A circular loop of area \( 0.02\,\mathrm{m^{2}} \) carrying a current of 5 A is held with its plane normal to a magnetic field of 0.2 T. The torque acting on the loop is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 0
The torque on a current loop is \( \vec{\tau} = \vec{m}\times\vec{B} \), of magnitude
\( \displaystyle \tau = NIAB\sin\theta \)
where \( \theta \) is the angle between the magnetic moment \( \vec{m} \) (along the normal to the plane) and \( \vec{B} \). The plane of the loop is normal to the field, so its normal is parallel to \( \vec{B} \): \( \theta = 0 \).
\( \displaystyle \tau = 1\times5\times0.02\times0.2\times\sin0^\circ = 0 \)
The loop is in stable equilibrium and feels no torque.
Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.
NCERT (2026-27):
Torque on Current Loop, Magnetic Dipole – “… From Eq. (4.23), we see that the torque τ vanishes when m is either parallel or antiparallel to the magnetic field B. This indicates a state of equilibrium as there is no torque on the coil (this also applies to any object with a magnetic moment m).”