Question

A short bar magnet of magnetic moment $5.25\times {10}^{-2}J{T}^{-1}$ is placed with its axis perpendicular to the earths field direction. At what distance from the centre of the magnet, the resultant field is inclined at ${45}^{\circ }$ with earths field on
(a) its normal bisector and (b) its axis. Magnitude of the earths field at the place is given to be $0.42G$. Ignore the length of the magnet in comparison to the distances involved

Answer 1

(a) According to question, the resultant field is inclined at ${45}^o$ with earth's magnetic field.

$tan\theta =\frac{B_H}{B}$

$\theta ={45}^{o}tan\theta =1=\frac{B_H}{B}$

$B_H=B=\frac{{\mu }_{o}M}{4\pi r^3}$

Given, B= $0.42\times {10}^{-4}T$

$M=5.25\times {10}^{-2}J/T$

Therefore,$0.42+{10}^{-4}={10}^{-7}\times \frac{5.25\times {10}^{-2}}{r^3}{r}^3=12.5\times {10}^{-5}=125\times {10}^{-6}r=5\times {10}^{-2}m$ = 5 cm

(b) At axis of magnet, $tan45°=1=\frac{B}{B_H}$

$B_H=B=\frac{{\mu }_{o}2M}{4\pi r^3}$

=>$0.42\times {10}^{-4}={10}^{-7}\times \frac{2\times 5.25\times {10}^{-2}}{r^3}=>r^3=25\times {10}^{-5}m=>r=6.3cm$