A current loop has a magnetic moment of \(0.5\ \mathrm{A\,m^{2}}\). It is placed in a uniform magnetic field of 0.4 T. The plane of the loop makes an angle of \(30^\circ\) with the field. The magnitude of the torque on the loop is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(0.1\sqrt{3}\ \mathrm{N\,m}\)
The torque on a current loop is \(\tau=mB\sin\theta\). Here \(\theta\) is the angle between the field and the normal to the loop. The normal is at \(90^\circ\) to the plane, so the normal makes an angle of \(60^\circ\) with the field. That is, \(\theta=60^\circ\).
\( \begin{aligned} \tau &= mB\sin\theta \\ &= 0.5\times0.4\times\sin60^\circ \\ &= 0.2\times\frac{\sqrt{3}}{2} \\ &= 0.1\sqrt{3}\ \mathrm{N\,m} \end{aligned} \)
NCERT (2026-27):
Torque on a rectangular current loop in a uniform magnetic field – “We next consider the case when the plane of the loop, is not along the magnetic field, but makes an angle with it. We take the angle between the field and the normal to the coil to be angle θ (The previous case corresponds to θ = π/2). … The magnitude of the torque on the loop is,”
\( \tau=IAB\sin\theta \)\( \qquad (4.21) \)
Torque on a rectangular current loop in a uniform magnetic field – “… We define the magnetic moment of the current loop as,”
\( m=IA \)\( \qquad (4.22) \)