A short bar magnet has a magnetic moment of 0.5 J `T^(-1)`. Calculate magnitude and direction of the magnetic field produced by the bar magnet which is kept at a distance of 0.1 m from the centre of the bar magnet (a) axial line of the bar magnet and (b) normal bisector of the bar magnet.

Explanation

Given magnetic 0.5 J `T^(-1)` and distance r = 0.1 m
(a) When the point lies on the axial line of the bar magnet, the magnetic field for short magnet is given by .
`vec(B)_("axial") = (mu_(0))/(4 pi) ( (2p_(m))/(r^(3)) ) hat(i)`
`vec(B)_("axial") = 10^(-7) xx ( (2xx 0.5)/((0.1)^(3)) ) hat(i) = 1 xx 10^(-4) T hat(i)`
hence, the magnitude of the magnetic field along axial is `B_("equatorial ") = 0.5 xx 10^(-4) `T and direction is towards North to South.
(b) When the piont lies on the normal bisector (equatorial ) line of the bar magnet, the magnetic field for short magnet is given by
`vec(B)_("axial") = (mu_(0))/(4 pi) (p_(m))/(r^(3)) hat(i)`
`vec(B)_("axial") = 10^(-7) xx ( ( 0.5)/((0.1)^(3)) ) hat(i) = -0.5 xx 10^(-4) T hat(i)`
Note that magnitude of `B_("axial")` is twice that of magnitude of `B_("equatorial")` and the direction of `B_("axial") and B_("equatorial") ` are opposite.