JEE Mains · Physics · STD 12 - Moving charges and magnetism
- JEE Mains 2024
- Hard
- Numerical
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 \(1 \mathrm{~T}\). Initially the magnetic dipole moment \((\vec{M})\) was directed along \(\vec{B}\). Amount of work, required to rotate the coil through \(90^{\circ}\) from its initial orientation such that \(\overrightarrow{\mathrm{M}}\) becomes perpendicular to \(\vec{B}\) is _______ \(\mu \mathrm{J}\).
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(5\)
Step-by-step Solution
Detailed explanation
We know
\(\mathrm{W}_{\text {ext }} =\Delta \mathrm{U}+\Delta \mathrm{KE} \quad(\mathrm{P} . \mathrm{E} .=-\overrightarrow{\mathrm{M}} \cdot \overrightarrow{\mathrm{B}})\)
\(=-\overrightarrow{\mathrm{M}} \cdot \overrightarrow{\mathrm{B}}_{\mathrm{f}}+\overrightarrow{\mathrm{M}} \cdot \overrightarrow{\mathrm{B}}_{\mathrm{i}}+0\)
\(=-\mathrm{MB} \cos 90+\mathrm{MB} \cos 0\)
\(=\mathrm{MB}\)
\(=\mathrm{NIAB}\)
\(=200 \times 100 \times 10^{-6} \times \frac{5}{2} \times 10^{-4} \times 1=5 \mu\)
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