Question

A solenoid of radius 2 cm and with 125 turns is kept in a uniform magnetic field of 0.4 T carries a current of 1A. The axis of the solenoid makes an angle of \( 30^\circ \) with the field. The torque acting on the solenoid will be:

• A. \( \pi \times 10^{-6} \) N-m
• B. \( \pi \times 10^{-2} \) N-m
• C. \( 2\pi \times 10^{-6} \) N-m
• D. \( 2\pi \times 10^{-2} \) N-m

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
A current-carrying solenoid behaves like a magnetic dipole. When placed in an external magnetic field, it experiences a torque that tends to align its axis with the field.
Step 2: Key Formula or Approach:
1. Magnetic Moment (\( M \)) = \( NIA \).
2. Torque (\( \tau \)) = \( MB \sin \theta \).
Area (\( A \)) = \( \pi r^2 \).
Step 3: Detailed Explanation:
Given: \( N = 125 \), \( I = 1 \, \text{A} \), \( r = 2 \, \text{cm} = 0.02 \, \text{m} \), \( B = 0.4 \, \text{T} \), \( \theta = 30^\circ \). Calculate Area: \[ A = \pi (0.02)^2 = 4\pi \times 10^{-4} \, \text{m}^2 \] Calculate Magnetic Moment: \[ M = 125 \times 1 \times (4\pi \times 10^{-4}) = 500\pi \times 10^{-4} = 5\pi \times 10^{-2} \, \text{A-m}^2 \] Calculate Torque: \[ \tau = (5\pi \times 10^{-2}) \times 0.4 \times \sin 30^\circ \] \[ \tau = (2\pi \times 10^{-2}) \times \frac{1}{2} = \pi \times 10^{-2} \, \text{N-m} \] (Note: Re-checking the turn density/current if the answer deviates. For the values given, \( \pi \times 10^{-2} \) is correct).
Step 4: Final Answer:
The torque acting on the solenoid is \( \pi \times 10^{-2} \) N-m.