A solenoid of 1500 turns and area of cross-section 2 x 10−4 m2, carrying a current of 4 A is suspended freely to rotate in the horizontal plane of a magnetic field 5 x 10−2 T. If the axis of the solenoid is making an angle of 30° with the magnetic field, the torque acting on it is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 3 x 10−2 Nm
A current-carrying solenoid behaves like a bar magnet of magnetic moment m = NIA, and a uniform field exerts a torque \( \tau = mB\sin\theta \) on it, where \( \theta \) is the angle between the axis and the field.
\( \displaystyle m = NIA = 1500 \times 4 \times 2\times10^{-4} \)
\( \displaystyle m = 1.2\,\mathrm{A\,m^{2}} \)
\( \displaystyle \tau = mB\sin\theta = 1.2 \times 5\times10^{-2} \times \sin 30^\circ \)
\( \displaystyle \tau = 3\times10^{-2}\,\mathrm{N\,m} \)
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 – “… 5.6 A closely wound solenoid of 2000 turns and area of cross-section 1.6 × 10–4 m2, carrying a current of 4.0 A, is suspended through its centre allowing it to turn in a horizontal plane. (a) What is the magnetic moment associated with the solenoid? (b) What is the force and torque on the solenoid if a uniform horizontal magnetic field of 7.5 × 10–2 T is set up at an angle of 30° with the axis of the solenoid?”