Question

An iron bar of length $10\text{cm}$ and diameter $2\text{cm}$ is placed in a magnetic field of intensity $1000{\text{Am}}^{-1}$ with its length parallel to the direction of the field. The magnetic moment produced in the bar , if the permeability of its material is $6.3\times {10}^{-4}{\text{TmA}}^{-1}$ is $x\pi {\text{Am}}^2$. Find $x$

Answer 1

$H=1000{\text{Am}}^{-1};\mu =6.3\times {10}^{-5}{\text{TmA}}^{-1}$

$l=10\text{cm},d=2\text{cm};$

Radius $r=1\text{cm},r={10}^{-2}\text{m}$

We know , ${\mu }_r=1+{\chi }_m$

$\frac{\mu }{{\mu }_0}=1+{\chi }_m$

${\chi }_m=\frac{\mu }{{\mu }_0}-1=\frac{6.3\times {10}^{-4}}{4\pi \times {10}^{-7}}-1\approx 500$

Intensity of magnetisation , $I={\chi }_{m}H$

$=500\times 1000=5\times {10}^5{\text{Am}}^{-1}$

Magnetic moment , $M=IV$ where $\text{" V "}$ is volume.

$M=I\left(\pi r^{2}l\right)$

$\Rightarrow M=5\times {10}^5\times \pi \times {10}^{-4}\times 10\times {10}^{-2}$

$=5\pi {\text{Am}}^2$

Comparing this with the data given in the question we get, $x=5$