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: Understanding the Concept:
A current-carrying loop placed in a magnetic field experiences a torque. The magnitude of this torque depends on the magnetic moment of the loop, the strength of the magnetic field, and the orientation of the loop with respect to the field.
Step 2: Key Formula or Approach:
The torque (\(\tau\)) on a coil is given by the vector product of its magnetic dipole moment (\(\vec{m}\)) and the magnetic field (\(\vec{B}\)):
\[ \vec{\tau} = \vec{m} \times \vec{B} \] The magnitude of the torque is:
\[ \tau = mB \sin\theta \] where:
- \(m = NIA\) is the magnitude of the magnetic moment.
- \(N\) is the number of turns.
- \(I\) is the current.
- \(A\) is the area of the coil.
- \(\theta\) is the angle between the magnetic moment vector \(\vec{m}\) and the magnetic field vector \(\vec{B}\).
Step 3: Detailed Explanation:
First, let's list the given values in SI units:
- Side of the square coil, \(s = 10 \text{ cm} = 0.1 \text{ m}\).
- Area of the coil, \(A = s^2 = (0.1 \text{ m})^2 = 0.01 \text{ m}^2\).
- Number of turns, \(N = 200\).
- Magnetic field strength, \(B = 2 \text{ T}\).
- Current, \(I = 3 \text{ mA} = 3 \times 10^{-3} \text{ A}\).
Next, determine the angle \(\theta\). The problem states "the plane of the coil is in the direction of magnetic field". This means the magnetic field lines are parallel to the surface of the coil. The magnetic moment vector \(\vec{m}\) is, by definition, perpendicular to the plane of the coil. Therefore, the angle between \(\vec{m}\) and \(\vec{B}\) is \(\theta = 90^\circ\).
Now, calculate the magnetic moment \(m\):
\[ m = NIA = (200) \times (3 \times 10^{-3} \text{ A}) \times (0.01 \text{ m}^2) = 600 \times 10^{-5} \text{ A}\cdot\text{m}^2 = 6 \times 10^{-3} \text{ A}\cdot\text{m}^2 \] Finally, calculate the torque \(\tau\):
\[ \tau = mB \sin\theta = (6 \times 10^{-3} \text{ A}\cdot\text{m}^2) \times (2 \text{ T}) \times \sin(90^\circ) \] Since \(\sin(90^\circ) = 1\), the torque is at its maximum value.
\[ \tau = (6 \times 10^{-3}) \times 2 \times 1 = 12 \times 10^{-3} \text{ Nm} \] Step 4: Final Answer:
The torque acting on the coil is \(12 \times 10^{-3}\) Nm. Therefore, option (A) is correct.