Question

A square loop with each side 1 cm, carrying a current of 10 A, is placed in a magnetic field of 0.2 T. The direction of the magnetic field is parallel to the plane of the loop. The torque experienced by the loop is .
Fill in the blank with the correct answer from the options given below

• A. \(2 × 10^{-4} Nm\)
• B. zero
• C. \(2 × 10^{-2} Nm\)
• D. 2Nm

Answer 1

The Correct Option is A

The torque on a current-carrying loop in a magnetic field is calculated using \(\tau = n \cdot I \cdot A \cdot B \cdot \sin\theta\). For this square loop, \(n=1\), \(I=10\) A, and \(B=0.2\) T. The magnetic field is parallel to the loop's plane, meaning \(\theta = 90^\circ\) between the field and the loop's normal, so \(\sin\theta = 1\).

The loop's side length is 1 cm (0.01 m), giving an area \(A = (0.01)^2 = 0.0001 \, \text{m}^2\).

Substituting these values yields \(\tau = 1 \cdot 10 \cdot 0.0001 \cdot 0.2 \cdot 1 = 2 \times 10^{-4} \, \text{Nm}\).

The resulting torque is \(2 \times 10^{-4} \, \text{Nm}\).