Electromagnetic Induction

Every free charge $q$ inside the moving rod feels a magnetic force $\vec{F} = q\vec{v}\times\vec{B}$, pushing positive charge toward one end of the rod. Charge piles up there until the resulting electric field inside the rod exactly balances the magnetic force, at equilibrium:

The potential difference between the rod’s ends is this field times the rod’s length:

If the rod is a distance $x$ from the closed end of the rails, the enclosed area is $Lx$ and the flux is $\Phi = BLx$. Applying Faraday’s law:

Both routes agree exactly, as they must.

In a moving-coil galvanometer, the coil is wound on a light aluminium frame (or swings around a fixed non-magnetic metal core). Whenever the coil swings, the flux through that metal changes, and the eddy currents drag against the motion, so instead of overshooting and oscillating about its final reading, the pointer settles almost at once. Crucially, the drag is proportional to the coil’s speed, so it vanishes the moment the pointer stops: it damps the motion without shifting the reading itself.

The same effect, scaled up, brakes trains. Switch on strong electromagnets near the rails (or near metal discs fixed to the axles), and the moving metal carries eddy currents whose drag slows the train with nothing touching, so there is no wear and no jolt. The speed dependence is the catch as well as the virtue: the braking force is strongest at high speed, but it fades away as the train slows and can never hold it stationary, so ordinary friction brakes complete the stop. Many roller coasters use permanent magnets and metal fins for the same job, which needs no power at all and so keeps working even in a power cut.

Here the heating is the whole point. In an induction furnace, a coil carrying high-frequency alternating current surrounds a crucible of metal. The coil’s field reverses thousands of times a second, so the flux through the metal changes very rapidly, inducing large EMFs and large eddy currents inside the metal itself. Their $I^2R$ heating melts the metal from within, with no flame and no contact, which also keeps the melt clean, ideal for preparing alloys. The high frequency matters because the induced EMF is the rate of change of flux: reversing the same field twice as often doubles the EMF, and the heating, which goes as $\varepsilon^2/R$, grows faster still.

An induction cooktop is the same idea in a kitchen: a flat coil under a ceramic-glass top carries high-frequency AC, and eddy currents induced in the iron or steel base of the pan heat the pan directly. The glass top is an insulator, with no free charges to set circulating, so it carries no eddy currents of its own and gets warm only from contact with the hot pan.

The traditional analogue speedometer turns speed into a needle position using eddy currents. A permanent magnet, driven by a cable from the wheels, spins inside a light aluminium cup that is free to pivot but held back by a hairspring; the needle is fixed to the cup. As the magnet spins, its field sweeps across the cup, the flux through each patch of aluminium keeps changing, and eddy currents flow. By Lenz’s law they oppose the relative motion between magnet and cup, which they do by dragging the cup round in the same direction as the magnet. The cup turns until the hairspring’s restoring torque, $k\phi$, balances that drag. Because the drag torque is proportional to the magnet’s speed (the §6.5 scaling once more), the deflection $\phi$ is proportional to the vehicle’s speed too, giving an evenly spaced dial. It is the galvanometer’s balance-against-a-spring logic, $k\phi = NIAB$, with an eddy-current drag standing in for the torque on a current loop.

Start the clock ($t=0$) when the coil’s area vector points along $\vec{B}$, that is, when the plane of the coil is perpendicular to the field. After a time $t$ the coil has turned through an angle $\theta = \omega t$, so the flux through each turn is:

All $N$ turns are threaded by this same flux, so the coil’s total flux linkage is $N\Phi = NBA\cos(\omega t)$. Faraday’s law, with $N$, $B$ and $A$ all constant:

By the chain rule, $\frac{d}{dt}\left[\cos(\omega t)\right] = -\omega\sin(\omega t)$, and the two minus signs cancel:

Since $\sin(\omega t)$ swings between $+1$ and $-1$, the EMF swings between $+\varepsilon_0$ and $-\varepsilon_0$. The peak value $\varepsilon_0 = NBA\omega$ grows with each of the four things you can turn up: more turns, a stronger field, a bigger coil, or faster spinning. Writing $\omega = 2\pi\nu$, with $\nu$ the number of revolutions per second, the same result reads $\varepsilon = \varepsilon_0\sin(2\pi\nu t)$.

Take the coil to be a rectangle, with its two sides of length $a$ parallel to the rotation axis and its two sides of length $b$ perpendicular to it, so $A = ab$. Each $a$-side sits a distance $b/2$ from the axis and moves in a circle at speed $v = \omega b/2$. Its velocity is perpendicular both to the axis and to the arm joining it to the axis (which lies in the coil’s plane), so the velocity points along the area vector and makes the same angle $\omega t$ with $\vec{B}$. Only the part of the velocity perpendicular to $\vec{B}$, $v\sin(\omega t)$, pushes charges along the wire, so by §6.4.1 the motional EMF along each $a$-side is:

The two $a$-sides move in opposite directions, so their EMFs point opposite ways in space, but that is exactly the arrangement that drives current the same way round the loop (up one side, down the other), so the two add. The $b$-sides contribute nothing: the magnetic force on their charges points along the axis direction, perpendicular to these wires, so it cannot push charge along them. Adding the two sides, then the $N$ turns in series:

The same result, as it must be, now from the force on the charges rather than the flux rule, just as §6.4 found two routes to $BLv$. This picture also shows physically why the EMF vanishes at $\theta=0$: there the $a$-sides are moving straight along the field lines, and a charge moving parallel to $\vec{B}$ feels no magnetic force at all.

Put the flux and the EMF side by side and a result appears that surprises most people the first time:

• At $\theta = 0^\circ$ and $180^\circ$ the coil faces the field squarely and its flux linkage is at its largest, $\pm NBA$, yet the EMF is zero: the flux has reached its peak (or trough) and, for that instant, is not changing at all.
• At $\theta = 90^\circ$ and $270^\circ$ the coil is edge-on to the field and the flux through it is zero, yet the EMF is at its peak, $\pm\varepsilon_0$: the flux is sweeping through zero at its steepest.

It is §6.2’s lesson again: EMF follows how fast the flux is changing, never how much flux there is. Because $\varepsilon$ changes sign every half revolution, any current it drives through an external circuit reverses direction twice per turn. This is alternating current, and for this simple one-coil machine its frequency is just the coil’s rotation frequency $\nu$; India’s mains supply alternates at 50 Hz.

This sinusoid is exactly the source voltage $V = V_0\sin(\omega t)$ that the Alternating Current chapter takes as its starting point: its peak $V_0$ is this $NBA\omega$, and its $\omega$ is literally the generator’s angular speed.

The field a coil produces is proportional to the current in it (double the current and, by the Biot–Savart law, every element of wire contributes double the field), so the flux $\Phi$ through each turn is proportional to $I$ too. For a coil of $N$ turns all threaded by the same flux, the total flux linkage is $N\Phi$, and its constant of proportionality with the current is the self-inductance $L$:

Feeding this into Faraday’s law, with $L$ constant as long as the coil keeps its shape:

The argument also runs backwards. If a coil obeys $\varepsilon = -L\,dI/dt$, Faraday’s law says $-d(N\Phi)/dt = -L\,dI/dt$, so:

and the constant must be zero, because a coil carrying no current makes no field of its own ($N\Phi = 0$ when $I = 0$). So $N\Phi = LI$ and $\varepsilon = -L\,dI/dt$ are two statements of the same fact: $L$ is the flux linkage per ampere of current, and equally the back-EMF per ampere-per-second of current change.

Take a long solenoid of length $l$ and cross-sectional area $A$, with $n$ turns per unit length, so $N = nl$ turns in all. Three steps, each using a result already in hand.

Step 1, the field. Ampere’s law gives the field inside a long solenoid (derived in Moving Charges and Magnetism, §4.7.1), uniform across the interior and directed along the axis:

Step 2, the flux through one turn. That uniform field crosses each turn’s area $A$ at right angles ($\theta = 0$), so:

Step 3, the flux linkage. All $N = nl$ turns are threaded by this same flux (neglecting the weaker field near the ends):

Comparing with $N\Phi = LI$, the current cancels out:

Exactly as promised, $L$ contains only geometry (turns per length, area, length) and the constant $\mu_0$, and no current anywhere. The $n^2$ is worth understanding rather than memorising: one factor of $n$ comes from the field (more turns per metre make a stronger $B$), the other from the linkage (more turns per metre means more turns catching that flux). And since $Al$ is the solenoid’s volume, a longer or fatter solenoid with the same winding density has proportionally more inductance. Writing $n = N/l$ gives an equivalent form in total turns, which §6.9.1 will use:

If the solenoid is filled with a core of relative permeability $\mu_r$ (soft iron, say), the magnetised core multiplies the field inside, for the same current, by $\mu_r$. Every step above simply carries $B$ through to $L$, so $L$ picks up the same factor, $L = \mu_r\mu_0 n^2 Al$, which is why practical inductors are wound on iron or ferrite cores.

The cleanest case to calculate: two coils wound one over the other on the same long former, of length $l$ and cross-sectional area $A$, coil 1 with $N_1$ turns and coil 2 with $N_2$ turns, each spread evenly along the full length.

Flux from coil 1 through coil 2. Coil 1 is a long solenoid with $N_1/l$ turns per unit length, so a current $I_1$ in it sets up a uniform field inside:

Each turn of coil 2 wraps the same cross-section $A$, so the flux through one of its turns, and the flux linkage of all $N_2$ of them, are:

Comparing with $N_2\Phi_2 = MI_1$:

Notice this forward calculation never used the fact that coil 2 runs the full length: each of its turns only has to sit inside coil 1’s uniform field. So $M = \mu_0(N_1/l)N_2A$ holds just as well for a short coil 2 wound round the middle of a long coil 1.

The reverse direction, as a check. Now run a current $I_2$ through coil 2 instead and ask for the flux linkage of coil 1. The roles simply swap: $B_2 = \mu_0(N_2/l)I_2$, and

with the identical coefficient. So $M_{12} = M_{21}$: coil 1 induces in coil 2 exactly as strongly as coil 2 induces in coil 1, which is why a pair of coils has a single $M$. (The equality holds for any pair of coils, not just this symmetric one, though proving it in general is beyond CBSE.)

If the two solenoids have different radii, as in NCERT’s version (an inner solenoid of radius $r_1$ inside a wider outer one), the area that matters in both directions is the inner one’s, $A = \pi r_1^2$. The outer coil’s field fills the inner coil, but each inner turn encloses only $\pi r_1^2$ of it; and the inner coil’s field is confined to its own interior (a long solenoid’s field outside is negligible), so each outer turn catches flux only through that same $\pi r_1^2$. Writing $N_1 = n_1l$ and $N_2 = n_2l$ turns the result into the textbook form:

Finally, for the same-former pair, compare $M$ with each coil’s own inductance from §6.8.2, $L_1 = \mu_0N_1^2A/l$ and $L_2 = \mu_0N_2^2A/l$:

This holds because the coupling here is perfect: every line of one coil’s flux threads every turn of the other. When some of the flux leaks past the second coil, $M$ falls below $\sqrt{L_1L_2}$.