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 $2\, A$. If the number of turns is $1000$ per metre, calculate $H$ and $M$.

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

Answer 1

Answer: (A)

(i) The field $H$ is dependent of the material core and is
$H=n I=1000 \times 2.0=2 \times 10^{3} Am ^{-1}$
(ii) Magnetisation, $M=\left(B-\mu_{0} H\right) / \mu_{0}$
So, $B=\mu_{r} \mu_{0} H$
$=400 \times 4 \pi \times 10^{-7}\left( NA ^{-2}\right) \times 2 \times 10^{3}\left( Am ^{-1}\right)$
$=1.0\, T$
$ I =\left(B-\mu_{0} H\right) / \mu_{0}$
$=\left(\mu_{r} \mu_{0} H-\mu_{0} H\right) / \mu_{0}$
$=\left(\mu_{r}-1\right) H=399 \times H=8 \times 10^{5} Am ^{-1}$