Motional EMF

A bicycle dynamo is a small magnet spinning near a coil. It costs you nothing when the lamp is switched off, and the moment you switch the lamp on the pedals get heavier. Nothing about the magnet changed. Nothing about the coil changed. The only thing you did was give the current somewhere to go.

That is the whole subject of this lesson. In Faraday's law you had a fixed circuit and a changing field. Here the field is constant and it is the circuit that moves, so the flux changes because the area does. The algebra is easier than Faraday's, but there is a second half that Faraday alone does not give you: the moving conductor gets pushed back, and working out how hard is what most problems are really asking.

The rod on rails

Take two long parallel rails a distance $L$ apart, joined at one end by a resistor $R$. Lay a conducting rod across them so it can slide, and fill the region with a uniform magnetic field $B$ pointing perpendicular to the plane the rails lie in. Push the rod along at speed $v$.

The loop of circuit is a rectangle with one side $L$ and the other side growing. In a time $\Delta t$ the rod moves $v\,\Delta t$, so the enclosed area grows by $\Delta A = Lv\,\Delta t$. The flux through the loop is $\Phi = BA$, so

$\frac{\Delta \Phi}{\Delta t} = B\frac{\Delta A}{\Delta t} = BLv$

and Faraday's law says the magnitude of the emf around the loop is that rate. So the rod behaves like a battery of voltage

$\mathcal{E} = BLv$

Each symbol: $B$ in teslas, $L$ the length of rod actually between the rails in metres (not the whole rod, only the part carrying the circuit), $v$ the speed in metres per second, and $\mathcal{E}$ the emf in volts. Double any one of them and the emf doubles. If the rod stops, the emf is zero: a static circuit in a static field produces nothing, no matter how strong $B$ is.

There is a second route to $BLv$, and it is worth having because it survives into cases where "the area swept out" is awkward to define. The rod is full of mobile charges, and they are being carried sideways at speed $v$ along with it. A charge $q$ moving at $v$ through a field $B$ feels the magnetic force $q\vec v \times \vec B$. With $\vec v$ along the rails and $\vec B$ out of the plane, that force points along the rod, and it has magnitude $qvB$.

So charge gets shoved towards one end of the rod and piles up there, leaving the other end positive. The pile-up makes an electric field inside the rod that pushes back, and the shoving stops when the two balance: $qE = qvB$, so $E = vB$. The potential difference between the ends is that field times the length, $\mathcal{E} = vBL$. Same answer, arrived at without mentioning flux at all. The general statement is

$\mathcal{E} = \int (\vec v \times \vec B)\cdot d\vec{\ell}$

which just says: walk along the conductor, and at each point add up how hard the magnetic force pushes charge in the direction you are walking.

Worked example 1: emf, current, force, power

A copper rod of length $0.50$ m slides at $3.0$ m/s on horizontal rails in a uniform $0.40$ T field perpendicular to the plane of the rails. The rails are joined by a $2.0\ \Omega$ resistor and nothing else in the circuit has any resistance.

Start with the emf, because everything else hangs off it:

$\mathcal{E} = BLv = (0.40)(0.50)(3.0) = 0.60\ \text{V}$

The rod is a $0.60$ V battery driving current round a $2.0\ \Omega$ loop, so by Ohm's law

$I = \frac{\mathcal{E}}{R} = \frac{0.60}{2.0} = 0.30\ \text{A}$

Now the part that Faraday's law alone does not hand you. The rod is a current-carrying wire sitting in a magnetic field, so it feels a force $F = BIL$:

$F = (0.40)(0.30)(0.50) = 0.060\ \text{N}$

Which way? Lenz's law: the induced effects oppose the change that caused them, so this force points backwards along the rod's motion. It brakes. To keep the rod at a steady $3.0$ m/s you must push with $0.060$ N, and the power you spend is

$P = Fv = (0.060)(3.0) = 0.18\ \text{W}$

Check that against the resistor: $P = I^2R = (0.30)^2(2.0) = 0.18$ W. The same number, and it has to be. Your muscles are the only energy source in the problem.

That pattern is worth writing once in general. Substituting $I = BLv/R$ into $F = BIL$,

$F = \frac{B^2L^2v}{R}, \qquad P = Fv = \frac{B^2L^2v^2}{R} = \frac{\mathcal{E}^2}{R}$

The braking force is proportional to speed. That is exactly the form of linear drag, and it means every result you know about drag transfers over: terminal velocity, exponential decay, the lot.

Terminal speed

If something steady drives the rod (gravity, a hanging mass, a constant pull) then since the braking force grows with $v$, sooner or later it matches the drive and the acceleration goes to zero. That speed is the terminal speed, and you find it by setting the two forces equal. You never have to solve a differential equation for it.

Worked example 2. Frictionless rails $0.30$ m apart lie on a plane inclined at $30^\circ$. A $0.050$ kg rod slides down them, a $0.40\ \Omega$ resistor closes the circuit, and a $0.50$ T field points perpendicular to the inclined plane. Take $g = 9.8$ m/s². Find the terminal speed.

Work in coordinates along the incline, as you did for blocks on ramps. The component of gravity down the slope is $mg\sin\theta$:

$mg\sin\theta = (0.050)(9.8)(0.5) = 0.245\ \text{N}$

The magnetic force up the slope is $B^2L^2v/R$. At terminal speed they are equal:

$mg\sin\theta = \frac{B^2L^2v_t}{R} \;\Rightarrow\; v_t = \frac{mgR\sin\theta}{B^2L^2}$

Put the numbers in. The denominator first: $B^2L^2 = (0.50)^2(0.30)^2 = (0.25)(0.09) = 0.0225$. The numerator is $0.245 \times 0.40 = 0.098$. So

$v_t = \frac{0.098}{0.0225} = 4.36\ \text{m/s}$

Two variations show up constantly. If the rails are vertical and the field is horizontal, it is the same problem with $\sin\theta = 1$. If there is friction between rod and rails, the normal force is $mg\cos\theta$ and you subtract $\mu mg\cos\theta$ from the driving force before balancing: the magnetic part is untouched. And if a hanging block of mass $M$ pulls the rod over a pulley, the drive is $Mg$ and the balance is $Mg = B^2L^2v_t/R$.

Coasting to a stop: time, distance, heat

Now take the drive away. A rod given speed $v_0$ on frictionless rails, with nothing but the resistor to slow it, obeys Newton's second law with the magnetic force as the only horizontal force:

$m\frac{dv}{dt} = -\frac{B^2L^2}{R}v$

Rate of change proportional to the thing itself, with a minus sign: that is an exponential. Writing $\tau = mR/B^2L^2$, the solution is $v(t) = v_0 e^{-t/\tau}$. The rod never formally stops, but after a few $\tau$ it might as well have.

Three questions get asked about this, and each has a shortcut.

Time to half speed. Set $v_0 e^{-t/\tau} = v_0/2$, so $t = \tau \ln 2$.

Total distance. You could integrate the exponential, but the impulse route is cleaner and works even when you have not solved for $v(t)$. Multiply Newton's law by $dt$ and add up:

$m\,\Delta v = -\frac{B^2L^2}{R}\int v\,dt = -\frac{B^2L^2}{R}\,\Delta x$

Since the rod goes from $v_0$ to zero, $\Delta x = mv_0R/B^2L^2 = v_0\tau$.

Total heat. No integration at all. The rod's kinetic energy has nowhere else to go, so the resistor dissipates $\tfrac12 mv_0^2$.

Worked example 3. A $0.050$ kg rod of length $0.50$ m is launched at $4.0$ m/s along frictionless rails in a perpendicular $0.40$ T field, and coasts to rest through induction alone. How much total charge circulates?

You are not told the resistance, which is the hint. Charge is $q = \int I\,dt$, and the impulse equation above already contains that integral. The braking force is $F = BIL$, so the total impulse on the rod is

$\int F\,dt = BL\int I\,dt = BLq$

and that impulse equals the momentum lost, $mv_0$. So

$q = \frac{mv_0}{BL} = \frac{(0.050)(4.0)}{(0.40)(0.50)} = \frac{0.20}{0.20} = 1.0\ \text{C}$

Resistance genuinely does not appear: a smaller $R$ gives a bigger current for a shorter time, and the product is fixed.

The other charge formula worth keeping is the flux version. Since $I = \mathcal{E}/R = (B L v)/R$ and $v\,dt = dx$,

$q = \frac{BL\,\Delta x}{R} = \frac{\Delta\Phi}{R}$

Use this one when you are told the distance and the resistance. Remember to include the rod's own resistance in $R$ when the problem gives it: the rod and the external element are in series.

Rotating rods, disks and generators

Not every conductor slides in a straight line. Pin one end of a rod and spin it at angular speed $\omega$, with its far end brushing a circular rail and the field parallel to the rotation axis. Now different parts of the rod move at different speeds, so $BLv$ cannot be used as it stands. The integral version handles it. A little piece of rod at radius $r$, of length $dr$, moves at $v = \omega r$ and contributes $B\omega r\,dr$ to the emf. Add them up from the pivot out to the tip:

$\mathcal{E} = \int_0^L B\omega r\,dr = \tfrac12 B\omega L^2$

The factor of a half is the average of $\omega r$ over the rod. A spinning metal disk with brushes at its axle and rim gives the same expression with $L$ the radius: each radial line of the disk is one of these rods, all in parallel. For a disk of radius $0.10$ m spinning at $30$ revolutions per second, $\omega = 2\pi(30) = 188.5$ rad/s and in a $0.20$ T field the emf is $\tfrac12(0.20)(188.5)(0.10)^2 = 0.188$ V.

A generator is the third case: a flat coil of $N$ turns and area $A$ spinning at $\omega$ about an axis perpendicular to $B$. The flux through one turn is $BA\cos\omega t$, and Faraday's law differentiates it, so

$\mathcal{E} = NBA\omega \sin\omega t$

The peak value is $\mathcal{E}_0 = NBA\omega$. Because the output is a sine wave, the rms value (the steady voltage that would heat a resistor at the same average rate) is $\mathcal{E}_0/\sqrt2$, and the average power into a resistor $R$ is $\mathcal{E}_0^2/2R$. Watch the units on the spin rate: $1500$ rpm is $1500 \times 2\pi/60 = 157.1$ rad/s.

When the load is not a resistor

Replace the resistor with a capacitor or an inductor and the force law $F = BIL$ still holds, only the relation between emf and current changes.

With a capacitor $C$ across the rails, the charge is $q = C\mathcal{E} = CBLv$, so the current is $I = dq/dt = CBL\,a$ where $a$ is the rod's acceleration. The magnetic force is then $BIL = CB^2L^2a$, pointing backwards, and Newton's second law for an applied force $F$ reads $F - CB^2L^2a = ma$. So

$a = \frac{F}{m + CB^2L^2}$

The circuit acts like extra inertia, not like friction, and the motion is uniformly accelerated. No energy is lost.

With an inductor $L_{\text{ind}}$ across resistanceless rails, the loop equation is $BLv = L_{\text{ind}}\,di/dt$. Integrating from the start, $i = BL(x)/L_{\text{ind}}$ where $x$ is the distance travelled: current is proportional to displacement. Then $m\,dv/dt = -B^2L^2x/L_{\text{ind}}$, which is Hooke's law in disguise: the rod oscillates with $\omega = BL/\sqrt{mL_{\text{ind}}}$ and the farthest it gets is $x_{\max} = v_0/\omega$. Again nothing is dissipated; the kinetic energy goes into the magnetic field of the inductor and comes back.

And if two rods are free on the same rails, there is no external horizontal force on the pair, so their total momentum is conserved. Current flows only while their speeds differ, so the motion settles when they move together. Find the common speed from momentum conservation, then the heat is whatever kinetic energy went missing.

Recognising it

Reach for motional emf when the problem has a conductor physically moving through a field that is not itself changing. The tell-tales:

• A rod, bar, wire or wingspan with a length and a speed, in a field described as perpendicular. Start with $\mathcal{E} = BLv$.
• The words terminal speed, steady speed, constant velocity. Balance the drive against $B^2L^2v/R$ and solve. No calculus needed.
• Total distance, total charge, total heat with no time given. These are the three conserved-quantity shortcuts: distance from impulse, charge from $\Delta\Phi/R$ or $mv_0/BL$, heat from the kinetic energy lost.
• Time to half speed, how long until. That is the exponential, with $\tau = mR/B^2L^2$.
• A pivot rather than rails, or a spinning disk. Use $\tfrac12 B\omega L^2$. A coil with a number of turns means $NBA\omega$ instead.
• A capacitor or inductor instead of a resistor. Nothing is dissipated: expect constant acceleration or oscillation, not a terminal speed.

The habit that saves the most time: whenever you compute a current here, ask what force it makes and where the energy is going. Most of the hard problems in this bank are energy bookkeeping wearing a magnetic costume.

The inductor case above was only a sketch. Circuits where the current itself resists being changed have their own arithmetic, and that is inductance and LR/LC circuits.