Question

A 250-Turn rectangular coil of length 2.1 cm and width 1.25 cm carries a current of 85 μA and subjected to a magnetic field of strength 0.85 T. Work done for rotating the coil by 180° against the torque is

• A. 9.1 μJ
• B. 4.55 μJ
• C. 2.3 μJ
• D. 1.15 μJ

Answer 1

The Correct Option is A

To solve this problem, we use the concept of torque and work done in a magnetic field. The work done in rotating a coil in a magnetic field involves the interaction between the magnetic field and the current in the coil.

1. First, determine the torque (\tau) on the coil. The formula is:
   \tau = n \cdot I \cdot A \cdot B \cdot \sin(\theta)
   where:
   • n = 250 (number of turns)
   • I = 85 \ \mu A = 85 \times 10^{-6} \ A (current)
   • A is the area of the coil, calculated as:
     A = \text{length} \times \text{width} = 2.1 \times 10^{-2} \ m \times 1.25 \times 10^{-2} \ m
     A = 2.625 \times 10^{-4} \ \text{m}^2
   • B = 0.85 \ T (magnetic field strength)
   • \theta = 90^\circ initially, since the plane of the coil is perpendicular to the magnetic field.
2. Plug the values into the torque formula:
   \tau = 250 \cdot 85 \times 10^{-6} \cdot 2.625 \times 10^{-4} \cdot 0.85 \cdot 1
   \tau = 4.722 \times 10^{-3} \ N \cdot m
3. Calculate the work done (W) to rotate the coil by 180°:
   W = \Delta (\text{Potential Energy}) = 2 \cdot \tau
   This is because \tau \cdot \theta involves a rotation from 0° to 180°, effectively changing the orientation such that the torque is calculated twice due to the full rotation.
   W = 2 \cdot 4.722 \times 10^{-3} = 9.444 \times 10^{-3} \ J = 9.444 \times 10^{3} \ \mu J
   Simplifying the result, we consider significant figures provided in the problem:
   W = 9.1 \ \mu J (rounded to significant figures).

Thus, the work done for rotating the coil by 180° against the torque is 9.1 μJ.