A solenoid a core of a material with relative permeability 400. The windings of the solenoid are insulated from the core and carry a current of 2A. If the number of turns is 1000 per metre, calculate (a) H, (b) B and (c) the magnetising current `I_m`.

Explanation

(a) The field H is dependent of the material of the core, and is `H=nI=100 xx 2.0=2 xx 10^(3) A//m`
(b) The magnetic field B is given by
`B=mu_(r) mu_(0)H`
`=400 xx 4pi xx10^(-7) (N//A^(2)) xx 2 xx 10^(3) (A//m)`
=1.0T
(c) Magnetisation is given by
`M=(B-mu_(0) H)//mu_(0)`
`=(mu_(r) mu_(0) H-mu_(0) H)=(mu_(r)-1)=399 xx H`
`=8 xx 10^(5) A//m`
(d) The magnetising current `I_(M)` is the additional current that needs to be passed through the winding of the solenoid in the absence of the core which would give a B value as in the presence of the core. Thus `B=mu_(r) n(I+I_(M))`. Using I=2A, B=1 T, we get `I_(M)=494A`.