Question

The absolute magnetic permeability $\mu$ of a specimen of magnetic material is related to magnetic intensity $H$ according to the relation as
$\mu=\frac{0.6}{H}+8.0 \times 10^{-4} T m A { }^{-1}$
Find the value of $H$ (in $Am ^{-1}$ ) for which magnetic induction of $0.22\, T$ can be produced.

Answer 1

Here, $B=0.22\, T$,
$\mu=\frac{0.6}{H}+8.0 \times 10^{-4} T m A ^{-1}$
Now, $\mu=\frac{B}{H}$
$\therefore \frac{B}{H}=\frac{0.6}{H}+8.0 \times 10^{-4} $
or $\frac{0.22}{H}=\frac{0.6}{H}+8.0 \times 10^{-4}$
or $\frac{0.16}{H}=8.0 \times 10^{-4} $
or $ H=\frac{0.16}{8.0 \times 10^{-4}}=200\, A \,m ^{-1}$