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(D) The magnetic field $B$ at a distance $r$ from a long straight wire carrying current $I$ is given by $B = \frac{\mu_0 I}{2\pi r}$.For a vertical wi

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

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(D) The magnetic field $B$ at a distance $r$ from a long straight wire carrying current $I$ is given by $B = \frac{\mu_0 I}{2\pi r}$.For a vertical wire segment of length $a$ moving with velocity $V$ in a magnetic field,the induced $emf$ is $\varepsilon = B l V$.The frame has two vertical sides at distances $r_1 = x - a/2$ and $r_2 = x + a/2$ from the wire.The induced $emf$ in the left side $(1)$ is $\varepsilon_1 = B_1 a V = \frac{\mu_0 I}{2\pi (x - a/2)} a V = \frac{\mu_0 I a V}{\pi (2x - a)}$.The induced $emf$ in the right side $(2)$ is $\varepsilon_2 = B_2 a V = \frac{\mu_0 I}{2\pi (x + a/2)} a V = \frac{\mu_0 I a V}{\pi (2x + a)}$.The net induced $emf$ in the frame is $\varepsilon_{net} = |\varepsilon_1 - \varepsilon_2| = \frac{\mu_0 I a V}{\pi} \left( \frac{1}{2x - a} - \frac{1}{2x + a} \right)$.Simplifying the expression: $\varepsilon_{net} = \frac{\mu_0 I a V}{\pi} \left( \frac{(2x + a) - (2x - a)}{(2x - a)(2x + a)} \right) = \frac{\mu_0 I a V}{\pi} \left( \frac{2a}{(2x - a)(2x + a)} \right)$.Thus,$\varepsilon_{net} \propto \frac{1}{(2x - a)(2x + a)}$.Therefore,the correct option is $D$.

$A$ square conducting frame of side $a$ and a long straight wire carrying current $I$ are placed as shown in the figure. If the frame is moved to the right with a constant velocity $V$,then the induced $emf$ in the frame is proportional to:

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