Question

A current carrying circular loop is produced a magnetic field of $8\text{mT}$ on the axis of the loop at a distance of $0.5\text{m}(>>\text{radius of loop})$ from its centre. The magnetic moment (in ${\text{Am}}^2$) of the given loop will be

Answer 1

Given:
$B=8\text{mT}=8\times {10}^{-3}\text{T};d=0.5\text{m}$
As we know, the current carrying loop is equivalent to a bar magnet, so replacing the loop with a bar magnet of magnetic moment $M$ as shown above.

The magnetic field due to a bar magnet at an axial point is given by

$B_{ax}=\frac{{\mu }_0}{4\pi }\frac{2Mr}{(d^2-r^2{)}^2}$

Because, $r<<d$ is given,

$\therefore B_{ax}=\frac{{\mu }_0}{4\pi }\frac{2M}{d^3}$

putting the given data, we get

$8\times {10}^{-3}=\frac{4\pi \times {10}^{-7}}{4\pi }\frac{2M}{{0.5}^3}$

$\therefore M=5000{\text{Am}}^2$