Question

A circular coil of radius 0.1 m is placed in the X-Y plane and a current 2 A is passed through the coil in the clockwise direction when looking from above. Find the magnetic dipole moment of the current loop.

• A. \( 0.02\pi \, \text{Am}^2 \) in the \(-\hat{x}\) direction
• B. \( 0.02\pi \, \text{Am}^2 \) in the \(-\hat{z}\) direction
• C. \( 0.02\pi \, \text{Am}^2 \) in the \(+\hat{y}\) direction
• D. \( 0.02\pi \, \text{Am}^2 \) in the \(+\hat{z}\) direction

Hint:

When a current flows in a coil, the magnetic dipole moment points along the axis of the coil. For clockwise current, the magnetic dipole moment points in the negative \( z \)-direction.

Answer 1

The Correct Option is B

The magnetic dipole moment \( \vec{m} \) of a current loop is given by: \[ \vec{m} = I \cdot A \cdot \hat{n} \] Where: - \( I = 2 \, \text{A} \) is the current, - \( A = \pi r^2 \) is the area of the loop, and - \( \hat{n} \) is the unit vector normal to the plane of the loop. Here, the radius of the coil \( r = 0.1 \, \text{m} \), so the area is: \[ A = \pi (0.1)^2 = 0.01\pi \, \text{m}^2 \] The magnetic dipole moment is: \[ \vec{m} = 2 \times 0.01\pi \times \hat{n} = 0.02\pi \, \text{Am}^2 \] Since the current flows in the clockwise direction when viewed from above, the unit vector \( \hat{n} \) points in the \(-\hat{z}\) direction. 
Thus, the magnetic dipole moment is \( 0.02\pi \, \text{Am}^2 \) in the \(-\hat{z}\) direction.