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An iron bar having a cross sectional area of $2 \times 10^{-5} \mathrm{~m}^2$ and magnetising field of $2400 \mathrm{~A} / \mathrm{m}$ produce a magnetic flux of $2.4 \pi \times 10^{-5} \mathrm{~Wb}$. What will be the value of permeability $(\mu)$ and susceptibility $(\chi)$ of the bar (Given $\left.\mu_0=4 \pi \times 10^{-7}\right)$
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Verified Answer
The correct answer is:
$\mu=5 \pi \times 10^{-4}, \chi=1249$
$\phi=\mathrm{BA}$
$\mathrm{B}=\frac{\phi}{\mathrm{A}}=\frac{2.4 \pi \times 10^{-5}}{2 \times 10^{-5}}=1.2 \pi \mathrm{T}$
Now,
$\begin{aligned}
& \mathrm{B}=\mu \mathrm{H} \\
& \mu=\frac{B}{H}=\frac{1.2 \pi}{2400}=5 \pi \times 10^{-4} \\
& \mu_{\mathrm{r}}=\frac{\mu}{\mu_0}=\frac{5 \pi \times 10^{-4}}{4 \pi \times 10^{-7}}=1250 \\
& \text {as, } \mu_{\mathrm{r}}=1+\chi \Rightarrow \chi=1249
\end{aligned}$
$\mathrm{B}=\frac{\phi}{\mathrm{A}}=\frac{2.4 \pi \times 10^{-5}}{2 \times 10^{-5}}=1.2 \pi \mathrm{T}$
Now,
$\begin{aligned}
& \mathrm{B}=\mu \mathrm{H} \\
& \mu=\frac{B}{H}=\frac{1.2 \pi}{2400}=5 \pi \times 10^{-4} \\
& \mu_{\mathrm{r}}=\frac{\mu}{\mu_0}=\frac{5 \pi \times 10^{-4}}{4 \pi \times 10^{-7}}=1250 \\
& \text {as, } \mu_{\mathrm{r}}=1+\chi \Rightarrow \chi=1249
\end{aligned}$
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