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Q. 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 center and perpendicular to its length, allowing it to turn in a horizontal plane in a uniform magnetic field $5 \times 10^{-2} T$, making an angle of $30^{\circ}$ with the axis of the solenoid. The torque on the solenoid will be

AIIMSAIIMS 2014

Solution:

$\tau =N i B A \sin \theta$
$=2000 \times 2 \times 5 \times 10^{-2} \times 1.5 \times 10^{-4} \times \sin 30^{\circ}$
$=2000 \times 50 \times 10^{-6} \times \frac{1}{2}$
$=1.5 \times 10^{-2} N - m$