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A rectangular loop contains a sliding conducting rod of length l = 1.0 m in a uniform magnetic field B = 2 T perpendicular to the plane of the loop. The rod's resistance is r = 2 ohm, and two resistors of 6 ohm and 3 ohm are connected (in parallel with each other across the rod). Find the external force needed to move the rod at constant velocity v = 2 m/s.
- 2 N
- 0.5 N
- 4 N
- 8 N
Correct answer: 2 N
Solution
EMF = Blv = 2*1*2 = 4 V. The 6 ohm and 3 ohm in parallel give 2 ohm, in series with r = 2 ohm, so R_total = 4 ohm. Current through the rod I = 4/4 = 1 A. For constant velocity the external force balances the magnetic force: F = B*I*l = 2*1*1 = 2 N.
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