$A$ $100$ turn closely wound circular coil of radius $10\; cm$ carries a current of $3.2\; A$.
$(a)$ What is the field at the centre of the coil?
$(b)$ What is the magnetic moment of this coil?
The coil is placed in a vertical plane and is free to rotate about a horizontal axis which coincides with its diameter. $A$ uniform magnetic field of $2\; T$ in the horizontal direction exists such that initially the axis of the coil is in the direction of the field. The coil rotates through an angle of $90^{\circ}$ under the influence of the magnetic field.
$(c)$ What are the magnitudes of the torques on the coil in the initial and final position?
$(d)$ What is the angular speed acquired by the coil when it has rotated by $90^{\circ}$? The moment of inertia of the coil is $0.1\; kg\; m^{2}$.

$(a)$ The magnetic field at the centre is given by $B = \frac{\mu_{0} N I}{2 R}$.
Given $N = 100$, $I = 3.2\; A$, $R = 0.1\; m$.
$B = \frac{4 \pi \times 10^{-7} \times 100 \times 3.2}{2 \times 0.1} = 2 \times 10^{-3}\; T$.
$(b)$ The magnetic moment is $m = N I A = N I \pi R^{2}$.
$m = 100 \times 3.2 \times 3.14 \times (0.1)^{2} = 10\; A\; m^{2}$.
$(c)$ Torque $\tau = |\vec{m} \times \vec{B}| = m B \sin \theta$.
Initially, $\theta = 0^{\circ}$, so $\tau_{i} = 0\; N\; m$.
Finally, $\theta = 90^{\circ}$, so $\tau_{f} = m B = 10 \times 2 = 20\; N\; m$.
$(d)$ Using the work-energy theorem, the work done by the magnetic torque equals the change in rotational kinetic energy.
$W = \int_{0}^{\pi/2} m B \sin \theta \; d\theta = m B [-\cos \theta]_{0}^{\pi/2} = m B$.
$W = \frac{1}{2} I_{coil} \omega^{2} = m B$.
$\omega = \sqrt{\frac{2 m B}{I_{coil}}} = \sqrt{\frac{2 \times 10 \times 2}{0.1}} = \sqrt{400} = 20\; rad/s$.