Electro Magnetic Induction And Alternating Currents Question 335
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=4.0,T $ directed into the paper. A capacitor of capacity C = 10 mF is connected as . Then
Options:
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
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Answer:
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.
BETA
