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Physics-

General

Easy

Question

A rectangular loop of length l and breadth b is placed at distance of $x$ from infinitely long wire carrying current i such that the direction of current is parallel to breadth. If the loop moves away from the current wire in a direction perpendicular to it with a velocity $v comma$ the magnitude of the emf in the loop is $left parenthesis mu equals$ permeability of free space)

$\frac{μ_{0} i v}{2 π x} (\frac{1 + b}{b})$
$\frac{μ_{0} i^{2} v}{4 π^{2} x} \log (\frac{b}{l})$
$\frac{μ_{0} i l b v}{2 π x (l + x)}$
$\frac{μ_{0} i l v b}{2 π x (1 + x)}$

The correct answer is: $fraction numerator mu subscript 0 end subscript i l blank v blank b over denominator 2 blank pi blank x blank left parenthesis 1 plus x right parenthesis end fraction$

We can show the situation as

Since, loop is moving away from the wire, so the direction of current in the loop will be as shown in the figure.
Net magnetic field on the loop due to wire
$B equals fraction numerator mu subscript 0 end subscript i over denominator 2 pi end fraction open parentheses fraction numerator 1 over denominator x end fraction minus fraction numerator 1 over denominator l plus x end fraction close parentheses$
$equals blank fraction numerator mu subscript 0 end subscript i l over denominator 2 pi x left parenthesis l plus x right parenthesis end fraction$
So, the magnitude of the emf in the loop
$e equals v B b equals fraction numerator mu subscript 0 end subscript i l blank v blank b over denominator 2 blank pi blank x blank left parenthesis 1 plus x right parenthesis end fraction$

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