A magnet of magnetic moment \( 50\hat{\imath}\,\mathrm{A\,m^2} \) is placed along the X-axis in a magnetic field \( \vec{B} = (0.5\hat{\imath} + 3.0\hat{\jmath})\,\mathrm{T} \). The torque acting on it is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 150\hat{k}\,\mathrm{N\,m} \)
The torque on a magnetic dipole in a uniform field is the cross product of the magnetic moment and the field:
\( \displaystyle \vec{\tau} = \vec{m} \times \vec{B} = 50\hat{\imath} \times (0.5\hat{\imath} + 3.0\hat{\jmath}) \)
\( \displaystyle \vec{\tau} = 25\,(\hat{\imath}\times\hat{\imath}) + 150\,(\hat{\imath}\times\hat{\jmath}) \)
Since \( \hat{\imath}\times\hat{\imath} = 0 \) and \( \hat{\imath}\times\hat{\jmath} = \hat{k} \),
\( \displaystyle \vec{\tau} = 150\hat{k}\,\mathrm{N\,m} \)
Only the field component perpendicular to the moment (the \( 3.0\hat{\jmath} \) part) produces torque.
Note: A similar question was also asked in CUET (UG) 2026 on 19-05-2026 (Slot 1).