Question

A circular coil has moment of inertia $0.8 \,kg \,m ^{2}$ around any diameter and is carrying current to produce 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 $4T$ 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:

• A. $10 $ rad $s^{-1}$
• B. $20 \pi $ rad $s^{-1}$
• C. $10 \pi $ rad $s ^{-1}$
• D. $20$ rad $s ^{-1}$

Answer 1

Answer: (A)

$I _{\text{ dia }}=0.8\, kg / m ^{2}$
$M =20 \,Am ^{2}$

$U _{ i }+ K _{ i }= U _{ f }+ K _{ f }$
$0+0=- MB \cos 30^{\circ}+\frac{1}{2} I \omega^{2}$
$20 \times 4 \times \frac{\sqrt{3}}{2}=\frac{1}{2}(0.8) \omega^{2}$
$\omega=\sqrt{100 \sqrt{3}}=10(3)^{1 / 4}$