JEE Mains · Physics · STD 12 - Moving charges and magnetism
- JEE Mains 2024
- Hard
- Numerical
A coil having \(100\) turns, area of \(5 \times 10^{-3} \mathrm{~m}^2\), carrying current of \(1 \mathrm{~mA}\) is placed in uniform magnetic field of \(0.20 \mathrm{~T}\) such a way that plane of coil is perpendicular to the magnetic field. The work done in turning the coil through \(90^{\circ}\) is _______ \(\mu \mathrm{J}\).
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(100\)
Step-by-step Solution
Detailed explanation
\(W=\Delta \mathrm{U}=\mathrm{U}_{\mathrm{f}}-\mathrm{U}_{\mathrm{i}}\)
\(\mathrm{W}=(-\vec{\mu} \cdot \overrightarrow{\mathrm{B}})_{\mathrm{f}}-(-\vec{\mu} \cdot \overrightarrow{\mathrm{B}})_i\)
\(=0+(\vec{\mu} \cdot \overrightarrow{\mathrm{B}})_i\)
\(=\left(100 \times 5 \times 10^{-3} \times 1 \times 10^{-3}\right) \times 0.2 \mathrm{~J}\)
\(=1 \times 10^{-4} \mathrm{~J}=100 \mu \mathrm{J}\)
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