Question

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

Answer 1

Answer: 200

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