Energy and Power in an Induced Circuit

Slide a conducting bar along two rails sitting in a magnetic field and you induce an EMF, which drives a current through the circuit's resistance. That current, sitting in the same magnetic field, feels a force — and by Lenz's law that force always points backward, opposing the push. So every bit of mechanical work you do fighting that force has to go somewhere. It goes into the resistor as heat. Energy is conserved: mechanical power in equals electrical power dissipated out, with nothing left over once the bar moves at constant speed.

To match them, write both sides in terms of the same variables. The induced EMF is BLv, so the current is I = BLv/R. The retarding force on the bar is F = BIL, so the mechanical power you supply to keep it moving at constant velocity is P = Fv = BILv. Substitute I and you get P = B^2L2v2/R. That is also exactly I^2R, the power dissipated in the resistor. Same expression, two derivations — proof that mechanical input and electrical dissipation are one and the same quantity.

This is the payoff of the unit: induction is not just about generating a voltage, it is about where the energy to generate it comes from. If you stop pushing, the retarding force doesn't vanish — it is proportional to the bar's current speed, so it shrinks as the bar slows, giving an exponential-style coast to rest rather than an abrupt stop. That same idea, a damping force tied to velocity, reappears whenever induced currents oppose motion, including in the oscillating energy exchanges between inductors and capacitors coming next.