Faraday's Law & Lenz's Law

Drop a small bar magnet down a vertical copper pipe. It does not clatter down in a fifth of a second the way a pebble does. It drifts, taking several seconds, and when it finally appears at the bottom it is moving slowly and steadily. Copper is not magnetic. Nothing is touching the magnet. Nothing is holding it up except the field it carries with it and the metal it is passing.

That is induction, and it is the other half of magnetism. Ampère's law told you that a current makes a field: point your thumb along the current and your fingers curl the way $\vec B$ goes, giving $B = \mu_0 I / 2\pi r$ outside a long wire. This lesson runs the sentence backwards. A field that changes pushes charge around a loop, with no battery anywhere in the circuit. Get that right, and the copper pipe, the electric guitar pickup, and every power station on the grid are the same problem.

Flux: how much field goes through the loop

The first thing to build is a number that says how much magnetic field is threading a loop of wire. Field strength alone is not enough: a strong field sliding past the edge of a loop does nothing, and a loop twice as big catches twice as much of the same field. So the quantity has to involve the field, the area, and the angle between them.

That quantity is the magnetic flux:

$\Phi_B = \int \vec B \cdot d\vec A = BA\cos\theta \ \ \text{(uniform field, flat loop)}$

Here $A$ is the area of the loop in square metres, $B$ is the field in tesla, and $\theta$ is the angle between $\vec B$ and the normal to the loop: the direction sticking straight out of the loop's face, perpendicular to it. When the field runs straight through the face, $\theta = 0$ and $\Phi_B = BA$. When the field lies in the plane of the loop, skimming past it, $\theta = 90^\circ$ and the flux is zero. The unit is the weber: $1\ \text{Wb} = 1\ \text{T}\cdot \text{m}^2$.

A circular loop of radius $0.10$ m lying square-on to a $0.40$ T field has $A = \pi(0.10)^2 = 0.0314\ \text{m}^2$ and $\Phi_B = (0.40)(0.0314) = 0.0126$ Wb. That is the whole calculation. The integral sign in the definition only earns its keep when $\vec B$ varies across the loop, which happens later in this lesson and is worth waiting for.

Faraday's law

Now the experiment. Hold a loop still in a field that is getting stronger. A current flows. Hold the field steady and yank the loop out. A current flows. Spin the loop in place. A current flows, back and forth. In every case the one thing that changed was the flux, and the faster you changed it the bigger the current. Careful measurement gives the law exactly:

$\mathcal{E} = -N\frac{d\Phi_B}{dt}$

$\mathcal{E}$ is the EMF: electromotive force, which is not a force at all but a voltage, measured in volts. It is the push per unit charge that the loop supplies, the same role a battery plays in a circuit. $N$ is the number of turns, because each turn is its own loop in series with the rest and the pushes add. The minus sign is Lenz's law, which gets its own section; for now read the equation as "the EMF equals the rate at which flux changes".

Three things can make $\Phi_B = BA\cos\theta$ change, and every induction problem you will meet changes at least one of them: the field $B$, the area $A$, or the orientation $\theta$. Naming which one is moving is the first thing to write on blank paper.

If the loop is a closed circuit of total resistance $R$, the EMF drives a current the same way a battery would:

$I = \frac{\mathcal{E}}{R}$

Example 1. A coil of $200$ turns, each of area $0.0050$ m², lies perpendicular to a field changing at $1.2$ T/s. Find the EMF.

Perpendicular to the field means the field runs straight through the face, so $\cos\theta = 1$ and $\Phi_B = BA$ per turn. The area is fixed, so only $B$ is moving:

$|\mathcal{E}| = N A \frac{dB}{dt} = (200)(0.0050\ \text{m}^2)(1.2\ \text{T/s}) = 1.2\ \text{V}$

Two hundred turns of small coil produce more than a volt out of a field change you could not feel. That is why pickup coils have many turns.

Lenz's law: getting the direction right

The minus sign says: the induced current flows in whatever direction fights the change that produced it. Not the field: the change.

That is a statement about energy, not about sign conventions. If the induced current helped the change along, the change would drive more current, which would help more, and a nudge would run away into infinite energy. The universe does not offer that. So the loop always pushes back, and if you want to keep changing the flux you have to keep paying.

Here is a recipe that works every time. Three steps, in order:

1. Which way does the existing flux point through the loop, and is it growing or shrinking?
2. The induced current wants to make flux inside the loop that opposes that change: if the flux is growing, the induced field inside points the other way; if it is shrinking, the induced field inside points the same way, to prop it up.
3. Point your right thumb along the induced field you want inside the loop; your fingers curl the way the current goes.

Try it on the magnet in the pipe. As the magnet approaches a ring of copper, the downward flux through that ring grows, so the ring's current makes upward flux inside itself, and the ring becomes a magnet whose top face repels the magnet's approaching pole. After the magnet passes, the flux through the ring is shrinking, so the current reverses and the ring now attracts the magnet back. Approaching rings push up, departing rings pull up. Both retard the fall.

When the field is not uniform across the loop

Put a rectangular loop beside a long straight wire, with its long sides parallel to the wire. The wire's field is $B = \mu_0 I / 2\pi r$, which is larger at the near edge than at the far edge. There is no single $B$ to multiply by the area, so you go back to the integral definition and slice the loop into strips parallel to the wire, each at distance $r$, each of width $dr$ and length $L$, so each has area $L\,dr$ and a field that is constant across it.

Adding the strips from the near edge $a$ to the far edge $b$:

$\Phi_B = \int_a^b \frac{\mu_0 I}{2\pi r} L\,dr = \frac{\mu_0 I L}{2\pi}\ln\!\frac{b}{a}$

Everything in front of $I$ is fixed geometry, so if the current changes the flux follows it in proportion, and

$\mathcal{E} = \frac{\mu_0 L}{2\pi}\ln\!\frac{b}{a}\ \frac{dI}{dt}$

Example 2. A loop with sides $0.30$ m long parallel to a wire sits with its near edge $2.0$ cm and its far edge $8.0$ cm from the wire. The wire's current rises at $200$ A/s. Find the EMF.

Take the geometry factor first, so the arithmetic stays clean:

$\frac{\mu_0 L}{2\pi} = (2\times10^{-7})(0.30) = 6.0\times10^{-8}\ \text{T}\cdot\text{m/A}$

using $\mu_0/2\pi = 2\times 10^{-7}$. The log is $\ln(0.080/0.020) = \ln 4 = 1.386$. So

$|\mathcal{E}| = (6.0\times10^{-8})(1.386)(200) = 1.66\times10^{-5}\ \text{V}$

Small, and it should be. If instead of ramping the current you dragged the loop away from the wire at speed $v$, the same flux formula applies but now $a$ and $b$ both grow with time, and you differentiate with respect to them: $d\Phi/dt = \frac{\mu_0 I L}{2\pi}\left(\frac{1}{b} - \frac{1}{a}\right)v$.

Rods, rails, force and power

Now change the area instead of the field. A rod of length $L$ slides at speed $v$ along two rails in a perpendicular field $B$, closing a circuit of resistance $R$. In a time $dt$ the enclosed area grows by $L v\,dt$, so

$\mathcal{E} = \frac{d\Phi}{dt} = B L v$

and the current is $I = BLv/R$. But that current is sitting in the field, so the rod feels a force $F = BIL = B^2L^2v/R$, and by Lenz's law it points backwards, against the motion. To hold the rod at constant speed you must push with that force, doing work at a rate $Fv = B^2L^2v^2/R$, which is exactly $\mathcal{E}^2/R$: the power appearing as heat in the resistor. Nothing is free, and nothing is lost.

If nobody pushes, that drag is all there is, and $m\,dv/dt = -B^2L^2v/R$. The speed decays exponentially with time constant $\tau = mR/B^2L^2$. If instead gravity pulls a loop down through the edge of a field region, the loop speeds up until the drag matches the weight, and at that terminal speed

$mg = \frac{B^2L^2 v_t}{R} \quad\Longrightarrow\quad v_t = \frac{mgR}{B^2L^2}$

Example 3. A square loop of side $0.25$ m, mass $20$ g and resistance $0.60\ \Omega$ falls with its lower edge crossing the boundary of a $0.50$ T field region. Find its terminal speed, then say how much heat it makes while falling $0.80$ m at that speed.

Only the lower edge is in the field and only it contributes EMF, so $L = 0.25$ m.

$v_t = \frac{(0.020)(9.8)(0.60)}{(0.50)^2(0.25)^2} = \frac{0.1176}{0.015625} = 7.53\ \text{m/s}$

For the heat, resist the urge to compute $I^2R$ and multiply by a time you do not know. At terminal speed the kinetic energy is not changing, so every joule gravity delivers goes into the resistor:

$Q = mgh = (0.020)(9.8)(0.80) = 0.157\ \text{J}$

That trick (energy in equals heat out, whenever the speed is constant) turns several genuinely hard problems into one line.

Two relatives of the sliding rod are worth knowing. A flat coil of $N$ turns and area $A$ spinning at angular frequency $\omega$ in a field $B$ has $\Phi = BA\cos\omega t$, so $\mathcal{E} = NAB\omega\sin\omega t$: a sinusoid of peak value $NAB\omega$, and RMS value $NAB\omega/\sqrt2$. That is a generator. And a solid disk of radius $a$ spinning at $\omega$ about its axis in an axial field has each radial element contributing $B\omega r\,dr$, which integrates to $\mathcal{E} = \tfrac12 B\omega a^2$ between axle and rim. That is the Faraday disk. The next lesson, motional EMF, takes these apart from the force-on-a-charge side and shows why both pictures agree.

Charge, not current

Sometimes a problem asks not for a current but for the total charge that flows during some change. The current is changing throughout, so you have to integrate, and the integral collapses:

$q = \int I\,dt = \int \frac{N}{R}\frac{d\Phi}{dt}dt = \frac{N\,\Delta\Phi}{R}$

The time cancels out. Only the total flux change matters, not how fast you made it happen. Whip the magnet away or ease it away over a minute: the same charge flows.

A $120$-turn coil of area $0.0040$ m² sits perpendicular to a $0.25$ T field and is flipped over through $180^\circ$, in a circuit of resistance $8.0\ \Omega$. The flux per turn goes from $+BA$ to $-BA$, so $|\Delta\Phi| = 2BA = 2(0.25)(0.0040) = 0.0020$ Wb, and

$q = \frac{(120)(0.0020)}{8.0} = 0.030\ \text{C}$

Forgetting the factor of two on a flip is the single most common slip here.

The field with no wire in it

One last idea, and it is the deep one. The loop in Figure 1 does not know whether the wire is there. The changing magnetic field creates a circulating electric field in space, and the wire only reveals it by letting charge move. Written for the field itself, Faraday's law is

$\oint \vec E \cdot d\vec\ell = -\frac{d\Phi_B}{dt}$

For a region of circular symmetry (inside a long solenoid, say) the induced $\vec E$ runs in circles, so the integral around a circle of radius $r$ is $E(2\pi r)$ and you can solve for $E$ the way Ampère's law gives you $B$.

Inside a solenoid of radius $a$ with the field changing at $dB/dt$, a circle of radius $r < a$ encloses flux $B\pi r^2$, so $E(2\pi r) = \pi r^2\,dB/dt$ and $E = \tfrac{r}{2}\,dB/dt$. With $a = 6.0$ cm and $dB/dt = 0.80$ T/s, at $r = 3.0$ cm: $E = (0.015)(0.80) = 0.012$ V/m.

Outside the windings the enclosed flux stops growing with $r$, because there is no field out there to enclose: $E(2\pi r) = \pi a^2\,dB/dt$, so $E = \frac{a^2}{2r}\frac{dB}{dt}$. At $r = 10$ cm that is $\frac{(0.060)^2}{2(0.10)}(0.80) = 0.0144$ V/m. The field is nonzero in a region where $\vec B$ is zero. Compare this to the charge case: an electric field made this way has no start and no end, and pushing a charge once around a closed path gains it energy. There is no potential here.

Recognising it

Reach for Faraday's law when a problem mentions a loop, coil, ring, rail or disk and something in the setup is changing with time. The trigger words are "increasing at", "per second", "is pulled out of", "spins at", "collapses", "falls through".

Then sort it by what is changing, because that decides the algebra:

• $B$ changes, geometry fixed. $\mathcal{E} = NA\,dB/dt$. If the source is a current in another wire, find $B$ from Ampère or Biot-Savart first, then differentiate.
• Area changes. Sliding rod, collapsing loop, shrinking radius. Write $A(t)$ and differentiate: for a circle of shrinking radius, $d(\pi r^2)/dt = 2\pi r\,\dot r$.
• Orientation changes. Spinning coil, coil flipped over. $\cos\omega t$, a sinusoidal EMF, and a factor of two on a full flip.
• $B$ varies across the loop. Slice into strips and integrate, as with the long wire.

Finally, watch what is being asked for. If it is a charge, use $q = N\Delta\Phi/R$ and do not touch the timing. If it is a force or a terminal speed, go in order: EMF, then current, then $F = BIL$, then balance. If it is heat, ask whether the kinetic energy is constant: if it is, the work done against the drag is the heat, and you never need the current at all.