1. 8 μ J 2. 4 μ J 3. 4 mJ 4. 8 mJ

Answer: 1. 8 μ J Energy stored in an inductor: U=1/2LI² . U=1/2×(4×10⁻⁶)×(2)² = 8×10⁻⁶ J = 8 μ J.

The magnetic energy stored in an inductor of inductance 4 μH carrying a current of 2 A is :

1. \(8\,\mu\)J

2. \(4\,\mu\)J

3. 4 mJ

4. 8 mJ

Doubtly Asked 30 Oct 2026 NEET UG 2023 Practice test neet Save WhatsApp Report

Best answer

Answer: 1. \(8\,\mu\)J

Energy stored in an inductor: \(U=\tfrac12LI^2\).

\(U=\tfrac12\times(4\times10^{-6})\times(2)^2 = 8\times10^{-6}\) J = 8 \(\mu\)J.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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