Questions
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 F is connected as shown in figure. Then
detailed solution
Correct option is A
Q=CV=C(Bvl)=10×10−6×4×2×1=80μCAccording 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.Talk to our academic expert!
Similar Questions
A square wire loop, of side a has a long straight wire carrying a constant current i that passes through the centre of the loop and perpendicular to its plane as shown in Fig.
The current induced in the square loop is
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