# A short bar magnet of magnetic moment

Question:

A short bar magnet of magnetic moment 5.25 × 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 45º 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.

Solution:

Magnetic moment of the bar magnet, $M=5.25 \times 10^{-2} \mathrm{~J} \mathrm{~T}^{-1}$

Magnitude of earth's magnetic field at a place, $H=0.42 \mathrm{G}=0.42 \times 10^{-4} \mathrm{~T}$

(a) The magnetic field at a distance from the centre of the magnet on the normal bisector is given by the relation:

$B=\frac{\mu_{0} M}{4 \pi R^{3}}$

Where,

$\mu_{0}=$ Permeability of free space $=4 \pi \times 10^{-7} \mathrm{Tm} \mathrm{A}^{-1}$

When the resultant field is inclined at 45° with earth’s field, B = H

$\therefore \frac{\mu_{0} M}{4 \pi R^{3}}=H=0.42 \times 10^{-4}$

$R^{3}=\frac{\mu_{0} M}{0.42 \times 10^{-4} \times 4 \pi}$

$=\frac{4 \pi \times 10^{-7} \times 5.25 \times 10^{-2}}{4 \pi \times 0.42 \times 10^{-4}}=12.5 \times 10^{-5}$

$\therefore R=0.05 \mathrm{~m}=5 \mathrm{~cm}$

(bThe magnetic field at a distanced from the centre of the magnet on its axis is given as:

$B^{\prime}=\frac{\mu_{0} 2 M}{4 \pi \mathrm{R}^{3}}$

The resultant field is inclined at 45° with earth’s field.

$\therefore B^{\prime}=H$

$\frac{\mu_{0} 2 M}{4 \pi\left(R^{\prime}\right)^{3}}=H$

$\left(R^{\prime}\right)^{3}=\frac{\mu_{0} 2 M}{4 \pi \times H}$

$=\frac{4 \pi \times 10^{-7} \times 2 \times 5.25 \times 10^{-2}}{4 \pi \times 0.42 \times 10^{-4}}=25 \times 10^{-5}$

$\therefore R^{\prime}=0.063 \mathrm{~m}=6.3 \mathrm{~cm}$