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Q. Magnetic field at the centre of a circular loop of area A is B. The magnetic moment of the loop will be

BHUBHU 2011

Solution:

$ B=\frac{{{\mu }_{0}}}{4\pi }\frac{2\pi 1}{r}=\frac{{{\mu }_{0}}I}{2r} $
Or $ I=\frac{2Br}{{{\mu }_{0}}} $
Also, $ A=\pi {{r}^{2}} $
Or $ r={{\left( \frac{A}{\pi } \right)}^{1/2}} $
Magnetic moment, $ M=IA=2\frac{Br}{{{\mu }_{0}}}A $
$=2\frac{BA}{{{\mu }_{0}}}{{\left( \frac{A}{\pi } \right)}^{1/2}}=2\frac{B{{A}^{3/2}}}{{{\mu }_{0}}{{\pi }^{1/2}}} $