Question:medium

A uniform magnetic field of \( 2 \times 10^{-3} \, \text{T} \) acts along positive Y-direction. A rectangular loop of sides 20 cm and 10 cm with current of 5 A is in the Y-Z plane. The current is in anticlockwise sense with reference to the negative X-axis. Magnitude and direction of the torque is:

Updated On: Jan 13, 2026
  • \( 2 \times 10^{-4} \, \text{N} \, \text{m} \) along positive Z-direction
  • \( 2 \times 10^{-4} \, \text{N} \, \text{m} \) along negative Z-direction
  • \( 2 \times 10^{-4} \, \text{N} \, \text{m} \) along positive X-direction
  • \( 2 \times 10^{-4} \, \text{N} \, \text{m} \) along positive Y-direction
Show Solution

The Correct Option is B

Solution and Explanation

The problem requires calculating the torque on a single-turn rectangular current loop situated in a magnetic field. The torque \((\tau)\) is determined using the formula: \(\tau = n \cdot I \cdot A \cdot B \cdot \sin(\theta)\)

where:

  • \(n = 1\) (number of loops).
  • \(I = 5 \, \text{A}\) (current).
  • \(A\) (area of the loop).
  • \(B = 2 \times 10^{-3} \, \text{T}\) (magnetic field strength).
  • \(\theta\) (angle between the loop's normal and the magnetic field).

The loop dimensions are 20 cm by 10 cm. Converting to meters for area calculation: \(A = 0.2 \, \text{m} \times 0.1 \, \text{m} = 0.02 \, \text{m}^2\). The loop lies in the Y-Z plane, meaning its normal is along the X-axis. With the magnetic field along the positive Y-axis, \(\theta = 90^\circ\).

Substituting values into the torque equation:

\(\tau = 1 \times 5 \, \text{A} \times 0.02 \, \text{m}^2 \times 2 \times 10^{-3} \, \text{T} \times \sin(90^\circ) = 2 \times 10^{-4} \, \text{N} \cdot \text{m}\)

The direction of the torque is found using the right-hand rule. With the area vector along the X-axis and the magnetic field along the Y-axis, the torque is initially along the Z-axis. However, considering the anticlockwise current direction relative to the negative X-axis, the torque direction is reversed from the Y to Z curl, resulting in a direction along the negative Z-axis.

Therefore, the torque is \(2 \times 10^{-4} \, \text{N} \, \text{m}\) along the negative Z-direction.

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