A bar magnet has a pole strengh of `3.6 A m` and magnetic length `8cm`. Find the magnetic field at (a) `a` point on the axis at a distance of `6cm` from the centre towards the north pole and (b) a point on the perpendiuclar bisector at the same distance.

Explanation

The point in question is in end-on position, so the magnetic field is
`B = (mu_0)/(4pi) (2Md)/(d^(2)-l^(2))^(2))`
`=10^(-7) ™/(A) xx (2xx3.6Amxx0.08mxx0.06m)/([(0.06m)^(2)-(0.04)^(2)]^(2))`
`=8.6xx10^(-4)T`
The field will be away from the magnet.
(b) In this case the point is in broadside on position so that the filed is
`B = (mu_(0))/(4 pi) (M)/((d^(2)+l^(2))^(3//2))`
`=10^(-7) (Tm)/A xx (3.6Am xx 0.08m)/([0.06m)^(2)+(0.04m)^(2)]^(3//2))`
`=7.7 xx 10^(-5_T`
The field will be parallel to the magnet.