A current loop is placed in a uniform magnetic field. The angle \(\theta\) between its magnetic moment and the field is changed slowly from \(0^\circ\) to \(180^\circ\). On the graph of torque against \(\theta\), the largest torque is \(4\times10^{-3}\ \mathrm{N\,m}\). The torque is \(2\times10^{-3}\ \mathrm{N\,m}\) when \(\theta\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(30^\circ\) and \(150^\circ\)
The torque on the loop is \( \vec{\tau}=\vec{m}\times\vec{B} \), so its size is \( mB\sin\theta \). The graph has the shape of \( \sin\theta \). It is zero at \( 0^\circ \), largest at \( 90^\circ \) and zero again at \( 180^\circ \). So the largest torque, \( 4\times10^{-3}\ \mathrm{N\,m} \), is equal to \( mB \).
\( \begin{aligned} \sin\theta &= \frac{\tau}{mB} \\ &= \frac{2\times10^{-3}}{4\times10^{-3}} \\ &= \frac{1}{2} \end{aligned} \)
Between \( 0^\circ \) and \( 180^\circ \), \( \sin\theta=1/2 \) at \( 30^\circ \) and again at \( 150^\circ \). The graph is the same on both sides of \( 90^\circ \), so each value below the peak comes twice.
NCERT (2026-27):
Torque on a rectangular current loop in a uniform magnetic field – “… The magnitude of the torque on the loop is,”
\( \tau=IAB\sin\theta \)\( \qquad (4.21) \)