Question

A square coil of side 10 cm having 200 turns is placed in a uniform magnetic field of 2 T such that the plane of the coil is in the direction of magnetic field. If the current through the coil is 3 mA, then the torque acting on the coil is

• A. \(12 \times 10^{-3}\) Nm
• B. \(24 \times 10^{-3}\) Nm
• C. \(6 \times 10^{-3}\) Nm
• D. Zero

Hint:

The most common source of error in torque problems is the angle \(\theta\). Remember that \(\theta\) is the angle between the magnetic field and the normal to the coil's area (the direction of \(\vec{m}\)), not the angle between the field and the plane of the coil. If the field is parallel to the plane, \(\theta = 90^\circ\). If the field is perpendicular to the plane, \(\theta = 0^\circ\) and the torque is zero.

Answer 1

The Correct Option is A

Step 1: Formula for Torque. The torque \(\tau\) on a current-carrying coil in a magnetic field is given by: \[ \tau = N I A B \sin\theta \] Where: - \(N\) is the number of turns. - \(I\) is the current. - \(A\) is the area of the coil. - \(B\) is the magnetic field strength. - \(\theta\) is the angle between the normal to the coil area and the magnetic field.
Step 2: Identify the angle \(\theta\). The problem states "the plane of the coil is in the direction of magnetic field". This means the area vector (which is normal to the plane) is perpendicular to the magnetic field. Therefore, \(\theta = 90^\circ\), and \(\sin 90^\circ = 1\). The torque is maximum: \(\tau = N I A B\).
Step 3: Calculation. Given: \(N = 200\) \(I = 3 \, \text{mA} = 3 \times 10^{-3} \, \text{A}\) Side \(a = 10 \, \text{cm} = 0.1 \, \text{m} \implies \text{Area } A = a^2 = (0.1)^2 = 0.01 \, \text{m}^2\) \(B = 2 \, \text{T}\) \[ \tau = 200 \times (3 \times 10^{-3}) \times 0.01 \times 2 \] \[ \tau = 1200 \times 10^{-5} \, \text{Nm} \] \[ \tau = 12 \times 10^{-3} \, \text{Nm} \]