Question 103 of 153

Inductance of a coil with $10^{4}$ turns is 10 mH and it is connected to a dc source of 10 V with internal resistance of $10 \Omega$. The energy density in the inductor when the current reaches $\left(\frac{1}{e}\right)$ of its maximum value is $\alpha \pi \times \frac{1}{e^{2}} \mathrm{~J} / \mathrm{m}^{3}$. The value of $\alpha$ is $\_\_\_\_$. ($\mu_{0}=4 \pi \times 10^{-7} \mathrm{Tm} / \mathrm{A}$).

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Answer: 20

$\mathrm{L}=10 \times 10^{-3} \mathrm{H}$
$\mathrm{N}=10^4$

$I_0=\frac{10}{10}=1 \mathrm{~A} \text { (max current) }$
$I=\frac{1}{\mathrm{e}}$
$E_{\mathrm{d}}=\frac{B^2}{2 \mu_0}$
$B=\mu_0 n I$
$L=\mu_0 n^2 \pi R^2 \ell$
$E_d=\frac{\mu_0 n^2 I^2}{2}$
$=\frac{4 \pi \times 10^{-7} \times 10^8 \times \frac{1}{\mathrm{e}^2}}{2}$
$=\frac{20 \pi}{\mathrm{e}^2}$