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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

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

b

$\frac{1}{(2 x - a)^{2}}$

c

$\frac{1}{x^{2}}$

d

$\frac{1}{(2 x - a) (2 x + a)}$

answer is B.

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Detailed Solution

Here, P Q = R S = P R = Q S = a

Question Image

Emf induced in the frame

$\begin{array}{l} ε = B_{1} (PQ) V - B_{2} (RS) V \\ = \frac{μ_{0} I}{2 π (x - a / 2)} aV - \frac{μ_{0} I}{2 π (x + a / 2)} aV \\ = \frac{μ_{0} I}{2 π} [\frac{2}{(2 x - a)} - \frac{2}{(2 x + a)}] aV \\ = \frac{μ_{0} I}{2 π} \times 2 [\frac{2 a}{(2 x - a) (2 x + a)}] aV \end{array}$

$∴ ε \propto \frac{1}{(2 x - a) (2 x + a)}$

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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