Question

An iron rod of cross sectional area $4sq.cm$. is placed with its length parallel to a magnetic field of intensity $160A/m$. The flux through the rod is $4\times {10}^{-4}Wb$. Then the permeability of the material of the rod is :

[in $Wb/(A-m)$]

• A. $0.625$
• B. $6.25$
• C. $6.25\times {10}^{-3}$
• D. $0.0625$

Answer 1

Answer: (C)

$6.25\times {10}^{-3}$

$Area=4c{m}^2=4\times {10}^{-4}{m}^2$

$Flux=4\times {10}^{-4}Wb$

Therefore $B=\frac{Flux}{Area}=\frac{4\times {10}^{-4}}{4\times {10}^{-4}}=1\frac{Wb}{m^2}$

And we know, $B=\mu H$

Therefore, $\mu =\frac{B}{H}=\frac{1}{160}=6.25\times {10}^{-3}\frac{Wb}{A-m}$