Question

A magnetising field of $1600A{m}^{-1}$ produces a magnetic flux $2.4\times 10^{-5}Wb$ in an iron bar of cross-sectional area $0.2c{m}^2$. 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, $\varphi =2.4\times 10^{-5}Wb,$
$A=0.2\times 10^{-4}{m}^2$
$\therefore B=\frac{\varphi }{A}=\frac{2.4\times 10^{-5}}{0.2\times 10^{-4}}$
$=1.2Wb/m^2$
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.