Faraday Induction on a Sliding Bar

Slide a conducting bar along two rails through a magnetic field and a current appears from nothing. As the bar advances it sweeps out area, so the magnetic flux through the loop, $\Phi = B L x$, grows, and Faraday's law turns that change into an electromotive force $\mathcal{E} = -\,d\Phi/dt = -B L v$. The EMF drives a current $I = \mathcal{E}/R$ around the loop, and that current, sitting in the same field, feels a force. Lenz's law fixes its direction: it always opposes the motion that created it, so the bar is dragged backward by $F = B^2 L^2 v / R$. Push the bar with a steady force and it speeds up only until this magnetic drag grows to match the push, settling at the terminal velocity $v_t = F_{\text{app}} R / (B^2 L^2)$. Nothing is free: at terminal speed the mechanical power you supply, $F_{\text{app}} v_t$, is exactly the heat $I^2 R$ dumped in the resistor.

Figure 1. Faraday induction on a sliding-bar circuit. Top: the rail loop in a field into the page, the swept flux, the induced current and the opposing Lenz force. Bottom: the velocity rising to the terminal value where the push balances the drag. Method: closed-form motional EMF with a linear-drag equation of motion. Source: Griffiths, Introduction to Electrodynamics, 5e, Sec. 7.1-7.2.

B (T)1.00

L (m)1.00

R (Ohm)2.00

F_app (N)1.00

WHAT TO TRY

Raise the field $B$: the EMF and current jump, the Lenz force bites harder, and the bar settles at a much lower terminal velocity (it scales as $1/B^2$).
Raise the resistance $R$: less current flows for the same EMF, so the drag weakens and the bar runs faster, reaching $v_t = F_{\text{app}} R / (B^2 L^2)$.
Watch the velocity curve flatten onto the dashed terminal line, and the two power readouts converge: at terminal speed the push power equals the heat dissipated.
Increase the rail separation $L$: a taller loop sweeps flux faster, so the bar brakes sooner (the drag scales as $L^2$).