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
JEE MAIN 2020, Diffcult
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$I _{ dia }=0.8 kg / m ^{2}$
$M =20 Am ^{2}$
$U _{ i }+ K _{ i }= U _{ f }+ K _{ f }$
$0+0=- MB \cos 30^{\circ}+\frac{1}{2} I \omega^{2}$
$20 \times 4 \times \frac{\sqrt{3}}{2}=\frac{1}{2}(0.8) \omega^{2}$
$\omega=\sqrt{100 \sqrt{3}}=10(3)^{1 / 4}$
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