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 `xx 2.0 A= 8000 A//m`
the magnetization is `I=etaH`
`=2.3xx10^(-5)xx8000A//m=0.18A//m`
The magnetic field is `B=mu_(0)(H+I)`
`=(4pixx10^(-7)T-m//A)[800+0.18]A//m=3.2pixx10^(-4)T`
Note that H >>I in case of a paramagnetic core