A short bar magnet of magnetic movement `5.25xx10^(-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^(@)` 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.

Explanation

Figure shows the conditions of the problem. Suppose P is the point on the normal bisector of the magnet where the resultant of B (due to magnet) and H is inclined at `45^(@)` with H. This is possible if magnitudes of B and H are the same.

`B = (mu_0)/(4pi) ( M)/(d^(3))`......... for a short magnet
` or d^(3) = (mu_0)/(4pi) (M)/(B) = 10^(-7) xx (5.25 xx 10^(-2))/( 0.42 xx 10^(-4)) = 125 xx 10^(-6)`
`:. d = (125 xx 10^(-6))^(1//3) m = 5 xx 10^(-2) m = 5 cm `