Question

A solenoid of length 0.4 m and having 500 turns of wire carries a current of 3.0 A. A thin coil having 10 turns of wire and of radius 0.01 m carries a current of 0.4 A. Calculate the torque required to hold the coil in the middle of the solenoid with its axis perpendicular to the axis of the solenoid.

• A. $6\times10^6\,N$
• B. $5.94\times10^{-6}\,Nm$
• C. $9.54\times10^{6}\,Nm$
• D. $5.9\times10^{-8}\,Nm$

Answer 1

Answer: (B)

Magnetic field at the middle of the solenoid, B = $m_0 nI = m_0 \frac{N}{L} I$
= $4 \pi \times 10^{-7} \times \frac{500}{0.4} \times 3 = 4.713 \times 10^{-3}$T.
Magnetic dipole moment of the coil, M = NIA = NI$\pi r^2$
= $10 \times 0.4 \times 3.142 \times (0.01)^2 = 1.26 \times 10^{-3} \, Am^2$
Torque acting on the coil $\tau $ = MB sin $\theta$
= $1.26 \times 10^{-3} \times 4.713 \times 10^{-3} = 5.94 \times 10^{-6} $Nm