A short bar magnet of mangetic moment `5*25xx10^-2JT^-1` is placed with its axis perpendicular to earth's field direction. At what distance from the centre of the magnet, is the resultant field inclined at `45^@` with earth's field on (i) its normal bisector, (ii) its aixs? Magnitude of earth's field at the place `0*42G`. Ignore the length of the magnet in comparison to the distances involved.

Explanation

Given, magnetic moment of magnet
` M = 5.25 xx 10^(-3) JT^(-1)`
The earth.s magnetic field `B_E = 0.42 G = 0.42 xx 10^(-4) T`
(i) Let d be the distance of the point from the centre of the magnet.

Since the resultant `(B_R)` makes an angle of `45^@` with the Earth.s field `(B_E)` , the magnetic field due to the magnet must be equal to that of the earth.
` therefore B_("equatorial") = B_E`
`(mu_0)/(4pi) (M)/(d^3) = 0.42 xx 10^(-4) `
` (10^(-7) xx 5.25 xx 10^(-2) )/(d^2) = 0.42 xx 10^(-4)`
` d = 5 xx 10^(-2) m = 5 cm `
(ii)
`B_E = B_("axial")`
[ `because ` Resultant `(B_R)` is inclined at `45^@` with `B_E` ]
for a point on its axial line, the magnetic field is given by :
`B_e = (mu_0)/(4pi) (2M)/(d^2) ` (for small magnet )
` therefore (mu_0 )/(4pi) (2M)/(d^2) = 0.42 xx 10^(-4)`
On solving we get d = 6.3 cm