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

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

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

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

answer is B.

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Detailed Solution

Potential difference across AD
Question Image
$V_{D} - V_{A} = Bav = \frac{μ_{0} iav}{2 π (x - \frac{a}{2})}$
Potential difference across= BC
$\begin{array}{l} V_{C} - V_{B} = Bav = \frac{μ_{0} iav}{2 π (x + \frac{a}{2})} \\ V_{net} = (V_{D} - V_{A}) - (V_{C} - V_{B}) \\ = \frac{μ_{0} iav}{2 π} [\frac{a}{(x - \frac{a}{2}) (x + \frac{a}{2})}] = \frac{μ_{0} {ia}^{2} v}{π (2 x - a) (2 x + a)} \\ ∴ V_{net} \propto \frac{1}{(2 x - a) (2 x + a)} \end{array}$