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Physics Question on Moving charges and magnetism

A conducting square frame of side 'a' and a long straight wire carrying current II 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

A

1(2x+a)2\frac{1}{(2x+a)^2}

B

1(2x−a)(2x+a)\frac{1}{(2x-a)(2x+a)}

C

1(2x−a)\frac{1}{(2x-a)}

D

1(2x−a)2\frac{1}{(2x-a)^2}

Answer

1(2x−a)(2x+a)\frac{1}{(2x-a)(2x+a)}

Explanation

Solution

Here, PQ = RS = PR =QS = a

Emf induced in the frame
ε=B1(PQ)V−B2(RS)V\varepsilon =B_1(PQ)V-B_2(RS)V
=μ0I2π(x−a/2)aV−μ0I2π(x+a/2)aV=\frac{\mu_0 I}{2 \pi(x-{a/2})}aV-\frac{\mu_0 I}{2 \pi(x+{a/2})}aV
=μ0I2π[2(2x−a)−2(2x+a)]aV=\frac{\mu_0 I}{2 \pi}\big[\frac{2}{(2x-a)}-\frac{2}{(2x+a)}\big]aV
=μ0I2π×2[2a(2x−a)(2x+a)]aV=\frac{\mu_0 I}{2 \pi} \times 2\big[\frac{2a}{(2x-a)(2x+a)}\big]aV
∴ε∝1(2x−a)(2x+a)\therefore \varepsilon \propto \frac{1}{(2x-a)(2x+a)}