Question

A magnetising field of $1600A{m}^{-1}$ produces a magnetic flux of $2.4\times {10}^{-5}Wb$ in an iron bar of cross sectional area $0.2c{m}^2$. Calculate permeability and susceptibility of the bar.

• A. $7.5\times {10}^{-4}N/A^2;596$
• B. $2.7\times {10}^{-3}N/A^2;26$
• C. $2500N/A^2;68$
• D. $2.8\times {10}^{-6}N/A^2;625$

Answer 1

The correct option is A

$7.5\times {10}^{-4}N/A^2;596$

Magnetic Field can be written as$B=\frac{\varphi }{A}=\frac{2.4\times {10}^{-5}}{0.2\times {10}^{-4}}=1.2Wb/m^2\text{Permeability}\mu =\frac{B}{H}=\frac{1.2}{1600}=7.5\times {10}^{-4}N/A^{2}Now,\mu ={\mu }_0(1+X_m)and{X}_m=\frac{\mu }{{\mu }_0}-1\left)\right(Here{\mu }_o=4\pi \times {10}^{-7}and{X}_m\text{is Susceptibility}\left)\right)\Rightarrow X_m=\frac{7.5\times {10}^{-4}}{4\times 3.14\times {10}^{-7}}-1=596$