Electromagnetic Induction

Written and reviewed by Akanksha Tiwari, Physics & Chemistry faculty at UniDrill (M.Sc. Physics, Banaras Hindu University · 8 years teaching Class XI–XII science). Based on NCERT Physics Class XII, Chapter 6.

Why does a magnet fall slowly through a copper pipe?

Copper is not magnetic. A magnet does not stick to it. Yet if you drop a strong magnet down a vertical copper pipe, it drifts down far more slowly than free fall, as if the pipe were full of thick syrup. A plain piece of the same metal, dropped through the same pipe, falls freely.

There is no battery, no wire and no switch. The explanation is the subject of this chapter: a changing magnetic flux creates an electric field, which drives currents in any conductor nearby, and those currents push back on whatever caused them. Everything here, from the power station to the transformer to the phone charger, is a use of that one idea.

What Faraday and Henry saw

Around 1830, Michael Faraday in England and Joseph Henry in the USA found that a magnetic field can produce an electric current, provided the field is changing.

The key lesson: the current appears only while something is changing. Relative motion is one way to do it, but changing the current in a neighbouring coil does it without any motion at all. What changes in every case is the magnetic flux through the coil.

Magnetic flux

The magnetic flux through a flat area A in a uniform field B is

$$\Phi_B = BA\cos\theta$$

where θ is the angle between B and the area vector (the normal to the surface), not the angle between B and the plane. Flux is a scalar. Its SI unit is the weber: $1\ \text{Wb} = 1\ \text{T m}^2 = 1\ \text{V s}$. If the field is not uniform, add the contributions of small pieces: $\Phi_B = \sum B \cdot \Delta A$.

Flux depends on three things, and changing any one of them induces an emf.

Faraday's law

The emf induced in a circuit equals the rate of change of flux through it:

$\varepsilon = -d\Phi_B/dt$, and for a closely wound coil of N turns, $\varepsilon = -N\,d\Phi_B/dt$

The product $N\Phi_B$ is the flux linkage. If the circuit has resistance R, the induced current is $I = \varepsilon/R$. The total charge that flows while the flux changes by ΔΦ is

$$Q = I\Delta t = \dfrac{N\Delta\Phi}{R}$$

This charge depends only on the total flux change and the resistance, not on how quickly the change happens. A faster change gives a larger emf for a shorter time, with the same total charge.

Worked example: a changing field through a coil. A coil of 100 turns and area 0.020 m² has resistance 5.0 Ω. The field through it makes an angle of 60° with the area vector. The field falls steadily from 0.50 T to 0.10 T in 0.20 s. Find the induced emf and the total charge that flows.

Common wrong step: using sin 60° because the problem says "60° to the field". The 60° is measured from the area vector, so cos 60° is right. If the angle were measured from the plane of the coil, you would use sin. Forgetting N is the other usual slip: it makes the emf 100 times too small.

Pause and predict: a magnet is pushed into a coil in 1 s and then, in a second trial, in 0.5 s. Compare the induced emf and the total charge that flows through the galvanometer.

Check your answer: the emf is twice as large in the faster trial, because dΦ/dt doubles. But the total charge is the same, because $Q = N\Delta\Phi/R$ depends only on the flux change, and that is the same in both trials.

Lenz's law: the minus sign

The minus sign in Faraday's law is Lenz's law: the induced current flows in the direction that opposes the change in flux that produced it. Note that it opposes the change, not the flux itself. If the flux is increasing, the induced field points against it. If the flux is decreasing, the induced field points along it, trying to keep the flux up.

When a north pole moves towards a coil, the coil's nearer face becomes a north pole and repels the magnet. A current that makes the near face north flows anticlockwise as seen from the magnet. When the same north pole is pulled away, the near face becomes a south pole and attracts the magnet back, so the current reverses.

Why it must be so. Suppose the induced current helped the change instead of opposing it. A tiny push on the magnet would induce a current that pulled the magnet in faster, which induces a larger current, and so on. The magnet would speed up and the coil would heat up with no one doing any work. That would create energy from nothing. Lenz's law is therefore not an extra rule. It is conservation of energy. The work you do against the opposing force is what appears as heat in the coil.

This also explains the copper pipe. The falling magnet changes the flux through every ring of the pipe. Currents circulate in the pipe, and by Lenz's law their magnetic effect opposes the fall, so the magnet is slowed.

Beyond the textbook. For the exam: induced currents oppose the change of flux, and the work done against them becomes heat. In reality: currents induced in a solid piece of metal, called eddy currents, are put to use. They heat the base of the pan on an induction cooker, brake some trains without any contact, and make metal detectors work. They are a nuisance in transformer cores, which is why those cores are built from thin insulated sheets (laminations) that cut the paths for eddy currents.

Motional emf

Slide a straight rod of length l with speed v through a uniform field B, with v, l and B at right angles to each other. Each free charge in the rod moves with the rod, so the magnetic force qvB pushes positive charge towards one end. Charge piles up until the electric force balances it. The work done per unit charge along the rod is vBl, so the emf is

$$\varepsilon = Blv$$

The same answer comes from Faraday's law. In a closed circuit with a sliding rod, the area grows at the rate l v, so $d\Phi/dt = Blv$. In general, $\varepsilon = Blv\sin\theta$ for the component of v perpendicular to B, provided the rod is also perpendicular to the plane of v and B. If the rod slides along its own length, no emf is induced.

Now let the rod slide on parallel rails joined by a resistor so that the circuit is closed. The emf drives a current $I = Blv/R$. A current in a field means a force: the magnetic force on the rod is $F = BIl = B^2 l^2 v/R$, and by Lenz's law it points opposite to v. To keep the rod moving at constant speed you must push with this force, and the power you supply, $Fv = B^2 l^2 v^2/R$, equals the electrical power $I^2 R$ dissipated in the circuit.

Worked example: a rod on rails. A rod of length 0.50 m slides at 4.0 m s⁻¹ on rails in a uniform field of 0.30 T perpendicular to the rails. The total resistance of the circuit is 0.60 Ω. Find the emf, the current, the force needed to keep the speed constant and the power supplied.

Common wrong step: arguing that no force is needed because the speed is constant. The speed is constant only because you are pushing with 0.15 N to balance the magnetic drag. Take away your push and the rod slows down.

Rotating rod. A rod of length R rotating about one end at angular speed ω in a field B parallel to the axis has points moving at $v = \omega r$, which gives a different emf for each tiny element. Adding up the contributions from $r = 0$ to $r = R$:

$$\varepsilon = \dfrac{B\omega R^2}{2}$$

Worked example: a rotating rod. A metal rod 0.50 m long rotates about one end at 10 revolutions per second in a field of 0.20 T perpendicular to the plane of rotation. Find the emf between the ends.

Common wrong step: writing $\varepsilon = B\omega R^2$ and getting 3.14 V, or putting the frequency 10 into the formula instead of ω. A rod's average speed is half the speed of its tip, and that is where the ½ comes from. For a wheel with many spokes, the spokes act like batteries in parallel, so the number of spokes makes no difference.

The induced electric field

When a conductor is at rest and only B changes, there is no magnetic force on the charges, since $v = 0$. Something must still push them, and that something is an electric field. A changing magnetic field creates an electric field. This induced field is unlike the electrostatic field. Its field lines form closed loops around the changing flux, and the work done in carrying a charge once round a loop is not zero: $\oint E \cdot dl = -d\Phi/dt$. It is non-conservative, so it has no potential.

For a uniform field changing at rate dB/dt, through a circle of radius r that lies inside the field region, symmetry gives $E \times 2\pi r = \pi r^2\, dB/dt$, so

$$E = \dfrac{r}{2} \times \dfrac{dB}{dt}$$

Inductance

When the current in a coil changes, the flux through it, and through any neighbouring coil, changes too. Inductance is the constant that links flux linkage to current. It depends only on geometry and the material of the core.

Mutual inductance M. If a current I₂ in coil 2 produces flux linkage $N_1\Phi_1$ in coil 1, then $N_1\Phi_1 = M I_2$, and a changing current induces an emf in the other coil:

$$\varepsilon_1 = -M\,\dfrac{dI_2}{dt}$$

Mutual inductance is symmetric: $M_{12} = M_{21}$. For two long coaxial solenoids of common length l, the inner of radius r₁ with n₁ turns per unit length and the outer with n₂ turns per unit length:

$$M = \mu_0 n_1 n_2 \pi r_1^2 l$$

The radius is the inner one, since the flux only exists as far out as the inner solenoid's cross-section.

Self inductance L. A coil also links with its own flux: $N\Phi = LI$. A changing current induces an emf in the coil itself,

$$\varepsilon = -L\,\dfrac{dI}{dt}$$

This is called the back emf because it opposes any change in the current, whether it is rising or falling. The unit of inductance is the henry: $1\ \text{H} = 1\ \text{V s A}^{-1} = 1\ \text{Wb A}^{-1}$. For a long solenoid of cross-section A and length l with n turns per unit length ($N = nl$):

$L = \mu_0 n^2 A l$ in air, and $L = \mu_r \mu_0 n^2 A l$ with a core of relative permeability $\mu_r$

$L \propto N^2$ for a given length and area. Doubling the turns quadruples the inductance. An iron core can multiply it by hundreds.

Worked example: a solenoid. A long solenoid has 1000 turns uniformly wound over a length of 0.50 m with cross-sectional area 4.0 × 10⁻⁴ m². Find its self-inductance, the average emf induced if the current falls from 2.0 A to zero in 1.0 ms, and the energy stored at 2.0 A.

Common wrong step: putting $N = 1000$ where the formula needs $n = 2000$ turns per metre, which makes L wrong by a factor of 4 and the units inconsistent. In $L = \mu_0 n^2 A l$, n is per metre and A l is the volume. Equivalently, $L = \mu_0 N^2 A/l$.

Worked example: mutual inductance. An inner solenoid of radius 2.0 cm with 1000 turns per metre is placed inside a coaxial outer solenoid with 2000 turns per metre. Both are 0.50 m long. Find M.

Common wrong step: using the outer radius. The field of the outer solenoid is uniform over the inner one, but only the inner solenoid's cross-section links the flux.

Energy in an inductor. To build up a current I against the back emf, the source must do work, and it is stored in the magnetic field:

$W = 1/2\,LI^2$, and the energy per unit volume is $u_B = B^2/(2\mu_0)$

This matches the kinetic energy $1/2\,mv^2$. Inductance plays the role of mass and current plays the role of velocity: the inductor resists a change in current just as mass resists a change in velocity.

Beyond the textbook. For the exam: a coil opposes changes in current with a back emf $\varepsilon = -L\,dI/dt$. In reality: when a switch is opened in a circuit with a large inductance, the current is forced to fall very rapidly, so the induced emf becomes very large and can force a spark across the switch contacts. A car's ignition coil uses exactly this to turn a 12 V supply into thousands of volts for the spark plug.

The AC generator

An AC generator rotates a coil of N turns and area A at a steady angular speed ω in a uniform field B. At time t the angle between the area vector and B is $\theta = \omega t$, so the flux is $\Phi = BA\cos\omega t$, and Faraday's law gives

$\varepsilon = NBA\omega\sin\omega t = \varepsilon_0 \sin\omega t$, with peak emf $\varepsilon_0 = NBA\omega$

The emf alternates sinusoidally. In India the frequency is 50 Hz, so $\omega = 2\pi \times 50 \approx 314$ rad s⁻¹.

This is a favourite trap: the emf is largest when the flux is zero, because the flux is changing fastest there.

Worked example: a generator. A generator has a coil of 200 turns and area 0.050 m² rotating at 50 Hz in a field of 0.20 T. Find the peak emf.

Common wrong step: using $f = 50$ in place of $\omega = 314$, which gives only 100 V.

Exam traps

Formula sheet

60-second revision

Where this chapter connects

The magnetic fields and forces that make induction work come from Moving Charges and Magnetism, and the cores that boost inductance are the materials of Magnetism and Matter. The sinusoidal emf of the generator is the starting point of Alternating Current, where inductors and transformers are put to work. The idea that a changing magnetic field creates an electric field leads directly to Electromagnetic Waves.