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Thermal power in rod moving in field

Short answer

Induced emf ε = BLv, current I = BLv/R, thermal power P = I^2R = B^2 L^2 v^2 / R, which equals the mechanical power supplied by the external agent.

[Figure: a conducting rod of length L is pulled along horizontal, frictionless conducting rails at a constant velocity v; the rails are connected at one end with a metal strip; a uniform magnetic field B directed out of the page fills the region in which the rod moves.] Derive an expression for the rate of thermal energy being generated in the rod.

Let the closed circuit (rod + rails + strip) have total resistance R. The rod has length L and moves at constant speed v in field B (out of the page).

   . . . . . . . . . .
   ======+----------+====
   strip |  I       |-> v
   ======+----------+====
   . . . . . . . . . .   B out of page

Derivation

  1. Area enclosed grows as the rod moves: dA/dt = Lv, so the flux Φ = BLx changes at dΦ/dt = BLv.

  2. Faraday's law: induced emf ε = BLv.

  3. Induced current: I = ε/R = BLv/R. Lenz's law gives a clockwise current for B out of the page and the area increasing.

  4. Rate of thermal energy (Joule heating):

    P = I²R = (BLv/R)² R

P = B²L²v²/R

Check by energy conservation

  • Magnetic force on the rod: F = ILB = B²L²v/R, opposing the motion.
  • The agent pulling at constant speed must supply F_ext = F, so its power is F v = B²L²v²/R.

This equals I²R, so all the mechanical work appears as heat in the resistance. If only the rod has resistance, then R is the rod's resistance.

✓ Answer checked by Claude Opus 5.5 (high effort)

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