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JEE Main 20246 AprilMorning Shift
Exam
JEE Main
Year
2024
Date
6 April
Shift
Morning Shift
Subject
Physics
Chapter
Magnetic Effects of Current

JEE Main 2024 · 6 April · Morning Shift · Physics · Magnetic Effects of Current

A circular coil having 200 turns, 2.5 × 10^ -4 ~m^2 area and...

A circular coil having 200 turns, 2.5×10−4 m22.5 \times 10^{-4} \mathrm{~m}^2 area and carrying 100μA100 \mu \mathrm{A} current is placed in a uniform magnetic field of 1T. Initially the magnetic dipole moment (M⃗)(\vec{M}) was directed along B⃗\vec{B}. Amount of work, required to rotate the coil through 90∘90^{\circ} from its initial orientation such that M⃗\vec{M} becomes perpendicular to B⃗\vec{B}, is ______ μJ\mu \mathrm{J}.

Correct answer

0

Step-by-step solution

We know \begin{aligned} \mathrm{W}_{\text{ext }} & =\Delta \mathrm{U}+\Delta \mathrm{KE} \quad(\text{ P.E. }=-\overrightarrow{\mathrm{M}} \cdot \overrightarrow{\mathrm{B}}) \\ & =-\overrightarrow{\mathrm{M}} \cdot \overrightarrow{\mathrm{B}_{\mathrm{f}}+\overrightarrow{\mathrm{M}} \cdot \overrightarrow{\mathrm{B}_{\mathrm{i}}+0 \\ & =-\mathrm{MB} \cos 90+\mathrm{MB} \cos 0 \\ & =\mathrm{MB} \\ & =\mathrm{NIAB} \\ & =200 \times 100 \times 10^{-6} \times \frac{5}{2} \times 10^{-4} \times 1=5 \mu \mathrm{J} \end{aligned}
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