A short bar magnet of magnetic moment `5.25 xx 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 (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

Here `M = 5.25 xx 10^(-2) JT^(-1)`
r = ?
Earth's field `vec(B)_(e) = 0.42 G = 0.42 xx 10^(-4)T`
(a) At a point P distant r on normal bisector, fig, field due to the magnet is
`vec(B)_(2) = (mu_(0))/(4pi) (M)/(r^(3))` along PAI/NS.
The resultant field `vec(R)` will be inclined at `45^(@)` to the earth's field along PQ only when `|vec(B)_(2)| = |vec(B)_(e)|`
`(mu_(0))/(4pi).(M)/(r^(3)) = 0.42 xx 10^(-4)`
`(10^(-7) xx 5.25 xx 10^(-2))/(r^(3)) = 0.42 xx 10^(-4)`
which gives, r = 0.05m = 5 cm
(b) When the point P lies on axis of the magnet such that OP = r, field due to magnet [fig.] is `vec(B)_(1) = (mu_(0))/(4pi).(M)/(r^(3))` along PO, Earth's field `vec(B)_(e)` is along `vec(PA)`. The resultant field `vec(R)` will be inclined at `45^(@)` to Earth's field [Figure.] only when
`|vec(B)_(1)| = |vec(B)_(e)|`
`(mu_(0))/(4pi).(2M)/(r^(3)) = 0.42 xx 10^(-4)` which gives
`r = 6.3 xx 10^(-2) m = 6.3 cm`