# 15 (b). A short bar magnet of magnetic movement $5.25 \times 10 ^{-2} JT^{-1}$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° with earth’s field on its axis. The 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.

The magnetic field at a distance R from the centre of the magnet on its axis:

$B = \mu_{0}m/2\pi R^{3}$

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

$R^3 = \mu_{0}m/2\pi B$ = $4\pi\times10^{-7}\times5.25\times10^{-2}/2\pi\times0.42\times10^{-4}$= $25 \times 10^{-5}m^3$

Therefore, R = 0.063 m  = 6.3 cm

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