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A conducting square frame of side  ‘a’  and a long straight wire carrying current I are located in the same plane as shown in the figure. The frame moves to the right with a constant velocity ' V '. The emf induced in the frame will be proportional to

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a

1(2x+a)2

b

1(2x-a)(2x+a)

c

1x2

d

1(2x-a)2

answer is B.

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Detailed Solution

Here,  P Q = R  S = P R = Q S = aEmf induced in the frame ε=B1(PQ)V-B2(RS)V=μ0I2π(x-a/2)aV-μ0I2π(x+a/2)aV=μ0I2π2(2x-a)-2(2x+a)aV=μ0I2π×22a(2x-a)(2x+a)aV∴ ε∝1(2x-a)(2x+a)
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A conducting square frame of side  ‘a’  and a long straight wire carrying current I are located in the same plane as shown in the figure. The frame moves to the right with a constant velocity ' V '. The emf induced in the frame will be proportional to