Let the side of square loop be a.
The magnetic dipole is∴m=Ia2
a=Im
Radius of circular loop r related to a as circumference of circular loop is equal to perimeter of square loop.
2πr=4a=4Im
r=π2Im
∴ New dipole moment m′=π(π2Im)2I
=π4m
JEE Main 2019 — Physics Electromagnetism
A square loop is carrying a steady current I and the magnitude of its magnetic dipole moment is m . If this square loop is changed to a circular loop and it carries the same current, the magnitude of the magnetic dipole moment of circular loop will be:
Held on 10 Apr 2019 · Verified 6 Jul 2026.
π4m
π3m
π2m
πm
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