Question

A rectangular loop of sides \(a\) and \(b\) carrying current \(I\) is placed in a magnetic field \(\vec B\) such that its area vector \(\vec A\) makes an angle \(\theta\) with \(\vec B\). With the help of a suitable diagram, show that the torque \(\vec \tau\) acting on the loop is given by \[ \vec \tau=\vec m\times \vec B, \] where \[ \vec m=I\vec A \] is the magnetic dipole moment of the loop.

Hint:

For a current loop in a magnetic field: \[ \vec\tau=\vec m\times\vec B \] and \[ m=NIA. \] Maximum torque occurs when \[ \theta=90^\circ. \]

Answer 1

Torque on a Current Loop Concept: A current carrying loop behaves like a magnetic dipole when placed in an external magnetic field. The magnetic field exerts equal and opposite forces on opposite sides of the loop. These forces constitute a couple and produce a torque.

Step 1: Consider a rectangular loop. Let \[ \text{Length}=a, \qquad \text{Breadth}=b. \] Area of loop: \[ A=ab. \] Current flowing through the loop is \(I\). The area vector \(\vec A\) is normal to the plane of the loop.

Step 2: Calculate force on the sides. For a straight conductor of length \(l\), \[ F=BIl\sin\phi. \] The pair of opposite sides experiences equal and opposite forces. These forces form a couple.

Step 3: Calculate the torque of the couple. Magnitude of torque is \[ \tau=(BIb)(a\sin\theta). \] Since \[ ab=A, \] \[ \tau=BIA\sin\theta. \]

Step 4: Introduce magnetic dipole moment. Magnetic dipole moment of the loop is defined as \[ \vec m=I\vec A. \] Therefore, \[ m=IA. \] Substituting, \[ \tau=mB\sin\theta. \] The vector form becomes \[ \boxed{ \vec\tau=\vec m\times \vec B } \] which is the required result.