A short bar magnet has a magnetic moment of `0.48 JT^(-1)` . Give the direction and magnitude of the magnetic field produced by the magnet at a distance of 10 cm from the centre of the magnet on (a) the axis, (b) the equatorial lines (normal bisector) of the magnet.

Explanation

`m = 0.48 JT^(-1),r = 10 cm`
a . Field on axial line = `(mu_0)/(4pi)(2m)/r^3=(10^(-7)xx2xx0.48)/((10xx10^(-2))^3)=(10^(-7)xx2xx0xx0.48)/(10^(-3))`
`=0.96xx10^(-4)T,` in the direction of `barm`
b. Field on equatorial line = `=(mu_0)/(4pi)xxm/r^3=0.48xx10^(-4)T, ` opposite to `barm`