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Question

In the given circuit, inductor of L=1 mH and resistance R=1 Ω are connected in series to the ends of two parallel conducting rods as shown. Now a rod of length 10 cm is moved with constant velocity of 1 cm/s in magnetic field B=1 T. If rod starts moving at t=0, then the current in circuit after 1 millisecond is x×103 A. Then the value of x is (given e1=0.37)

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Solution

If this rod will move then there will be a change in magnetic flux and if magnetic flux will change then emf will develop.
So,
emf=Bvl, where v is the velocity of rod
E=1×102×101=103 V
This will behave as LR circuit as shown in figure
As we switch on the circuit there will be growth of current in it.
i=imax(1etRL)
imax=ER=1031=103 A
imax=103(1e103×1103)
imax=103(1e1)=103(10.37)
imax=0.63×103 A
so, x=0.63

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