Question

A short bar magnet has a magnetic moment of $0.48J{T}^{-1}$ . Give the direction and magnitude of the magnetic field produced by the magnet at a distance of 10 cm from the centre of the magnet on (a) the axis, (b) the equatorial lines (normal bisector) of the magnet.

Answer 1

Here $M=0.48J{T}^{-1},d=10cm=0.1m$

(a)

On axis,

$B=\frac{{\mu }_0}{4\pi }\frac{2M}{d^3}={10}^{-7}\times \frac{2\times 0.48}{{0.1}^3}=0.96\times {10}^{-4}T$

Direction is away from magnet on the axis on the North-side and towards the magnet on the axis on the South-side

(b) On the equatorial line of the magnet

$B=\frac{{\mu }_0}{4\pi }\frac{M}{d^3}=0.48\times {10}^{-4}T$ along N-S direction.

Direction at equatorial line is same as direction from north pole to south pole.