A circular coil has moment of inertia $0.8 \mathrm{~kg} \mathrm{~m}^{2}$ around any diameter and is carrying current to produce a magnetic moment of $20 \mathrm{Am}^{2}$. The coil is kept initially in a vertical position and it can rotate freely around a horizontal diameter. When a uniform magnetic field of $4 \mathrm{~T}$ 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 :
Correct Option: 1
(1) Given,
Moment of inertia of circular coil, $I=0.8 \mathrm{~kg} \mathrm{~m}^{2}$
Magnetic moment of circular coil, $M=20 \mathrm{Am}^{2}$
Rotational kinetic energy of circular coil,
$\mathrm{KE}=\frac{1}{2} I \omega^{2}$
Here, $\omega=$ angular speed of coil
Potential energy of bar magnet $=-M B \cos \phi$
From energy conservation
$\frac{1}{2} I \omega^{2}=U_{\text {in }}-U_{f}=-M B \cos 60^{\circ}-(-M B)$
$\Rightarrow \frac{M B}{2}=\frac{1}{2} I \omega^{2}$
$\Rightarrow \frac{20 \times 4}{2}=\frac{1}{2}(0.8) \omega^{2}$
$\Rightarrow 100=\omega^{2} \Rightarrow \omega=10 \mathrm{rad}$
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