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: \[ \vec{m} = I \vec{A}, \] where \( \vec{A} \) is the area vector. For a rectangle of sides \( l \) and \( b \), the area is: \[ A = l \times b. \] The current is clockwise. Assuming the loop lies in the \( xy \)-plane, the right-hand rule gives the area vector in the negative \( z \)-direction (\( -\hat{k} \)): \[ \vec{A} = - l b \hat{k}. \] Thus: \[ \vec{m} = I (- l b \hat{k}) = - I l b \hat{k}. \] Diagram: The loop is in the \( xy \)-plane, with sides \( l \) (along \( x \)-axis) and \( b \) (along \( y \)-axis). The current \( I \) is clockwise, so \( \vec{m} = - I l b \hat{k} \) points along the negative \( z \)-axis. Step 2: Net force on the loop (Part ii).
The force on a current-carrying wire in a magnetic field is: \[ \vec{F} = I (\vec{L} \times \vec{B}), \] where \( \vec{L} \) is the length vector of the wire. For the rectangular loop: \begin{itemize} \item Side 1 (\( l \), along \( \hat{i} \)): \( \vec{L}_1 = l \hat{i} \), force \( \vec{F}_1 = I (l \hat{i} \times \vec{B}) \). \item Side 2 (\( b \), along \( \hat{j} \)): \( \vec{L}_2 = b \hat{j} \), force \( \vec{F}_2 = I (b \hat{j} \times \vec{B}) \). \item Side 3 (\( l \), along \( -\hat{i} \)): \( \vec{L}_3 = -l \hat{i} \), force \( \vec{F}_3 = I (-l \hat{i} \times \vec{B}) = -\vec{F}_1 \). \item Side 4 (\( b \), along \( -\hat{j} \)): \( \vec{L}_4 = -b \hat{j} \), force \( \vec{F}_4 = I (-b \hat{j} \times \vec{B}) = -\vec{F}_2 \). \end{itemize} Net force: \[ \vec{F}_{\text{net}} = \vec{F}_1 + \vec{F}_2 + \vec{F}_3 + \vec{F}_4 = 0. \] Step 3: Torque on the loop (Part ii).
The torque on a magnetic dipole is: \[ \vec{\tau} = \vec{m} \times \vec{B}. \] Using \( \vec{m} = - I l b \hat{k} \), the torque is: \[ \vec{\tau} = (- I l b \hat{k}) \times \vec{B}, \] which matches the given expression \( \vec{\tau} = \vec{m} \times \vec{B} \). The axis of rotation being perpendicular to \( \vec{B} \) is consistent with this torque.