# 5.15  (a). 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 normal bisector

Given,

The magnetic moment of the bar magnet, $m = 5.25 \times$ $\dpi{100} 10^{-2}\ JT^{-1}$

The magnitude of earth’s magnetic field at a place, $H = 0.42 G = 0.42 \times$ $\dpi{100} 10^{-4}$   T

The magnetic field at a distance R from the centre of the magnet on the normal bisector is:

$B = \mu_{0}m/4\pi R^3$

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

$B = H = 0.42\times$ $10^{-4}$ $T$

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

$= 12.5 \times 10^{-5 }m^3$

Therefore,  R = 0.05 m = 5 cm

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