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Question

If the resistances of values R1 and R2 are connected on both ends as shown in figure, current I, flowing through resistance R1 is given by
224740_a2d571ca0f864375be6f73bae8334662.png

A
BlR2(v1r2v2r1)R1R2(r1+r2)+r2r1(R1+R2)
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B
BlR2(v1r2+v2r1)R1R2(r1+r2)+r2r1(R1+R2)
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C
BlR2(v1r2v2r1)R1R2(r1r2)+r2r1(R1+R2)
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D
BlR2(v1r2v2r1)R1R2(r1+r2)r2r1(R1+R2)
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E
none
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Solution

The correct option is A BlR2(v1r2v2r1)R1R2(r1+r2)+r2r1(R1+R2)
As the conducting rods are moving in magnetic field with velocity v in opposite direction, thus an EMF is induced in each rod according to the Lenz law.
EMF induced in rod MN E1=Bv1l .........(a)
EMF induced in rod MN E2=Bv2l .......(b)

Combining the EMFs of the two rods, 1req=1r1+1r2 req=r1r2r1+r2 .......(1)
Also equivalent EMF Eeq=E2r2reqE1r1req
Putting the values and on solving we get, Eeq=E2r1E1r2r1+r2 ............(2)
From Kirchhoff current law i=i1+i2
Using Nodal analysis EeqVreq=V0R1+V0R2

V=EeqreqR1R2req[(R1+R2)req+R1R2] .............(3)

i1=VR1
From (1), (2) and (3), we get i1= (E2r1E1r2)r1r2R2r1r2[(R1+R2)r1r2+(r1+r2)R1R2]

Using (a) and (b), |i1|= |v2r1v1r2|BlR2[(R1+R2)r1r2+(r1+r2)R1R2]

480361_224740_ans_e454e7caebdf462ea533425f9fc4930e.png

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