Inductance of a coil with 10⁴ turns is 10 mH and it is connected to a… | JEE Main 2026 (22 Jan, Shift 1)

Q20JEE Main 2026 · 22 JAN · SHIFT 1Difficult

Self-Inductance & Energy

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 $\alpha \pi \times \frac{1}{e^2}\text{ J/m}^{3}$ . The value of $\alpha$ is

______.

$\left(\mu_0 = 4\pi \times 10^{-7}\text{ Tm/A}\right).$

Numerical answer type. Work it out, then open the answer.

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Answer: 20

Maximum current $I_0 = \dfrac{10}{10} = 1$ A, so $I = \dfrac{1}{e}$ A.
Energy density $u = \dfrac{B^2}{2\mu_0}$ with $B = \mu_0 n I$ , so $u = \dfrac{\mu_0 n^2 I^2}{2}$ .
Taking $n = 10^4$ turns per metre: $u = \dfrac{4\pi\times 10^{-7}\times 10^{8}}{2}\times\dfrac{1}{e^2} = 20\pi \times \dfrac{1}{e^2}$ J/m $^3$
So $\alpha = 20$ .

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