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 :

Engineering

Physics

Magnetic Moment

Question

$\frac{3 m}{π}$

$\frac{2 m}{π}$

$\frac{m}{π}$

$\frac{4 m}{π}$

LiveFree JEE Main Previous Year Online Papers Live Free JEE Advance Previous Year Online Papers

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

m = Iℓ² 2πr = 4ℓ

$m' = \frac{I 4 ℓ^{2}}{π} r = \frac{2 ℓ}{π}$

$\frac{m'}{m} = \frac{4}{π} π r^{2} = \frac{π 4 ℓ^{2}}{π^{2}} = \frac{4 ℓ^{2}}{π}$

$m' = \frac{4}{π} m$