A magnetizing field of $100 \mathrm{~A} / \mathrm{m}$ produces a magnetic flux of $2.4 \times 10^{-5}…
A magnetizing field of $100 \mathrm{~A} / \mathrm{m}$ produces a magnetic flux of $2.4 \times 10^{-5} \mathrm{~Wb}$ in an iron bar of cross-sectional area $0.3 \mathrm{~cm}^2$. The magnetic permeability of the iron bar in the SI unit is
$8 \times 10^{-4}$
$2.5 \times 10^{-4}$
$4 \times 10^{-4}$
$5 \times 10^{-4}$
Solution
The correct option is (A).
Concept: Magnetic induction $\mathrm{B}=\frac{\phi}{\mathrm{A}}$,
The permeability is the ratio of magnetic induction and the magnetic field $\mu=\frac{\mathrm{B}}{\mathrm{H}}$.
Therefore, permeability can be written as $\mu=\frac{\phi}{\mathrm{AH}}$
Given values are, $\mathrm{A}=0.3 \times 10^{-4} \mathrm{~m}^2$ and $\phi=2.4 \times 10^{-5} \mathrm{~Wb}$
$B=\frac{\phi}{A}=\frac{\left(2.4 \times 10^{-4} \mathrm{~Wb}\right)}{\left(0.3 \times 10^{-4} \mathrm{~m}^2\right)}=0.8 \mathrm{~Wb} / \mathrm{m}^2$
Therefore, $\mu=\frac{\mathrm{B}}{\mathrm{H}}=\frac{0.8}{1000}=8 \times 10^{-4}$
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