A rod sliding on rails is the simplest generator. It joins Motional emf, Ohm's law and the force on a current, with Lenz's law deciding every direction. Each problem here is one short scene and a chain of questions. Answer them in order: each later question uses what you found before, so a slip early on carries through. Keep full precision between steps and round only at the end.
Key idea
$$\varepsilon = BLv, \qquad I = \frac{\varepsilon}{R}, \qquad F = ILB = \frac{B^2L^2v}{R}.$$ The magnetic force always opposes the motion, so something must keep pulling; its power $Fv$ equals the heating $I^2R$. With a steady pull $F$, the rod settles at $v_t = FR/(B^2L^2)$.
Worked example: Pulling the rod
A rod $0.40\ \text{m}$ long slides at $3.0\ \text{m/s}$ on rails joined by $2.0\ \Omega$, in a $0.50\ \text{T}$ field. Find (1) the emf, (2) the current and (3) the pulling force needed.
- (1) $\varepsilon = (0.50)(0.40)(3.0) = 0.600\ \text{V}$.
- (2) $I = 0.600/2.0 = 0.300\ \text{A}$.
- (3) $F = (0.300)(0.40)(0.50) = 0.0600\ \text{N}$.
Answer: $0.600\ \text{V}$, $0.300\ \text{A}$, $0.0600\ \text{N}$
Common mistake
Getting the current's direction backward. The induced current must push back on the rod; if your current would speed the rod up, energy would come from nowhere. Check with $F = IL \times B$.