A square frame ABCD of side a and a long current carrying conductor PQ are arranged as shown in the figure. If the frame moves towards right with a velocity v, the resultant induced e.m.f in the loop will be

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$\frac{μ_{0} {ia}^{2} v}{2 π (x + a)}$

$\frac{μ_{0} iav}{2 πX}$

$\frac{μ_{0} i^{2} v}{2 πx (x + a)}$

$\frac{μ_{0} iav}{2 π (x + a)}$

answer is D.

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Detailed Solution

$\begin{array}{l} E = (B_{2} - B_{1}) lV; E = [\frac{μ_{0} i}{2 π (x + a)} ~ \frac{μ_{0} i}{2 πx}] aV \\ E = \frac{μ_{0} i}{2 π} [\frac{1}{X + a} ~ \frac{1}{x}] aV \\ E = \frac{μ_{0} iaV}{2 π} [\frac{a}{x (x + a)}] = \frac{μ_{0} {ia}^{2} V}{2 πx (x + a)} \end{array}$

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