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 $