Question

A magnetizing field of 1600A/m producers a magnetic flux of 2 x ${10}^{-5}$ Wb in a bar of iron of cross section 0.2 x ${10}^{-4}{m}^2$. Calculate the susceptibility of the bar.

• A. 596.8
• B. 288.9
• C. 2
• D. 1328

Answer 1

The correct option is A 596.8

$H=1600{AM}^{-1}$

$\varphi =2.4\times {10}^{-5}$ Weber

$A=0.2{cm}^2=0.2\times {10}^{-4}\times m^2$

$B=\mu H$

$\frac{\varphi }{A}=\mu H$

$\mu =\frac{\varphi }{\left(A\right)\left(H\right)}$

$\mu =\frac{2.4\times {10}^{-5}\left(\frac{NM}{A}\right)}{(0.2\times {10}^{-4}{m}^2)\left(1600{AM}^{-1}\right)}$

$\mu =7.5\times {10}^{-4}{NA}^{-2}$

Now, ${\mu }_r=1+\psi$

$\psi ={\mu }_r-1=\frac{\mu }{{\mu }_0}-1$

$\psi =\frac{7.5\times {10}^{-4}}{4\pi \times {10}^{-7}}-1=596.1$