If \( \vec{M} \) is the magnetic moment and \( \vec{B} \) is the magnetic field, then the torque is given by
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \vec{M}\times\vec{B} \)
The torque on a magnetic dipole in a uniform field is the cross product of the moment and the field; it has magnitude \( MB\sin\theta \) and acts to turn the dipole towards the field.
\( \displaystyle \vec{\tau} = \vec{M}\times\vec{B} \)
Option (1) \( \vec{M}\cdot\vec{B} \) — Incorrect. The dot product \( \vec{M}\cdot\vec{B} = MB\cos\theta \) is a scalar; it gives the potential energy (with a minus sign), not the torque.
Option (2) \( \displaystyle\frac{|\vec{M}|}{|\vec{B}|} \) — Incorrect. The torque is the product of \( M \) and \( B \) (with \( \sin\theta \)), not their ratio.
Option (3) \( \vec{M}\times\vec{B} \) — Correct. The torque is a vector, equal to \( \vec{M}\times\vec{B} \).
Option (4) \( \displaystyle\frac{|\vec{B}|}{|\vec{M}|} \) — Incorrect. The torque is a product, not the ratio of \( B \) to \( M \).
Note: A similar question was also asked in CUET (UG) 2025 on 22-05-2025 (Slot 1).
NCERT (2026-27):
Torque on Current Loop, Magnetic Dipole – “… The torques in Eqs. (4.20) and (4.21) can be expressed as vector product of the magnetic moment of the coil and the magnetic field.”
Torque on Current Loop, Magnetic Dipole – “… Then as the angle between m and B is \( \theta \), Eqs. (4.20) and (4.21) can be expressed by one expression
\( \displaystyle \boldsymbol{\tau} = \mathbf{m} \times \mathbf{B} \qquad (4.23) \)”