ElectroMagnetic InductionHard
Question
A rectangular loop has a sliding connector PQ of length l and resistance R Ω and it is moving with a speed v as shown. The set-up is placed in a uniform magnetic field going into the plane of the paper. The three currents I1, I2 and I are
Options
A.
B.
C.
D.
Solution
A moving conductor is equivalent to a battery of emf = v B l (motion emf)Equivalent circuit
I = I1 + I2
applying Kirchoff′s law
I1 R + IR - v B l = 0 .........(1)
I2 R + IR - v B l = 0 .........(2)
adding (1) & (2)
2IR + IR = 2vB l
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