Question

A magnetising field of $1600 \,Am ^{-1}$ produces a magnetic flux $2.4 \times 10^{-5} Wb$ in an iron bar of cross-sectional area $0.2 \,cm ^{2}$. The susceptibility of an iron bar will be :

• A. 1788
• B. 1192
• C. 596
• D. 298

Answer 1

Answer: (C)

When an iron-bar is placed in a magnetising field, it becomes magnetised. The intensity $I$ of magnetisation is given by
$I=\frac{B}{\mu_{0}}-H$
where, $B$ is total magnetic flux density and magnetic intensity is $H$.
Also $I=\chi_{m} H$ where $\chi_{m}$ is the susceptibility of the substance.
$\therefore B=\mu_{0} H\left(1+\chi_{m}\right)$
Also $B=\mu_{0} H, $
$\Rightarrow \mu_{0}=\frac{B}{H}$
$\therefore \chi_{m}=\frac{\mu}{\mu_{0}}-1$
Given, $\phi=2.4 \times 10^{-5} Wb ,$
$ A=0.2 \times 10^{-4} \,m ^{2}$
$\therefore B=\frac{\phi}{A}=\frac{2.4 \times 10^{-5}}{0.2 \times 10^{-4}}$
$=1.2\, Wb / m ^{2}$
and $\mu=\frac{B}{H}=\frac{1.2}{1600} $
$=7.5 \times 10^{-4} \,N / m ^{2}$
Hence, $\chi_{m}=\frac{7.5 \times 10^{-4}}{4 \pi \times 10^{-7}}-1=596$
Note : Magnetic susceptibility is a pure number.
Its value for vacuum is zero as there can be no magnetisation in vacuum.