Omega$ is perpendicular to a field that falls at a steady $0.020$ T/s. Find the induced emf, the current and the rate at which heat is produced.
  • 8 A search coil of $50$ turns, area $1.5\ \text{cm}^2$, in a circuit of total resistance $30\ \Omega$, is pulled quickly out of the gap between two magnet poles (where its plane was perpendicular to the field) to a place where the field is negligible. A charge of $2.5\times10^{-5}$ C flows. Find the field in the gap.
Check: average emf, current and charge
$N\Delta\Phi_B$—
average emf—
current—
charge $Q$—
1 The emf induced in a circuit equals the negative rate of change of the magnetic flux through it, $\varepsilon=-N\,d\Phi_B/dt$. The flux $BA\cos\theta$ can be changed by changing $B$ (moving a magnet, switching a current), the area $A$ (squeezing a loop, sliding a rod) or the angle $\theta$ (rotating the coil).
2 $|\varepsilon|=\dfrac{NA\,\Delta B}{\Delta t}$ $=\dfrac{500\times4.0\times10^{-3}\times0.30}{0.15}$ $=4.0$ V.
3 $\varepsilon=-\dfrac{d\Phi_B}{dt}=-(6t^2-3)$ mV $=-(24-3)=-21$ mV at $t=2.0$ s; magnitude $21$ mV.
4 $Q=\dfrac{|\Delta(N\Phi_B)|}{R}=\dfrac{0.040}{5.0}=8.0\times10^{-3}$ C. No: the charge depends only on the change of flux linkage.
5 $\varepsilon=-d\Phi_B/dt$: $0$–$2$ s: $-\dfrac{6.0}{2.0}=-3.0$ mV; $2$–$4$ s: $0$; $4$–$5$ s: $-\dfrac{(0-6.0)}{1.0}=+6.0$ mV. The graph is a set of flat steps: $-3.0$ mV, then zero, then $+6.0$ mV. (The steeper the flux graph, the taller the step.)
6 $\varepsilon=-NA\dfrac{dB}{dt}=-NAB_0\omega\cos\omega t$, so $\varepsilon_\text{max}=NAB_0\omega$ $=100\times\pi(0.030)^2\times0.050\times2\pi(50)$ $\approx4.4$ V.
7 $\varepsilon=A\,\dfrac{dB}{dt}=0.10\times0.020=2.0\times10^{-3}$ V; $I=\dfrac{2.0\times10^{-3}}{0.50}=4.0$ mA; $P=I^2R=\dfrac{\varepsilon^2}{R}=8.0\times10^{-6}$ W.
8 $Q=\dfrac{NBA}{R}$, so $B=\dfrac{QR}{NA}$ $=\dfrac{(2.5\times10^{-5})(30)}{50\times1.5\times10^{-4}}$ $=0.10$ T.
III
Lenz's law: directions and energy.
  • 1 State Lenz's law and explain why it is a consequence of the conservation of energy.
  • 2 The N pole of a bar magnet is pushed towards one face of a coil. What polarity does that face acquire? Is the current anticlockwise or clockwise as seen from the magnet?
  • 3 The magnet of part 2 is now pulled away. Repeat part 2.
  • 4 Two coils are placed side by side on a common axis. A switch in the circuit of the first coil (with a battery) is closed, left closed, and later opened. Describe the current in the second coil at each stage, relative to the current in the first.
  • 5 A rectangular loop lies flat on a table beside a long straight wire on the table, the wire parallel to the near side of the loop. The current in the wire is increasing. In the side of the loop nearer to the wire, is the induced current parallel or antiparallel to the current in the wire? Is the loop attracted or repelled by the wire?
  • 6 A horizontal copper ring is held below a bar magnet whose S pole points down. The magnet is dropped through the ring. Looking from above, give the direction of the current as the magnet approaches the ring. Is the magnet's acceleration greater than, equal to or less than $g$?
  • 7 A loop of wire lies in the plane of the screen in a field pointing into the screen, and the field is decreasing. Is the induced current clockwise or anticlockwise?
  • 8 A student suggests that if induced currents aided the change of flux, a magnet pushed gently into a coil would keep speeding up. Explain why this would violate energy conservation.
1 The induced current flows in the direction whose magnetic field opposes the change of flux that causes it. Because of this opposition, an external agent must do work (pushing a magnet against repulsion, turning a coil against a torque) to keep changing the flux; that work becomes the electrical energy dissipated as heat. A current that aided the change would deliver energy with no work done.
2 The face becomes an N pole (repelling the approaching N pole), so the current is anticlockwise as seen from the magnet.
3 The face becomes an S pole (attracting the receding N pole); the current is clockwise as seen from the magnet.
4 On closing: a brief current in the second coil opposite to that in the first (opposing the growing flux). While the current is steady: no induced current. On opening: a brief current in the same direction as the first coil's former current (opposing the collapse of flux).
5 The wire's growing field through the loop is opposed, so the induced current in the near side is antiparallel to the wire's current (it circulates so as to make the opposite field inside the loop). Antiparallel currents repel, and the near side is closer than the far side, so the loop is repelled — moving away reduces the flux increase, as Lenz's law requires.
6 Field lines enter the S pole, so below the magnet the field through the ring points upwards, and this upward flux increases as the magnet approaches. The induced field points downwards inside the ring: the current is clockwise seen from above (the top face is an S pole, repelling the S pole). The acceleration is less than $g$.
7 The current tries to keep the into-the-screen flux up, so it makes a field into the screen inside the loop: clockwise.
8 The coil's near face would attract the magnet, the magnet would speed up, the flux would change faster, the current and attraction would grow, and so on: kinetic energy and heat would increase without limit with no work done — energy created from nothing. Nature forbids it, so the induced current must oppose the change.
IV
Harder problems and applications. $\mu_0=4\pi\times10^{-7}\ \text{T m/A}$.
  • 1 A flexible circular loop of radius $10$ cm is perpendicular to a field of $0.50$ T. It is squeezed in $0.10$ s so that its area halves. Find the average emf.
  • 2 A single loop is wound round a long solenoid of radius $2.0$ cm with $1000$ turns per metre. The current in the solenoid is reversed from $2.0$ A to $-2.0$ A in $0.050$ s. Find the change in the field inside the solenoid and the average emf in the loop. Does the size of the loop matter?
  • 3 A coil of $50$ turns, each of area $0.010\ \text{m}^2$, with resistance $2.0\ \Omega$, is perpendicular to a field $B=(0.20t-0.020t^2)$ T. Find (a) the current at $t=2.0$ s, (b) the instant when the current is zero, and (c) the charge that flows between $t=0$ and that instant.
  • 4 Modelling. A wireless charging pad makes a field through a phone's receiving coil of $B=B_0\sin(2\pi ft)$, with $B_0=50\ \mu\text{T}$ and $f=100$ kHz (typical-order values). The receiving coil has $10$ turns of radius $2.0$ cm. Find the peak emf. Why are such pads driven at high frequency?
  • 5 Modelling. In an emergency shutdown, the $1.5$ T field of a hospital MRI scanner is reduced to zero in about $20$ s. Estimate the average emf round a loop of radius $10$ cm inside a patient's body, perpendicular to the field. Is it dangerous?
  • 6 A steady emf of $2.0$ V is induced in a coil of $100$ turns for $3.0$ s. By how much did the flux through each turn change?
1 $|\Delta\Phi_B|=B\,\dfrac{A}{2}$ $=0.50\times\dfrac{\pi(0.10)^2}{2}$ $\approx7.9\times10^{-3}$ Wb; $\varepsilon_\text{av}=\dfrac{7.85\times10^{-3}}{0.10}$ $\approx7.9\times10^{-2}$ V $=79$ mV.
2 $\Delta B=\mu_0n\,\Delta I$ $=(4\pi\times10^{-7})(1000)(4.0)$ $\approx5.0\times10^{-3}$ T. $\varepsilon_\text{av}=\dfrac{\pi(0.020)^2\times5.03\times10^{-3}}{0.050}$ $\approx1.3\times10^{-4}$ V. The loop's size does not matter (as long as it encloses the solenoid), because the field outside a long solenoid is negligible.
3 $\varepsilon=-NA\dfrac{dB}{dt}=-0.50(0.20-0.040t)$ V. (a) At $t=2.0$ s: $|\varepsilon|=0.50\times0.12=0.060$ V, $I=0.030$ A. (b) $0.20-0.040t=0$ at $t=5.0$ s ($B$ is then at its maximum, $0.50$ T). (c) $Q=\dfrac{NA\,\Delta B}{R}=\dfrac{50\times0.010\times0.50}{2.0}=0.125$ C.
4 $\varepsilon_0=NAB_0(2\pi f)$ $=10\times\pi(0.020)^2$ $\times(5.0\times10^{-5})\times2\pi(10^5)$ $\approx0.39$ V. The emf is proportional to the rate of change of flux, so a high frequency gives a useful emf from a weak, safe field (it is then rectified to charge the battery).
5 $\varepsilon\approx\dfrac{\pi(0.10)^2\times1.5}{20}\approx2.4\times10^{-3}$ V $=2.4$ mV — far too small to stimulate nerves; that is why the field is ramped down slowly rather than switched off suddenly.
6 $|\Delta\Phi_B|=\dfrac{\varepsilon\,\Delta t}{N}=\dfrac{2.0\times3.0}{100}=6.0\times10^{-2}$ Wb.

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 — flux, Faraday's law and Lenz's law BoardJEE MainJEE AdvancedNEET

E1. Board · 1 mark Assertion (A): When a magnet is pushed into a coil quickly, the galvanometer joined to the coil deflects more than when it is pushed in slowly. Reason (R): The charge that flows round the circuit is greater when the flux changes faster. Choose: (A) both A and R true and R is the correct explanation of A; (B) both true but R is not the correct explanation; (C) A true, R false; (D) A false, R true.

Concept The emf, and so the current (the deflection), is proportional to $d\Phi_B/dt$: A is true. The charge $Q=N|\Delta\Phi_B|/R$ does not depend on the rate: R is false.
Answer (C). Exam tip: current depends on the rate of change of flux; charge depends only on the change.

E2. Board · 3 marks State Faraday's law and Lenz's law of electromagnetic induction. A coil of $100$ turns, each of area $0.020\ \text{m}^2$, is held perpendicular to a field of $0.50$ T and is pulled out of the field in $0.10$ s. Find the average induced emf, and explain why work must be done to pull it out if the coil is part of a closed circuit.

Laws Faraday: the induced emf equals the negative rate of change of flux linkage, $\varepsilon=-N\,d\Phi_B/dt$. Lenz: the induced current opposes the change of flux that produces it.
emf $\Delta(N\Phi_B)=NBA$ $=100\times0.50\times0.020$ $=1.0$ Wb, so $\varepsilon_\text{av}=\dfrac{1.0}{0.10}=10$ V.
Work The induced current flows so as to keep the flux up, i.e. its field attracts the coil back into the field; a force must be applied through a distance, and this work appears as heat in the circuit (energy conservation). Exam tip: mark schemes reward "opposes the change" — not "opposes the field".

E3. JEE Main (Numerical value.) The flux through each turn of a $50$-turn coil varies as $\Phi_B=(5t^2-3t+2)$ mWb, with $t$ in seconds. The coil is connected to a resistor so that the total resistance is $2.5\ \Omega$. Find the current, in mA, at $t=1.0$ s.

Rate $\dfrac{d\Phi_B}{dt}=10t-3=7.0$ mWb/s at $t=1.0$ s.
Answer $|\varepsilon|=50\times7.0\times10^{-3}=0.35$ V; $I=\dfrac{0.35}{2.5}=0.14$ A $=\mathbf{140}$ mA. Exam tip: the constant term in $\Phi_B$ never matters.

E4. NEET The magnetic flux linked with a coil of resistance $20\ \Omega$ increases from $3.0$ mWb to $9.0$ mWb. The charge that flows through the coil is (A) $1.2\times10^{-4}$ C (B) $3.0\times10^{-4}$ C (C) $0.30$ C (D) impossible to find without the time taken.

Concept $Q=\dfrac{|\Delta(N\Phi_B)|}{R}$, independent of time.
Answer (B) $Q=\dfrac{6.0\times10^{-3}}{20}=3.0\times10^{-4}$ C. Exam tip: option (D) is the trap.

E5. JEE Advanced (One or more correct.) A bar magnet is held vertically, N pole down, above a horizontal conducting ring and released; it falls along the axis of the ring and passes through it. Air resistance is negligible. Which statements are correct? (A) While approaching the ring, the magnet's acceleration is less than $g$. (B) After passing through, while moving away below the ring, its acceleration is less than $g$. (C) Seen from above, the current in the ring has the same sense in both stages. (D) The ring warms up.

Approach Downward flux increasing → anticlockwise current from above; the ring's top face is N and repels the magnet: force upward, $a\lt g$ (A) ✓.
Leaving Downward flux decreasing → clockwise current from above; the ring's bottom face is N and attracts the magnet's upper S pole: force upward again, $a\lt g$ (B) ✓. The sense reverses, so (C) ✗.
Energy The current dissipates energy $I^2R$ taken from the magnet's mechanical energy (D) ✓.
Answer (A), (B), (D). Exam tip: the force on the magnet opposes its motion in both stages, although the current reverses.

E6. Board · Case study A laboratory measures the field of an electromagnet with a search coil of $200$ turns, each of area $1.0\ \text{cm}^2$, connected to a charge-measuring meter; the total resistance is $50\ \Omega$. The coil is placed in the gap with its plane perpendicular to the $0.25$ T field and is flipped over through $180^\circ$. (a) Find the charge that passes through the meter. (2 marks) (b) If the flip is done twice as fast, what happens to this charge? (1 mark) (c) If the flip takes $0.020$ s, find the average emf. (1 mark)

(a) Flipping reverses the flux: $|\Delta\Phi_B|=2BA$ $=2\times0.25\times1.0\times10^{-4}$ $=5.0\times10^{-5}$ Wb per turn. $Q=\dfrac{N|\Delta\Phi_B|}{R}$ $=\dfrac{200\times5.0\times10^{-5}}{50}$ $=2.0\times10^{-4}$ C.
(b) Unchanged, $2.0\times10^{-4}$ C: the current doubles but lasts half as long.
(c) $\varepsilon_\text{av}=\dfrac{N|\Delta\Phi_B|}{\Delta t}=\dfrac{1.0\times10^{-2}}{0.020}=0.50$ V. Exam tip: a $180^\circ$ flip doubles the flux change — the most common lost mark in this problem.
🔬 Research–Teaching NexusResearch-tutored · optional, about 40 minutes

Research question. When a magnet falls through a coil, how does the peak induced emf depend on the height from which it is dropped — and what stays the same however fast it falls?

Why it matters. The same physics is used to measure speed without contact: the wheel-speed sensors of anti-lock brakes, the pickups that time the crankshaft in a car engine, the coils that record the speed of trains, and the seismometers that detect earthquakes all turn the motion of a magnet (or a piece of steel) near a coil into a voltage pulse. Engineers need to know how the pulse height and shape depend on the speed.

What to do.

  1. The experiment: a small magnet is dropped down a vertical plastic tube through a coil of a few hundred turns; the coil is connected to a data logger (or a phone/computer sound-card input) that records the emf. The magnet is released from rest a height $h$ above the coil. Illustrative results for the first (entry) peak: $h=5$ cm: $50$ mV; $10$ cm: $74$ mV; $20$ cm: $101$ mV; $40$ cm: $147$ mV; $80$ cm: $203$ mV.
  2. Faraday's law gives $\varepsilon=-N\dfrac{d\Phi_B}{dz}\,v$, and the speed at the coil is $v=\sqrt{2gh}$. So the theory predicts $\varepsilon_\text{peak}\propto\sqrt h$. Plot $\varepsilon_\text{peak}$ against $\sqrt h$. Is it a straight line through the origin? Find its slope.
  3. Each recording has two peaks of opposite sign — one as the magnet enters, one as it leaves. Explain the signs with Lenz's law, and explain why the second peak is a little taller and narrower than the first.
  4. The area under each peak, $\int\varepsilon\,dt$, equals the change of flux linkage $N\Delta\Phi_B$. Predict how this area depends on $h$, and how you could use it to compare the strengths of two magnets.

Tools: Spreadsheet or graph paper; optionally a real coil, magnet, tube and a free data-logging or audio-recording app.

Hand in: The table of $h$, $\sqrt h$ and $\varepsilon_\text{peak}$; the graph with your best line; a sketch of one pulse with both peaks labelled; and a paragraph on what the pulse area tells you.

Going further. Drop the magnet through a copper tube instead of a plastic one and it falls noticeably more slowly. Use Lenz's law to explain why, and find out how this effect is used in the brakes of some roller coasters and trains.

Instructor note: the speeds at the coil are $0.99,\ 1.40,\ 1.98,\ 2.80,\ 3.96$ m/s. A least-squares line through the origin of $\varepsilon_\text{peak}$ against $\sqrt h$ (cm) has slope about $22.9$ mV per $\sqrt{\text{cm}}$, predicting $51,\ 72,\ 102,\ 145,\ 204$ mV — within about $3\%$ of every point. The second peak is taller because the magnet has gained speed while passing through, narrower for the same reason; the two peak areas are equal and opposite ($\pm N\Phi_\text{max}$), independent of $h$, which is the basis for comparing magnet strengths. The copper-tube question previews eddy currents, which lie beyond the board syllabus (Section 6.2 has a box on them).

🧪 Self-check quiz

0/5 correct

Q1. The SI unit of magnetic flux is the

Q2. A plane coil lies in a uniform magnetic field with its plane parallel to the field. The flux through it is

Q3. According to Faraday's law, the emf induced in a coil depends on

Q4. Lenz's law is a consequence of the law of conservation of

Q5. The charge that flows round a closed coil when the flux through it changes does not depend on

← Chapter 6 — Electromagnetic Induction6.2 Motional EMF and Energy Considerations →