A circular coil has moment of inertia
Question:

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 :

  1. (1) $10 \mathrm{rad} \mathrm{s}^{-1}$

  2. (2) $10 \pi \mathrm{rad} \mathrm{s}^{-1}$

  3. (3) $20 \pi \mathrm{rad} \mathrm{s}^{-1}$

  4. (4) $20 \mathrm{rad} \mathrm{s}^{-1}$


Correct Option: 1

Solution:

(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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