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Q.

A thin circular wire carrying a current $i$ has a magnetic moment M. The shape of the wire is changed to a square and it carries the same current. It will have a magnetic moment

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a

$\frac{4}{{{\pi ^2}}}M$

b

$\frac{4}{\pi }M$

c

M

d

$\frac{\pi }{4}M$

answer is D.

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Detailed Solution

Initially for circular coil $L = 2\pi r\,and\,M = i \times \pi {r^2}$
$= i \times \pi {\left( {\frac{L}{{2\pi }}} \right)^2} = \frac{{i{L^2}}}{{4\pi }}$ ..... (i)
Finally for square coil $M' = i \times {\left( {\frac{L}{4}} \right)^2} = \frac{{i{L^2}}}{{16}}$ ..... (ii)
Question Image
Solving equation (i) and (ii) $M' = \frac{{\pi M}}{4}$

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A thin circular wire carrying a current i has a magnetic moment M. The shape of the wire is changed to a square and it carries the same current. It will have a magnetic moment