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Q.

A square frame ABCD of side o and a long current carrying conductor PQ arc arranged as shown in fig.(7). If the frame moves towards right with a
velocity ‘v’, the resultant induced e.m.f. in the loop will be

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a

$\frac{μ_{0} {ia}^{2} v}{2 πx (x + a)}$

b

$\frac{μ_{0} iav}{2 πx}$

c

$\frac{μ_{0} iav}{2 πx (x + a)}$

d

$\frac{μ_{0} iav}{2 π (x + a)}$

answer is A.

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Detailed Solution

The induced e.m.f. in AD is given by
$e_{AD} = \frac{μ_{0} i}{2 πx} av Similarly, e_{BC} = \frac{μ_{0} i}{2 π (x + a)} av$
$∴$ Resultant e.m.f. = e_AD -e_BC
$or e = \frac{μ_{0} i a V}{2 π} [\frac{1}{x} - \frac{1}{x + a}] = \frac{μ_{0} {ia}^{2} v}{2 πx (x + a)}$

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A square frame ABCD of side o and a long current carrying conductor PQ arc arranged as shown in fig.(7). If the frame moves towards right with avelocity ‘v’, the resultant induced e.m.f. in the loop will be