Question

A circular loop of area 0.04 m\(^2\) carrying a current of 10 A is held with its plane perpendicular to a magnetic field induction of 0.4 T. Then the torque acting on the circular loop is:

• A. 0.004 Nm
• B. zero
• C. 0.04 Nm
• D. 0.02 Nm

Hint:

When the plane of a current-carrying loop is perpendicular to the magnetic field, the torque acting on the loop is zero because the angle between the normal to the plane of the loop and the magnetic field is zero.

Answer 1

The Correct Option is B

The torque \( \tau \) acting on a current-carrying loop in a magnetic field is given by the formula: \[ \tau = N \cdot B \cdot A \cdot I \cdot \sin(\theta) \] where: - \( N \) is the number of turns in the coil (which is 1 in this case), - \( B \) is the magnetic field induction, - \( A \) is the area of the coil, - \( I \) is the current flowing through the coil, - \( \theta \) is the angle between the normal to the plane of the coil and the magnetic fiel(D) Given: - The area \( A = 0.04 \, \text{m}^2 \), - The magnetic field induction \( B = 0.4 \, \text{T} \), - The current \( I = 10 \, \text{A} \), - The angle \( \theta = 0^\circ \) (because the plane of the loop is perpendicular to the magnetic field, so the normal to the loop is parallel to the magnetic field). Since the loop is perpendicular to the magnetic field, \( \sin(0^\circ) = 0 \). Therefore, the torque acting on the loop is: \[ \tau = N \cdot B \cdot A \cdot I \cdot \sin(0^\circ) = 0 \] Thus, the torque acting on the circular loop is zero. Therefore, the correct answer is Option (B): zero.