Question

A square current carrying loop of side 'a' produced magnetic field at its centre is 'B' find magnetic moment of this loop in terms of 'B'.

Answer 1

Magnetic field $B=4\times \frac{{\mu }_{0}I}{4\pi \frac{a}{2}}(\cos 45-\cos 135)$

$=\frac{2{\mu }_{0}I}{\pi a}\times \frac{2}{\sqrt{2}}$

$=\frac{4{\mu }_{0}I}{\sqrt{2}\pi a}$

$I=\frac{B^{\prime }\pi a}{2\sqrt{2}{\mu }_0}$

$\mu =i\times A$

magnetic moment

$\mu =\frac{B\pi aA}{2\sqrt{2}{\mu }_0}$