A square loop of side ‘a’ is placed in the same plane as a long straight wire carrying a current i. The nearer side of the loop is at a distance ‘a’ from the wire.
The loop is moved away from the wire with a constant velocity v. The induced e.m.f in the loop is

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$\frac{μ_{0}}{4 π} iv$

$\frac{μ_{0}}{2 π} iv$

$\frac{μ_{0}}{4 π} \frac{iv}{a^{2}}$

$\frac{μ_{0} iv}{2 πa}$

answer is A.

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Detailed Solution

$\begin{array}{l} dϕ = \frac{μ_{0} i}{2 πx} (adx); dϕ = \frac{μ_{0} ia}{2 π} (\frac{1}{x} dx) \\ ϕ = \int_{a}^{2 a} \frac{μ_{0} ia}{2 π} (\frac{1}{x} dx); ϕ = \frac{μ_{0} ia}{2 π} [lnx]_{a}^{2 a} \\ E = \frac{dϕ}{dt} = \frac{μ_{0} i}{4 π} v; alternate method \\ E = (B_{2} - B_{1}) lv; E = [\frac{μ_{0} i}{4 πa} - \frac{μ_{0} i}{2 πa}] aV \\ E = \frac{μ_{0} i}{4 πa} aV; V = \frac{μ_{0} iv}{4 π} \end{array}$