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Physics-

General

Easy

Question

A conducting rod PQ of length L = 1.0 m is moving with a uniform speed v = 2 m/s in a uniform magnetic field $B equals 4.0 T$ directed into the paper. A capacitor of capacity C = 10 mF is connected as shown in figure. Then

q_A = + 80 mC and q_B = – 80 mC
q_A = – 80 mC and q_B = + 80 mC
q_A = 0 = q_B
Charge stored in the capacitor increases exponentially with time

The correct answer is: q_A = + 80 mC and q_B = – 80 mC

Q = CV = C (Bvl) = 10 ´ 10^{– 6} ´ 4 ´ 2 ´ 1 = 80 mC
According to Fleming's right hand rule induced current flows from Q to P. Hence P is at higher potential and Q is at lower potential. Therefore A is positively charged and B is negatively charged.

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