A circular coil of \( 50 \) turns and radius \( 8.0 \) cm carrying a current of \( 5.0 \) A is suspended vertically in a uniform horizontal magnetic field of magnitude \( 2.0 \) T. The field lines make an angle \( 30^\circ \) with the normal to the plane of coil. The magnitude of counter torque that must be applied to prevent the coil from turning is, approximately
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 5 N m
A current loop in a uniform field experiences a torque \( \tau = NIAB\sin\theta \), where \( \theta \) is the angle between the field and the normal to the plane of the coil. The counter torque must be equal in magnitude.
\( N = 50 \)
\( I = 5.0\,\mathrm{A} \)
\( r = 8.0\,\mathrm{cm} = 0.08\,\mathrm{m} \)
\( B = 2.0\,\mathrm{T} \)
\( \theta = 30^\circ \)
\( A = \pi r^{2} = 3.14\times(0.08)^{2} = 2.01\times10^{-2}\,\mathrm{m^{2}} \)
\( \tau = (50)(5.0)(2.01\times10^{-2})(2.0)\sin30^\circ \)
\( \tau = 10.05\times0.5 \approx 5\,\mathrm{N\,m} \)
Note: NTA picked this question from an NCERT (2026-27) exercise and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Exercises – …
