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A current carrying wire is bent in the form of a circular ring and magnetic dipole moment of the ring is M. If the wire is bent in the form of a square loop of single turn, Its magnetic dipole moment will be

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$\frac{4 M}{π}$

$\frac{2 M}{π}$

$\frac{πM}{4}$

$4 πM$

answer is C.

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Detailed Solution

If r be the radius of the ring, $r = \frac{l}{2 π}$

$∴ M = i \times π r^{2} = i \times π {(\frac{l}{2 π})}^{2} = \frac{i l^{2}}{4 π}$

For the square loop if ‘a’ be the length of side

Then $M' = i \times {(a)}^{2} = i \times {(\frac{l}{4})}^{2} = \frac{1}{16} \times 4 π M = \frac{π M}{4}$

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