An ideal solenoid having 40 turns `cm^(-1)` has an aluminium core and carries a current of `2.0 A`. Calculate the magnetization `I` developed in the core and the magnetic field `B` at the centre. The susceptibility `chi` of aluminium `= 2.3 xx 10^(-5)`.

Explanation

The magnetic intensity `H` at the centre of the solenoid is
`H = ni = 4000 turns m^(-1) xx 2.0 A`
= 8000 A m^(-1)`
The magnetization is `I = chi H`
`= 2.3 xx 10^(-5) xx 8000 A m^(-1) = 0.18 A m^(-1)`.
The magnetic field is `B = mu_0 (H+I)`
`= (4pi xx 10^(-7) Tm A^(-1) [800 + 0.18] A m^(-1) = 3.2pi xx 10^(-4) T`.
Note that `H gtgt I` in case of a paramagnetic core.