Question

Torque on a coil carrying current \( I \) having \( N \) turns and area of cross section \( A \) when placed with its plane perpendicular to a magnetic field \( B \) is:

• A. \( 2NBI A \)
• B. \( \frac{NBI A}{3} \)
• C. 0
• D. \( \frac{NBI A}{2} \)
• E. \( NBI A \)

Hint:

Remember that when the plane of the coil is perpendicular to the magnetic field (\( \theta = 90^\circ \)), the torque simplifies to \( \tau = NBI A \).

Answer 1

The Correct Option is C

The torque \( \tau \) on a coil with \( N \) turns, carrying current \( I \), placed in a magnetic field \( B \) and having an area of cross section \( A \), is given by the formula: \[ \tau = NIBA \sin \theta \] where: \( N \) is the number of turns, \( I \) is the current in the coil, \( B \) is the magnetic field strength, \( A \) is the area of the coil's cross-section, \( \theta \) is the angle between the plane of the coil and the magnetic field. If the plane of the coil is perpendicular to the magnetic field, the angle \( \theta = 90^\circ \). Since \( \sin 90^\circ = 1 \), the torque is: \[ \tau = NIBA \] Thus, the correct formula for the torque when the plane of the coil is perpendicular to the magnetic field is: \[ \tau = NBI A \] Therefore, the correct answer is option (E), \( NBI A \).