Question

An iron rod is placed parallel to magnetic field of intensity $2000\, A/m$. The magnetic flux through the rod is $6 × 10^{-4}\, Wb$ and its cross-sectional area is $3 \,cm^2$. The magnetic permeability of the rod in $wb/A-m$ is

• A. $10^{-1}$
• B. $10^{-2}$
• C. $10^{-3}$
• D. $10^{-4}$

Answer 1

Answer: (C)

Cross-sectional area $A =3 \times 10^{-4} m ^{2}$
Magnetic flux through the rod $\phi=6 \times 10^{-4} Wb$
Magnetic field intensity $H =2000\, A / m$
Using $\phi=B A$ and $B=\mu H$
We get magnetic permeability $\mu=\frac{\phi}{ AH }$
$\therefore \mu=\frac{6000}{3 \times 10^{-4} \times 2000}=10^{-3} \frac{ Wb }{ A - m }$