A magnetic intensity of $500 \mathrm{~A} / \mathrm{m}$, produces a magnetic flux of $2.4 \times 10^{-5} .…

A magnetic intensity of $500 \mathrm{~A} / \mathrm{m}$, produces a magnetic flux of $2.4 \times 10^{-5} . \mathrm{Wb}$ in an iron bar of cross-sectional area $0.4 \mathrm{~cm}^2$. The magnetic permeability of the iron bar is
  1. $2.4 \times 10^{-3} \mathrm{~N} / \mathrm{m}^2$
  2. $1.2 \times 10^{-3} \mathrm{~N} / \mathrm{m}^2$
  3. $2.4 \times 10^{-4} \mathrm{~N} / \mathrm{m}^2$
  4. $1.2 \times 10^{-4} \mathrm{~N} / \mathrm{m}^2$

Solution

Permeability of iron bar, $\begin{aligned} & \mu=\frac{B}{H}=\frac{(\phi / \mathrm{A})}{\mathrm{H}}=\frac{\phi}{\mathrm{HA}}=\frac{2.4 \times 10^{-5}}{500 \times\left(0.4 \times 10^{-4}\right)} \\ \therefore \quad \mu & =1.2 \times 10^{-3} \mathrm{~N} / \mathrm{m}^2 \end{aligned}$

Asked in: MHT CET 2024 (04 May Shift 2)

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