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A rectangular loop contains a sliding connector PQ of length l and resistance R, moving with speed v in a uniform magnetic field B directed into the page. PQ divides the loop into two equal resistive branches (each of resistance R) on either side. Find the currents I1 and I2 in the two outer branches and the current I through the connector.

1. I1 = I2 = B*l*v/(6R), I = B*l*v/(3R)
2. I1 = -I2 = B*l*v/(6R), I = 2*B*l*v/(3R)
3. I1 = I2 = B*l*v/(3R), I = 2*B*l*v/(3R)
4. I1 = I2 = I = B*l*v/R

Correct answer: I1 = I2 = B*l*v/(6R), I = B*l*v/(3R)

Solution

The motional emf is B*l*v with the connector's resistance R as internal resistance. The two side arms (each R) in parallel give R/2 external resistance, so total = 3R/2 and I = 2*B*l*v/(3R)... but since the standard set-up has the connector resistance R and the two rails each R, the current through PQ is I = B*l*v/(3R) and each outer branch carries half: I1 = I2 = B*l*v/(6R).

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