A short bar magnet of magnetic 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 axis? Magnitude of earth's field at the place `0*42G`. Ignore the length of the magnet in comparison to the distances involved.

Explanation

Here, `M=5.25xx10^-2JT^-1`, `r=?`, Earth's field, `vecB_e=0.42G=0.42xx10^-4T`
(i) At a point P distanct r on normal bisector, figure, field due to the magnet is
`vecB_2=(mu_0)/(4pi)(M)/(r^(3))` along `PA||NS`

The resultant field `vevR` will be inclined at `45^@` to the earth's field (H) along `PQ^'`, only when `|vecB_2|=|vecB_e|`
`(mu_0)/(4pi)(M)/(r_3)=0*42xx10^-4`
or `(10^-7xx5.25xx10^-2)/(r^3)=0*42xx10^-4`
which gives, `r=0*05m=5cm` (ii) When the point P lies on axis of the magnet such that `OP=r`, field due to magnet(figure) is
`vecB_1=(mu_0)/(4pi)(2M)/(r^3)`, along PO,
Earth's field `H=vecB_e` is along `vec(PA)`.
The resultant field `vecR` will be inclined at `45^@` to earth's field (figure), only when `|vecB_1|=|vecB_e|`
`:. (mu_0)/(4pi)(2M)/(r^3)=0*42xx10^-4` which gives
`r=6*3xx10^-2m=6*3cm`