Q. The torque required to hold a small circular coil of 10 turns, $2 \times 10^{- 4}\ m^{2}$ area and carrying 0.5 A current in the middle of a long solenoid of $10^{3}\ turnsm^{- 1}$ carrying 3 A current. With its axis perpendicular to the axis of the solenoid is
A) $12\pi\ \times 10^{- 7}\ Nm\ $
B) $6\pi\ \times 10^{- 7}\ Nm$
C) $4\pi\ \times 10^{- 7}\ Nm$
D) $2\pi\ \times 10^{- 7}\ Nm$
Explanation
(a) Magnetic dipole moment of current loop is
$M = NiA = 10 \times 0.5\ \times 2 \times 10^{- 4} = 10^{- 3}\ Am^{2}$
Magnetic field the solenoid carrying current
$B = \ \mu_{0}ni = 4\pi \times 10^{- 7} \times 10^{- 2} \times 3$
$= 12\pi \times 10^{- 4}$ T
$\ \therefore\ \ $Torque, $\tau = MB\sin\theta$
$= 10^{- 3} \times 12\pi \times 10^{- 4} \times \sin{90{^\circ}}$
$= 12\ \pi\ \times 10^{- 7}$ Nm
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.