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 $e m f$ induced in the frame will be proportional to

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

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

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

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

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Solution

The correct option is A $\frac{1}{(2 x - a) (2 x + a)}$
Induced emf $ε = B_{1} a v - B_{2} a v = \frac{μ_{0} I}{2 π (x - \frac{a}{2}} a v - \frac{μ_{0} I}{2 π (x + \frac{a}{2})}) a v = \frac{μ_{0} I a v}{2 π} (\frac{1}{x - \frac{a}{2}} - \frac{1}{x + \frac{a}{2}}) = \frac{μ_{0} I a v}{2 π} \frac{a}{(2 x - a) (2 x + a)}$
$\Rightarrow B \propto \frac{1}{(2 x - a) (2 x + a)}$

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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 $e m f$ induced in the frame will be proportional to