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Question

Physics Question on torque

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

6×106 N6\times10^6\,N

B

5.94×10−6 Nm5.94\times10^{-6}\,Nm

C

9.54×106 Nm9.54\times10^{6}\,Nm

D

5.9×10−8 Nm5.9\times10^{-8}\,Nm

Answer

5.94×10−6 Nm5.94\times10^{-6}\,Nm

Explanation

Solution

Magnetic field at the middle of the solenoid, B = m0nI=m0NLIm_0 nI = m_0 \frac{N}{L} I = 4π×10−7×5000.4×3=4.713×10−34 \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 = NIπr2\pi r^2 = 10×0.4×3.142×(0.01)2=1.26×10−3 Am210 \times 0.4 \times 3.142 \times (0.01)^2 = 1.26 \times 10^{-3} \, Am^2 Torque acting on the coil τ\tau = MB sin θ\theta = 1.26×10−3×4.713×10−3=5.94×10−61.26 \times 10^{-3} \times 4.713 \times 10^{-3} = 5.94 \times 10^{-6} Nm