A solenoid of length 0.4 m having 500 turns of wire carries a current of 3 A. A coil of 10 turns and radius 0.01 m carries a current of 0.4 A. What will be the torque required to hold the coil in the middle of the solenoid with its axis perpendicular to the axis of the solenoid?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 5.9\times10^{-6}\,\mathrm{N\,m} \)
The coil is a magnetic dipole of moment \( m = NIA \) in the uniform field of the solenoid, \( B = \mu_0 n I \). Its axis is perpendicular to the field, so the torque \( \tau = mB\sin\theta \) is maximum, with \( \sin 90^\circ = 1 \). The holding torque must balance it.
\( \displaystyle n = \frac{500}{0.4\,\mathrm{m}} = 1250\,\mathrm{m^{-1}} \)
\( \displaystyle B = \mu_0 n I_{s} = 4\pi\times10^{-7} \times 1250 \times 3 = 4.71\times10^{-3}\,\mathrm{T} \)
\( \displaystyle m = N I A = 10 \times 0.4 \times \pi(0.01)^{2} = 1.26\times10^{-3}\,\mathrm{A\,m^{2}} \)
\( \displaystyle \tau = mB = 1.26\times10^{-3} \times 4.71\times10^{-3} \approx 5.9\times10^{-6}\,\mathrm{N\,m} \)