Question

A circular coil of radius 2 cm and 125 turns carries a current of 1 A. The coil is placed in a uniform magnetic field of magnitude 0.4 T. The axis of the coil makes an angle of \( 30^\circ \) with the direction of the magnetic field. The torque acting on the coil is \( \alpha \times 10^{-4} \text{ N.m} \). The value of \( \alpha \) is _______. (\( \pi = 3.14 \))

Answer 1

Correct Answer: 314

Step 1: Understanding the Concept:
A current-carrying loop in a magnetic field experiences a torque that depends on its magnetic dipole moment and the external field. The torque acts to align the magnetic moment with the field.

Step 2: Key Formula or Approach:
Torque: \( \tau = NIAB \sin \theta \), where \( A = \pi r^2 \) and \( \theta \) is the angle between the axis (normal) and the magnetic field.

Step 3: Detailed Explanation:
Given: \( N = 125 \), \( I = 1 \text{ A} \), \( r = 0.02 \text{ m} \), \( B = 0.4 \text{ T} \), \( \theta = 30^\circ \).
First, calculate the area \( A \):
\[ A = \pi r^2 = 3.14 \times (0.02)^2 = 3.14 \times 4 \times 10^{-4} = 12.56 \times 10^{-4} \text{ m}^2 \]
Now, calculate the torque \( \tau \):
\[ \tau = 125 \times 1 \times (12.56 \times 10^{-4}) \times 0.4 \times \sin(30^\circ) \]
\[ \tau = 125 \times 12.56 \times 0.4 \times 0.5 \times 10^{-4} \]
\[ \tau = 125 \times 12.56 \times 0.2 \times 10^{-4} \]
\[ \tau = 25 \times 12.56 \times 10^{-4} = 314 \times 10^{-4} \text{ N.m} \]
Comparing with \( \alpha \times 10^{-4} \), we find \( \alpha = 314 \).

Step 4: Final Answer:
The value of \( \alpha \) is 314.