A conducting square frame of side ‘a’ and along straight wire carrying current I are located in the same plane a shown in the figure. The fame moves to the right with constant velocity ‘v’. The emf induced in the frame will be proportional to:

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

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

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

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

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Solution

The correct option is A
$\frac{1}{(2 x - a) (2 x + a)}$

${¯ B}_{1} = \frac{μ_{0} I}{2 π [x - \frac{a}{2}]} = \frac{μ_{0} I}{π [2 x - a]} {¯ B}_{2} = \frac{μ_{0} I}{2 π [x + \frac{a}{2}]} = \frac{μ_{0} I}{π [2 x + a]} e m f = \frac{d ϕ}{d t} = \frac{d (B A)}{d t} (ϕ = B A) = \frac{A d B}{d t} = a^{2} [{¯ B}_{1} - {¯ B}_{2}] = a^{2} [\frac{μ_{0} I}{π [2 x - a]} - \frac{μ_{0} I}{π [2 x + a]}] = \frac{μ_{0} I}{π} a^{2} [\frac{1}{2 x - a} - \frac{1}{2 x + a}] = \frac{μ_{0} I}{π} a^{2} [\frac{2 x + a - (2 x - a)}{(2 x - a) (2 x + a)}] = \frac{μ_{0} I}{π} a^{2} [\frac{2 a}{(2 x - a) (2 x + a)}]$

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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 emf induced in the frame will be proportional to :