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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a
4Mπ
b
2Mπ
c
πM4
d
4πM
answer is C.
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Detailed Solution
If r be the radius of the ring, r=l2π∴M=i × πr2=i × πl2π2=il24πFor the square loop if ‘a’ be the length of side Then M'=i × (a)2=i × l42=116 × 4πM=πM4