A conducting rod of length L and mass m slides down frictionless vertical rails connected by a resistor R at the top in a horizontal magnetic field B. What is the terminal velocity v_t of the falling rod?
Answer and explanation
Correct answer: D. m g R / (B^2 * L^2)
- Step 1: Induced motional EMF in the rod: E = B L v.
- Step 2: Current in the circuit: I = E / R = B L v / R.
- Step 3: Upward magnetic Lorentz force on the rod: F_B = I L B = (B^2 L^2 v) / R.
- Step 4: At terminal velocity, magnetic force balances gravity: F_B = m*g.
- Step 5: (B^2 L^2 v_t) / R = mg, so v_t = m g R / (B^2 L^2).
Answer
m g R / (B^2 * L^2).
Why not the others: mg / (BL) gives current, not velocity; B^2L^2 / (mgR) is the reciprocal; mgR / (2B*L) has incorrect physical dimensions.
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