Physics · Electromagnetic Induction

NEET (UG) 2023 — Question 27

The magnetic energy stored in an inductor of inductance $4 \mu \mathrm{H}$ carrying a current of 2 A is :

Option A:
$4$ $mJ$
Option B:
$8$ $mJ$
Option C:
$8 \mu \mathrm{~J}$
Correct
Option D:
$4 \mu \mathrm{~J}$

Answer: C

Step-by-step solution

Magnetic energy stored in an inductor

\begin{aligned} U & =\frac{1}{2} L i^2 \\ & =\frac{1}{2} \times 4 \times 10^{-6} \times(2)^2 \\ & =8 \times 10^{-6} \mathrm{~J} \\ U & =8 \mu \mathrm{~J} \end{aligned}

Answer key and solution verified before publishing.

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Exam: NEET (UG) 2023
Paper: 2023 paper
Subject: Physics
Chapter: Electromagnetic Induction
Topic: Self-Inductance and Mutual Inductance and Energy Density

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The magnetic energy stored in an inductor of inductance 4 μ H… | NEET (UG) 2023 PYQ with Solution · DhiX AI