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A solenoid $30 \mathrm{~cm}$ long is made by winding 2000 loops of wire on an iron rod whose cross-section is $1.5 \mathrm{~cm}^{2}$.If the relative permeability of the iron is $6000 .$ What is the self-inductance of the solenoid?
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The correct answer is:
$1.5 \mathrm{H}$
As we know, Self-inductance of the solenoid
$\begin{array}{l}
\mathrm{L}=\frac{\mu_{\mathrm{r}} \cdot \mu_{0} \mathrm{~N}^{2} \mathrm{~A}}{\mathrm{I}} \\
=\frac{600 \times 4 \pi \times 10^{-7} \times(2000)^{2} \times(1.5) \times 10^{-4}}{0.3} \\
=1.5 \mathrm{H}
\end{array}$
$\begin{array}{l}
\mathrm{L}=\frac{\mu_{\mathrm{r}} \cdot \mu_{0} \mathrm{~N}^{2} \mathrm{~A}}{\mathrm{I}} \\
=\frac{600 \times 4 \pi \times 10^{-7} \times(2000)^{2} \times(1.5) \times 10^{-4}}{0.3} \\
=1.5 \mathrm{H}
\end{array}$
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