Question

A short bar magnet of magnetic moment $5.25\times {10}^{-2}J{T}^{-1}$ is placed with its axis perpendicular to the earth's field direction. At what distance from the centre of the magnet, the resultant field is inclined at ${45}^o$ with earth's field on (a) its normal bisector and (b) its axis. Magnitude of the earth's field at the place is given to be 0.42 G. Ignore the length of the magnet in comparison to the distances involved.

Answer 1

$M=5.25\times {10}^{-2}J{T}^{-1}$
$B=0.42G=0.42\times {10}^{-4}T$
(i) On its normal bisector:
Fig(!) shows the point $P$ on the normal bisector of the magnet, where the resultant magnetic field $B$ is inclined at ${45}^o$ , with the earth's field.
It can happen only, if $B_{equal}$ is equal to the horizontal component of the earth's filed $\left(B_H\right)$
i.e.
$B_{equa}=B_H$
For a short magnet,
$B_{equal}=\frac{{\mu }_0}{4\pi }\times \frac{M}{r^3}$
$\therefore \frac{{\mu }_0}{4\pi }\times \frac{M}{r^3}=B_H$
Or $r={(\frac{{\mu }_o}{4\pi }\times \frac{M}{B_H})}^{1/3}$
=${\left(\frac{{10}^{-7}\times 5.25\times {10}^{-2}}{0.42\times {10}^{-4}}\right)}^{1/3}$
=$(125\times {10}^{-6}{)}^{1/3}$
=$5\times {10}^{-2}M$
=$5cm$
Fig $\left(2\right)$ sows the point $P$ on the axis pf the magnet, where the resultant field is inclined at ${45}^o$ with the earth's magnetic field.
It will be so, if
$B_{axial}=B_H$
For a sort magnet,
$B_{axial}=\frac{{\mu }_0}{4\pi }\times \frac{2M}{r^3}$
$\therefore \frac{{\mu }_0}{4\pi }\times \frac{2M}{r^3}=B_H$
Or $r={(\frac{{\mu }_o}{4\pi }.\frac{2M}{B_H})}^{1/3}$
=${\left(\frac{{10}^{-7}\times 5.25\times {10}^{-2}}{0.42\times {10}^{-4}}\right)}^{1/3}$
=$(250\times {10}^{-6}{)}^{1/3}$
=$6.3\times {10}^{-2}m$
=$6.3cm.$