Question

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

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

Answer 1

Answer: (D)

Here, $H = 1500\,A\,m^{-1}, \phi = 2.4 \times 10^{-5}$ weber
$A = 0.5\, cm^{2} = 0.5 \times 10^{-4} \,m^{2}$
$\therefore \quad B = \frac{\phi}{A} = \frac{2.4 \times 10^{-5}}{0.5 \times 10^{-4}}$
$ = 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$