Torque Acting on Circular Coil in Magnetic Field

To find the torque acting on the circular coil, we use the formula for the torque on a current-carrying loop in a uniform magnetic field:

$\tau = M B \sin\theta$

where:

• $M$ is the magnitude of the magnetic dipole moment of the coil,
• $B$ is the magnitude of the magnetic field,
• $\theta$ is the angle between the normal to the plane of the loop (axis of the coil, along which the magnetic dipole moment vector $\vec{M}$ points) and the magnetic field vector $\vec{B}$.

Step 1: Calculate the area of the circular coil ($A$) Given radius $r = 2\text{ cm} = 2 \times 10^{-2}\text{ m}$: $A = \pi r^2 = \pi \times (2 \times 10^{-2}\text{ m})^2 = 4\pi \times 10^{-4}\text{ m}^2$

Step 2: Calculate the magnetic dipole moment ($M$) The magnetic dipole moment of a coil with $N$ turns carrying a current $I$ is given by: $M = N I A$

Given:

• $N = 125$
• $I = 1\text{ A}$

Substituting the values: $M = 125 \times 1\text{ A} \times (4\pi \times 10^{-4}\text{ m}^2) = 500\pi \times 10^{-4}\text{ A}\cdot\text{m}^2$

Step 3: Calculate the torque ($\tau$) Given:

• $B = 0.4\text{ T}$
• $\theta = 30^\circ$

Substituting $M$, $B$, and $\theta$ into the torque equation: $\tau = (500\pi \times 10^{-4}) \times 0.4 \times \sin(30^\circ)$

Since $\sin(30^\circ) = \frac{1}{2}$: $\tau = 500\pi \times 10^{-4} \times 0.4 \times 0.5$ $\tau = 100\pi \times 10^{-4}\text{ N}\cdot\text{m}$

Step 4: Substitute $\pi = 3.14$ and solve for $\alpha$ $\tau = 100 \times 3.14 \times 10^{-4}\text{ N}\cdot\text{m} = 314 \times 10^{-4}\text{ N}\cdot\text{m}$

Comparing this with the given expression $\tau = \alpha \times 10^{-4}\text{ N}\cdot\text{m}$: $\alpha = 314$