Question

A closely wound solenoid of 2000 turns and area of cross-section $1.5\times {10}^{-4}{m}^2$ carries a current of 2.0 A. It is suspended through its centre and perpendicular to its length, allowing it to turn in a horizontal plane in a uniform magnetic field $5\times {10}^{-2}$ making an angle of ${30}^o$ with the axis of the solenoid. The torque on the solenoid will be

• A. $3\times {10}^{-3}N-m$
• B. $1.5\times {10}^{-3}N-m$
• C. $1.5\times {10}^{-2}N-m$
• D. $3\times {10}^{-2}N-m$

Answer 1

Answer: (C)

Given, $N=2000,A=1.5\times {10}^{-4}{m}^2$ $i=2.0A,B=5\times {10}^{-2}T,$ and $\theta ={30}^o$ Torque, $\tau =NiBA\sin \theta$ $=2000\times 2\times 5\times {10}^2\times 1.5\times {10}^{-4}\times \sin {30}^o$ $=2000\times 50\times {10}^{-6}\times \frac{1}{2}$ $=1.5\times {10}^{-2}Nm$