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JEE Mains · Physics · STD 12 - Moving charges and magnetism
- JEE Mains 2020
- Hard
- MCQ
A circular coil has moment of inertia \(0.8 \,kg m ^{2}\) around any diameter and is carrying current to roduce a magnetic moment of \(20\, Am ^{2}\). The coil is kept initially in a vertical position and it can rotate freely around a horizontal diameter. When a uniform magnetic field of \(4\, T\) is applied along the vertical, it starts rotating around its horizontal diameter. The angular speed the coil acquires after rotating by \(60^{\circ}\) will be
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Answer & Solution
Correct Answer
(A) \(10\, rad \,s ^{-1}\)
Step-by-step Solution
Detailed explanation
\(I _{ dia }=0.8 kg / m ^{2}\)
\(M =20 Am ^{2}\)
\(U _{ i }+ K _{ i }= U _{ f }+ K _{ f }\)
\(0+0=- MB \cos 30^{\circ}+\frac{1}{2} I \omega^{2}\)
\(20 \times 4 \times \frac{\sqrt{3}}{2}=\frac{1}{2}(0.8) \omega^{2}\)
\(\omega=\sqrt{100 \sqrt{3}}=10(3)^{1 / 4}\)
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