Question

(i) A rectangular loop of sides \( l \) and \( b \) carries a current \( I \) clockwise. Write the magnetic moment \( \vec{m} \) of the loop and show its direction in a diagram.
(ii) The loop is placed in a uniform magnetic field \( \vec{B} \) and is free to rotate about an axis which is perpendicular to \( \vec{B} \). Prove that the loop experiences no net force, but a torque \( \vec{\tau} = \vec{m} \times \vec{B} \).

Hint:

For magnetic moment and torque problems: - Magnetic moment \( \vec{m} = I \vec{A} \), direction by right-hand rule. - In a uniform field, net force on a closed loop is zero, but torque is \( \vec{\tau} = \vec{m} \times \vec{B} \).

Answer 1

Step 1: Magnetic moment of the loop (Part i).
The magnetic moment \( \vec{m} \) of a current loop is given by \( \vec{m} = I \vec{A} \), where \( \vec{A} \) is the area vector. For a rectangular loop with sides \( l \) and \( b \), the area is \( A = l \times b \). Since the current is clockwise and the loop is in the \( xy \)-plane, the right-hand rule dictates that the area vector points in the negative \( z \)-direction (\( -\hat{k} \)). Therefore, \( \vec{A} = - l b \hat{k} \). Consequently, the magnetic moment is \( \vec{m} = I (- l b \hat{k}) = - I l b \hat{k} \). Diagram: The loop is situated in the \( xy \)-plane, with sides oriented along the \( x \)-axis (\( l \)) and \( y \)-axis (\( b \)). The clockwise current \( I \) results in a magnetic moment \( \vec{m} = - I l b \hat{k} \) directed along the negative \( z \)-axis. Step 2: Net force on the loop (Part ii).
The force on a current-carrying wire within a magnetic field is defined by \( \vec{F} = I (\vec{L} \times \vec{B}) \), where \( \vec{L} \) is the length vector of the wire. For the rectangular loop, the forces on each side are:

• Side 1 (\( l \), along \( \hat{i} \)): \( \vec{L}_1 = l \hat{i} \), and the force is \( \vec{F}_1 = I (l \hat{i} \times \vec{B}) \).
• Side 2 (\( b \), along \( \hat{j} \)): \( \vec{L}_2 = b \hat{j} \), and the force is \( \vec{F}_2 = I (b \hat{j} \times \vec{B}) \).
• Side 3 (\( l \), along \( -\hat{i} \)): \( \vec{L}_3 = -l \hat{i} \), resulting in a force \( \vec{F}_3 = I (-l \hat{i} \times \vec{B}) = -\vec{F}_1 \).
• Side 4 (\( b \), along \( -\hat{j} \)): \( \vec{L}_4 = -b \hat{j} \), yielding a force \( \vec{F}_4 = I (-b \hat{j} \times \vec{B}) = -\vec{F}_2 \).

The net force on the loop is the sum of these individual forces: \[ \vec{F}_{\text{net}} = \vec{F}_1 + \vec{F}_2 + \vec{F}_3 + \vec{F}_4 = 0. \] Step 3: Torque on the loop (Part ii).
The torque on a magnetic dipole is given by \( \vec{\tau} = \vec{m} \times \vec{B} \). Substituting the calculated magnetic moment \( \vec{m} = - I l b \hat{k} \), the torque becomes: \[ \vec{\tau} = (- I l b \hat{k}) \times \vec{B}. \] This result is consistent with the provided expression \( \vec{\tau} = \vec{m} \times \vec{B} \). The fact that the axis of rotation is perpendicular to \( \vec{B} \) aligns with this torque calculation.