Question

A solenoid has 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 $H$ and $M$.

• A. $2\times 10^{3}A{m}^{-1},8\times 10^{5}A{m}^{-1}$
• B. $1.0A{m}^{-1},1.5\times 10^{5}A{m}^{-1}$
• C. $6.8\times 10^{-5}A{m}^{-1},1.2\times 10^{-5}A{m}^{-1}$
• D. $2.1\times 10^{-4}A{m}^{-1},-2.6\times 10^{5}A{m}^{-1}$

Answer 1

Answer: (A)

(i) The field $H$ is dependent of the material core and is
$H=nI=1000\times 2.0=2\times 10^{3}A{m}^{-1}$
(ii) Magnetisation, $M=\left(B-{\mu }_{0}H\right)/{\mu }_0$
So, $B={\mu }_r{\mu }_{0}H$
$=400\times 4\pi \times 10^{-7}\left(N{A}^{-2}\right)\times 2\times 10^3\left(A{m}^{-1}\right)$
$=1.0T$
$I=\left(B-{\mu }_{0}H\right)/{\mu }_0$
$=\left({\mu }_r{\mu }_{0}H-{\mu }_{0}H\right)/{\mu }_0$
$=\left({\mu }_r-1\right)H=399\times H=8\times 10^{5}A{m}^{-1}$