question_answer

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 \[B=4.0\,T\] directed into the paper. A capacitor of capacity C = 10 mF is connected as shown in figure. Then

A)           qA = + 80 mC and qB = ? 80 mC

B)           qA = ? 80 mC and qB = + 80 mC

C)           qA = 0 = qB

D)           Charge stored in the capacitor increases exponentially with time

Correct Answer: A

Solution :

                   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.