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A solenoid of 500 turns $/ \mathrm{m}$ is carrying a current of 3 A. Its core is made of iron which has…
A solenoid of 500 turns $/ \mathrm{m}$ is carrying a current of 3 A. Its core is made of iron which has relative permeability 5001 . The magnitude of magnetization is
$4.5 \times 10^6 \mathrm{Am}^{-1}$ $6.0 \times 10^{-6} \mathrm{Am}^{-1}$ $7.5 \times 10^6 \mathrm{Am}^{-1}$ $9.0 \times 10^6 \mathrm{Am}^{-1}$ Solution
Given: $\mathrm{n}=500$ turns $/ \mathrm{m}, \mathrm{I}=3 \mathrm{~A}$
$\begin{aligned}
\mu_{\mathrm{r}}=5001 & \\
\therefore \quad \mu=\mathrm{nI} & =500 \times 3 \mathrm{~A} \\
& =1500 \mathrm{~A} / \mathrm{m} \\
\text { But, } \chi_{\mathrm{m}} & =\mu_{\mathrm{r}}-1 \\
& =5001-1 \\
& =5000
\end{aligned}$
$\begin{aligned}
\therefore \quad \text { Magnetization } \mathrm{M} & =\chi_{\mathrm{m}} \mathrm{H} \\
& =5000 \times 1500 \\
& =7.5 \times 10^6 \mathrm{Am}^{-1}
\end{aligned}$
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Asked in: MHT CET 2023 (10 May Shift 2)
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