Question

A magnetising field of $1500A{m}^{-1}$ produces flux of $2.4\times 10^{-5}$ weber in a iron bar of the cross-sectional area of $0.5c{m}^2$. The permeability of the iron bar is

• A. $245$
• B. $250$
• C. $252$
• D. $255$

Answer 1

Answer: (D)

Here, $H=1500A{m}^{-1},\varphi =2.4\times 10^{-5}$ weber
$A=0.5c{m}^2=0.5\times 10^{-4}{m}^2$
$\therefore B=\frac{\varphi }{A}=\frac{2.4\times 10^{-5}}{0.5\times 10^{-4}}$
$=4.8\times 10^{-1}T$
and $\mu =\frac{B}{H}$
$=\frac{4.8\times 10^{-1}}{1500}=3.2\times 10^{-4}$
So relative permeability
${\mu }_r=\frac{\mu }{{\mu }_0}=\frac{3.2\times 10^{-4}}{4\pi \times 10^{-7}}$
$=0.255\times 10^3=255$