A metal rod sliding along two rails makes a loop whose area grows as the rod moves. By Faraday's Law, that growing flux makes an emf. And because a current then flows in the rod, the force on a current acts on the rod too.
Why it works
Let the rails be $L$ apart and the rod move at speed $v$ in a field $B$ at right angles to the loop. In a time $\Delta t$ the rod moves $v\Delta t$, so the area grows by $\Delta A = Lv\Delta t$ and the flux by $B\,Lv\Delta t$. Faraday's law gives $$\varepsilon = \frac{\Delta\Phi}{\Delta t} = BLv.$$ The same answer comes from the charges in the rod: each moves at $v$ through $B$, so each is pushed along the rod with force $qvB$. Over the length $L$ that is $qvBL$ of work per charge $q$, an emf of $BLv$.
Key idea
A rod of length $L$ moving at speed $v$ across a field $B$ has an emf $\varepsilon = BLv$. With a circuit of resistance $R$, the current is $I = BLv/R$, and the field pushes on that current with $F = ILB$, always against the motion.
Common mistake
Forgetting the force on the induced current. That force always opposes the rod's motion (Lenz's law again), so to keep the rod moving at constant speed you must keep pushing with a force equal to $ILB$. Check with energy: your power $Fv$ must equal the electrical power $I^{2}R$. If the force helped the motion, the rod would speed up and make energy from nothing.
Worked example: Rod on rails
A rod $0.50\ \text{m}$ long slides at $4.0\ \text{m/s}$ on rails in a $0.80\ \text{T}$ field at right angles to the rails. The circuit resistance is $2.0\ \Omega$. Find the emf, the current and the force needed to keep the rod moving steadily.
- $$\varepsilon = BLv = (0.80)(0.50)(4.0) = 1.6\ \text{V}.$$
- $I = \varepsilon/R = 1.6/2.0 = 0.80\ \text{A}$.
- $$F = ILB = (0.80)(0.50)(0.80) = 0.32\ \text{N},$$ so you must push with $0.32\ \text{N}$.
- Energy check: $Fv = (0.32)(4.0) = 1.28\ \text{W}$, and $I^{2}R = (0.80)^{2}(2.0) = 1.28\ \text{W}$.
Answer: $1.6\ \text{V}$, $0.80\ \text{A}$, $0.32\ \text{N}$
Worked example: Which way does the current flow?
The field points into the page and the rod moves to the right, away from the resistor at the left end of the rails. Which way does the current flow in the rod?
- The loop's area grows, so the flux into the page grows.
- The induced field must point out of the page, so the current circulates counterclockwise.
- On the right side of a counterclockwise loop, the current runs toward the top of the page.
- Check with the force on a positive charge in the rod: fingers to the right, curl into the page, thumb toward the top. It agrees.
Answer: Toward the top of the page in the rod