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 4 T is applied along the vertical, it starts rotating around its horizontal diameter. The angular speed the coil acquires after rotating by 60° will be:

We have a flat circular coil whose plane is kept vertical at the beginning, so the magnetic moment vector $$\vec \mu$$ is horizontal while the external uniform magnetic field $$\vec B$$ is vertical. Hence the initial angle between the two vectors is $$\theta_0 = 90^{\circ}$$.

The coil can rotate freely about one of its horizontal diameters. When the field is suddenly switched on, the magnetic torque tries to turn the coil so that $$\vec \mu$$ may align with $$\vec B$$. We are told that the coil actually rotates through an angle of $$60^{\circ}$$ about the horizontal diameter. Therefore the new angle between $$\vec \mu$$ and $$\vec B$$ becomes

$$\theta = 90^{\circ} - 60^{\circ} = 60^{\circ}.$$

The magnetic potential energy of a magnetic dipole in a uniform magnetic field is given first - we state the formula explicitly:

$$U = -\,\vec \mu \cdot \vec B = -\,\mu B \cos\theta.$$

Initially (at $$\theta_0 = 90^{\circ}$$),

$$U_0 = -\,\mu B \cos 90^{\circ} = -\,\mu B \,(0) = 0.$$

After the coil has turned (at $$\theta = 60^{\circ}$$),

$$U = -\,\mu B \cos 60^{\circ} = -\,\mu B \left(\dfrac12\right) = -\dfrac{\mu B}{2}.$$

Thus the change in potential energy experienced by the coil is

$$\Delta U = U - U_0 = \left(-\dfrac{\mu B}{2}\right) - 0 = -\dfrac{\mu B}{2}.$$

The negative sign shows that the energy of the system has decreased; that lost potential energy appears as rotational kinetic energy. By conservation of mechanical energy:

$$\dfrac12 I \omega^{2} = -\Delta U = \dfrac{\mu B}{2}.$$

We now insert the numerical values given in the problem:

Moment of inertia: $$I = 0.8 \text{ kg m}^2,$$
Magnetic moment: $$\mu = 20 \text{ A m}^2,$$
Magnetic field: $$B = 4 \text{ T}.$$

Substituting these values into the energy-conservation equation, we get

$$\dfrac12 \,(0.8)\,\omega^{2} = \dfrac{(20)(4)}{2}.$$

Simplifying the right-hand side first,

$$\dfrac{(20)(4)}{2} = \dfrac{80}{2} = 40,$$

so the equation becomes

$$0.4\,\omega^{2} = 40.$$

Solving step by step:

$$\omega^{2} = \dfrac{40}{0.4} = 100,$$

and therefore

$$\omega = \sqrt{100} = 10 \text{ rad s}^{-1}.$$

Hence, the correct answer is Option A.