Moving Charges and Magnetism - Result Question 66
67. 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:
(a) $3 \times 10^{-2} N-m$
(b) $3 \times 10^{-3} N-m$
(c) $1.5 \times 10^{-3} N-m$
(d) $1.5 \times 10^{-2} N-m$
[2010]
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Answer:
Correct Answer: 67. (d)
Solution:
- (d) Torque on the solenoid is given by $\tau=M B \sin \theta$
where $\theta$ is the angle between the magnetic field and the axis of solenoid.
$M=$ niA
$\therefore \tau=n i A B \sin 30^{\circ}$
$=2000 \times 2 \times 1.5 \times 10^{-4} \times 5 \times 10^{-2} \times \frac{1}{2}$
$=1.5 \times 10^{-2} N-m$
