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

Magnetic moment of the bar magnet, `M=5.2 5 x x10^(-2)J T^(-1)`
Magnitude of earht's magnetic field at a place, `H=0.42G=0.42xx10^(-4)T`
(a) The magnetic field at a distance R from the centre of the magnet on the normal bisector is given by the relation:
`B=(mu_(@)M)/(rpiR^(3))`
Where,
`mu_(@)=` permeability of free space `=4nxx10^(-7)Tm A^(-1)`
when the resultant field is inclined at `45^(@)` with earth's field B=H
`therefore (mu_(@)M)/(4piR^(3))=H=0.42xx10^(-4)`
`R^(3)=(mu_(@)M)/(0.42xx10^(-4)xx4pi)`
`=(4pixx10^(-7)xx5.25xx10^(-2))/(4pixx0.42xx10^(-4))=12.5xx10^(-5)`
`therefore R=0.05m=5cm `
(b) The magnetic field at a distanced `R'` from the centre of the magnet on its axis is given as:
`B'=(mu_(@)2M)/(4piR^(3))`
The resultant field is inclined at `45^(@)` with earth's field.
`thereforeB'=H`
`(R])^(3)=(mu_(@)2M)/(4pixxH)`
`=(4pixx10^(-7)xx2xx5.25xx10^(-2))/(4pixx0.42xx10^(-4))=25xx10^(-5)`
`therefore R=0.063m=6.3cm`