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 Ω . The energy density in the inductor when the current reaches (frac 1 e ) of its maximum value is α π × frac 1 e 2 J / m 3 . The value of α is \ \ \ \ . (μ 0 =4 π × 10 -7 Tm / A ) .

Physics · Electromagnetic Induction

JEE Main 2026 — 22 January, Morning Shift — Question 48

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 $\left(\frac{1}{\mathrm{e}}\right)$ of its maximum value is $\alpha \pi \times \frac{1}{\mathrm{e}^{2}} \mathrm{~J} / \mathrm{m}^{3}$ . The value of $\alpha$ is $\_\_\_\_$ . $\left(\mu_{0}=4 \pi \times 10^{-7} \mathrm{Tm} / \mathrm{A}\right)$ .

Answer: 20

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{L}=10 \times 10^{-3} \mathrm{H}$ $\mathrm{N}=10^{4}$ $\mathrm{I}_{0}=\frac{10}{10}=1 \mathrm{~A}$ (max current) $\mathrm{I}=\frac{1}{\mathrm{e}}$ $\mathrm{E}_{\mathrm{d}}=\frac{\mathrm{B}^{2}}{2 \mu_{0}}$ $\mathrm{B}=\mu_{0} \mathrm{nI}$ $\mathrm{L}=\mu_{0} \mathrm{n}^{2} \pi \mathrm{R}^{2} \ell$ $\mathrm{E}_{\mathrm{d}}=\frac{\mu_{0} \mathrm{n}^{2} \mathrm{I}^{2}}{2}$ $=\frac{4 \pi \times 10^{-7} \times 10^{8} \times \frac{1}{\mathrm{e}^{2}}}{2}$ $=\frac{20 \pi}{\mathrm{e}^{2}}$

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Exam: JEE Main 2026
Paper: 22 January, Morning Shift
Subject: Physics
Chapter: Electromagnetic Induction
Topic: Self-Inductance and Mutual Inductance and Energy Density

Answer: 20

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