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A closely wound solenoid of 2000 turns and area of cross-section $1.5\times10^{-4}\ m^2$ 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 $5\times10^{-2}$ tesla making an angle of $30^\circ$ with the axis of the solenoid. The torque on the solenoid will be
A
$3\times10^{-3}$ N m
B
$1.5\times10^{-3}$ N m
C
$1.5\times10^{-2}$ N m
D
$3\times10^{-2}$ N m
Detailed Solution
Magnetic moment of the solenoid: $M = NIA = 2000\times2.0\times1.5\times10^{-4} = 0.6\ A\,m^2$
The magnetic moment is along the axis of the solenoid, which makes $30^\circ$ with the field.
Torque: $\tau = MB\sin\theta$
$\tau = 0.6\times5\times10^{-2}\times\sin30^\circ = 0.6\times5\times10^{-2}\times\frac{1}{2}$
$\tau = 1.5\times10^{-2}$ N m
The magnetic moment is along the axis of the solenoid, which makes $30^\circ$ with the field.
Torque: $\tau = MB\sin\theta$
$\tau = 0.6\times5\times10^{-2}\times\sin30^\circ = 0.6\times5\times10^{-2}\times\frac{1}{2}$
$\tau = 1.5\times10^{-2}$ N m