# 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, calculatea) Hb) Mc) Bd)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 Im 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 + Im).
Using I = 2A, B = 1 T,
we get Im = 794 A.

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