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Q. Magnetic induction at the centre of a circular loop carrying a current is $B$ . If $A$ is the area of the coil, the magnetic dipole moment of the loop is

NTA AbhyasNTA Abhyas 2022

Solution:

$ B =\frac{\mu_0 i}{2 r}, A=\pi r^2, r=\sqrt{\frac{A}{\pi}} $
$m =i A=\frac{B \times 2 r}{\mu_o} \times \pi r^2$
$=\frac{2 B}{\mu_o} \pi r^3 $
$=\frac{2 B \pi}{\mu_o}\left(\frac{A}{\pi}\right)^{\frac{3}{2}}$
$=\frac{2 B A}{\mu_o} \sqrt{\frac{A}{\pi}} $