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

(D) The magnetic field intensity $H$ is independent of the material of the core and is given by:
$H = nI = 1000 \times 2.0 = 2 \times 10^3 \; A/m$.
$(b)$ The magnetic field $B$ is given by $B = \mu_r \mu_0 H$:
$B = 400 \times (4\pi \times 10^{-7} \; T \cdot m/A) \times 2 \times 10^3 \; A/m \approx 1.0 \; T$.
$(c)$ Magnetisation $M$ is given by $M = \chi_m H = (\mu_r - 1)H$:
$M = (400 - 1) \times 2 \times 10^3 = 399 \times 2000 = 7.98 \times 10^5 \; A/m \approx 8 \times 10^5 \; A/m$.
$(d)$ The magnetising current $I_m$ is the additional current required to produce the same magnetic field $B$ in the absence of the core. The relation is $B = \mu_0 n (I + I_m)$.
Since $B = \mu_r \mu_0 n I$,we have $\mu_r \mu_0 n I = \mu_0 n (I + I_m)$,
$\mu_r I = I + I_m \implies I_m = I(\mu_r - 1)$,
$I_m = 2 \times (400 - 1) = 2 \times 399 = 798 \; A$.