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Question: Answered & Verified by Expert
In the figure, a conducting rod of length l=1 meter and mass m=1 kg moves with an initial velocity, u =5 m s-1 . On a fixed horizontal frame containing inductor L=2 H and resistance R=1 Ω . PQ and MN are smooth, conducting wires. There is a uniform magnetic field of strength B=1 T. Initially, there is no current in the inductor. Find the total charge in coulomb, flown through the inductor by the time velocity of the rod becomes vf= 1 m s-1 and the rod has travelled a distance x=3 meter.

PhysicsElectromagnetic InductionJEE Main
Solution:
2554 Upvotes Verified Answer
The correct answer is: 1
Let i1 and i2 be the current through L and R at any time t

     i=i1+i2   BlvR=i2 and Blv=Ldi1dt



Force on conducting rod =mdvdt=-ilB=-i1+BlvRlB

    mdv= -lB i1dt-B2l2Rv dt 

    mdv=-lB i1dt-B2l2R v dt

    mvf-u=-lBQ-B2l2Rx

( v1 = velocity, when it has moved a distance 'x')

    Q=-B2 l2R x-mvf-uB l=1C

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