Find (a) the magnetization `I`, (b) the magnetic intensity `H` and (c) the magnetic field `B` at the centre of a bar magnet having pole strength `3.6 A m`, magnetic length `12cm` and cross-sectional are `0.90 cm^2`.

Explanation

(a) Magnetization `I = m/A = (3.6 A m^(-1))/(0.90 xx 10^(-4) m^2)`
`= 4 xx 10^4 A m^(-1)`.
The direction will be from the south pole to the north pole at the centre of the magnet. (b) Magnetic intensity `H_n` due to the north pole is
`H_n = 1/(4pi) m/d^2 = (3.6 A m^(-1))/(4pi xx (6 xx 10^(-2) m)^2) = 79.6 A m^(-1)`.
The direction will be towards the south pole. The magnetic intensity `H_s` at this point due to the south pole is also `79.6 A m^(-1)` in the same direction. the resultant magnetic intensity is
`H = H_n + H_s`
`= 159.2 A m^(-1) towards the south pole.
(c) The magnetic field `vec B` at the centre is
`vec B` `= mu_0 (vec H + vec I)`
or, `B = (4pi xx 10^(-7) Tm A^(-1)) (4 xx 10^4 - 159.2) A m^(-1)`
`= 5.0 xx 10^(-2) T`.
The field is towards the north pole.