A closely wound solenoid of 1000 turns and area of cross section \( 1.2\,\mathrm{cm^{2}} \) carries a current of 5 A. It is suspended horizontally in uniform magnetic field of \( 5\times10^{-2}\,\mathrm{T} \) making an angle of 30° with the axis of solenoid. The torque acting on the solenoid is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 1.5\times10^{-2}\,\mathrm{N\,m} \)
A current-carrying solenoid behaves as a magnetic dipole of moment \( m = NIA \) along its axis. In a uniform field B the torque is \( \tau = mB\sin\theta \), where \( \theta \) is the angle between the axis and the field.
\( A = 1.2\,\mathrm{cm^{2}} = 1.2\times10^{-4}\,\mathrm{m^{2}} \)
\( m = NIA = 1000\times5\,\mathrm{A}\times1.2\times10^{-4}\,\mathrm{m^{2}} = 0.6\,\mathrm{A\,m^{2}} \)
\( \tau = mB\sin30^\circ = 0.6\times5\times10^{-2}\times\frac{1}{2} \)
\( \tau = 1.5\times10^{-2}\,\mathrm{N\,m} \)
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?”