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:

Remember:

Magnetic dipole moment: \( \vec{m} = I\vec{A} \)
Torque on loop: \( \vec{\tau} = \vec{m} \times \vec{B} \)
Loop tends to align with magnetic field like a compass needle.

Answer 1

Concept: A current-carrying loop in a magnetic field experiences torque due to magnetic forces on its sides. Magnetic force on a current element: \[ \vec{F} = I (\vec{l} \times \vec{B}) \]
Step 1: Consider rectangular loop.

Lengths: \( a \) and \( b \)
Area: \( A = ab \)
Area vector \( \vec{A} \) normal to plane of loop
Let magnetic field \( \vec{B} \) make angle \( \theta \) with area vector.
Step 2: Forces on sides.

Two sides parallel to field → no force
Two sides perpendicular to field → equal and opposite forces
These forces form a couple producing torque.
Step 3: Magnitude of force. For side of length \( a \): \[ F = I a B \sin \theta \]
Step 4: Torque on loop. Torque = force × perpendicular distance between forces = \( b \) \[ \tau = F \cdot b = (I a B \sin \theta) b \] \[ \tau = IabB \sin \theta \] Since \( A = ab \): \[ \tau = IAB \sin \theta \]
Step 5: Magnetic dipole moment. Magnetic dipole moment of loop: \[ \vec{m} = I \vec{A} \] So magnitude of torque: \[ \tau = mB \sin \theta \]
Step 6: Vector form. The direction of torque is perpendicular to both \( \vec{m} \) and \( \vec{B} \). Thus: \[ \boxed{\vec{\tau} = \vec{m} \times \vec{B}} \] Conclusion: A current loop behaves like a magnetic dipole and experiences torque tending to align its magnetic moment with the magnetic field.