A solenoid has a core of material with relative permability 400. The windings of the slenoid are insulated from the core and carry a current of 2A. If the number of turns is 1000 per metre, calculate

a) H

b) M

c) B

d)the magnetising current I.

Given: $\mu$_r= 400; I = 2 A; n = 1000 turns/m
1. H= nI = 1000*2 = 2000 A/m
2. B= $\mu$_r*$\mu$_o*H
= 400*4$\pi$*10^-7 (N/A²)* 2 * 1000 (A/m)
= 1 T
3. M = (B-$\mu$_o*H)/$\mu$o
= ($\mu$_r-1)H= 399 *  H
= 8 * 10⁵ A/m
4. The magnetising current I_m is the additional current that needs to be passed through the windings 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_​0 (I + I_m).
Using I = 2A, B = 1 T,
we get I_m = 794 A.