A circular coil having 200 turns, 2.5 × 10 -4 m 2 area and carrying 100 μ A current is placed in a uniform magnetic field of 1 T . Initially the magnetic dipole moment ( overrightarrow M ) was directed along overrightarrow B . Amount of work, required to rotate the coil through 90 ° from its initial orientation such that vec M becomes perpendicular to vec B , is \ \ \ \ \ \ \ μ J .

Physics · Electromagnetic Induction

JEE Main 2024 — 6 April, Shift 1 — Question 57

A circular coil having 200 turns, $2.5 \times 10^{-4} \mathrm{~m}^{2}$ area and carrying $100 \mu \mathrm{~A}$ current is placed in a uniform magnetic field of 1 T . Initially the magnetic dipole moment ( $\overrightarrow{\mathrm{M}}$ ) was directed along $\overrightarrow{\mathrm{B}}$ . Amount of work, required to rotate the coil through $90^{\circ}$ from its initial orientation such that $\vec{M}$ becomes perpendicular to $\vec{B}$ , is $\_\_\_\_\_\_\_$ $\mu \mathrm{J}$ .

Answer: 5

Numerical answer — enter this value.

Step-by-step solution

Given $\; N=200,\; A=2.5\times10^{-4}\,\text{m}^2,\;$ $I=100\,\mu\text{A}=10^{-4}\,\text{A},\; B=1\,\text{T}.$

Magnetic dipole moment: $M = NIA = 5\times10^{-6}\,\text{A m}^2.$

Potential energy: $U=-MB\cos\theta.$

$W = U_f-U_i = [-MB\cos90^\circ]-[-MB\cos0^\circ].$

$W = MB = 5\times10^{-6}\,\text{J} = 5\,\mu\text{J}.$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 6 April, Shift 1
Subject: Physics
Chapter: Electromagnetic Induction
Topic: Magnetic Flux, Faraday's Law and Lenz's Law