Magnetic Effects Of Current Question 448
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 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}} $ tesla making an angle of $ 30{}^\circ $ with the axis of the solenoid. The torque on the solenoid will be;
Options:
A) $ 3\times {{10}^{-2}}N-m $
B) $ 3\times {{10}^{-3}}N-m $
C) $ 1.5\times {{10}^{-3}}N-m $
D) $ 1.5\times {{10}^{-2}}N-m $
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Answer:
Correct Answer: D
Solution:
[d] Torque on the solenoid is given by $ \tau =MB,\sin ,\theta $ where $ \theta $ is the angle between the magnetic field and the axis of solenoid. $ M=niA $
$ \therefore ,\tau =niA,B,\sin ,30^{o} $ $ =2000\times 2\times 1.5\times {{10}^{-4}}\times 5\times {{10}^{-2}}\times \frac{1}{2} $ $ =1.5\times {{10}^{-2}}N-m $
BETA
