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A rectangular coil of length 0.12 m and width 0.1 m having 50 turns of wire is suspended vertically in a uniform magnetic field of strength 0.2 Weber/$m^2$. The coil carries a current of 2 A. If the plane of the coil is inclined at an angle of $30^\circ$ with the direction of the field, the torque required to keep the coil in stable equilibrium will be:
A
0.12 Nm
B
0.15 Nm
C
0.20 Nm
D
0.24 Nm
Detailed Solution
The plane makes $30^\circ$ with $\vec B$, so the area vector (normal) makes $\theta = 60^\circ$ with $\vec B$.
$\vec\tau = \vec M\times\vec B$, $|\tau| = NIAB\sin\theta$
$|\tau| = 50\times 2\times(0.12\times0.1)\times 0.2\times\sin 60^\circ \approx 0.20$ Nm
$\vec\tau = \vec M\times\vec B$, $|\tau| = NIAB\sin\theta$
$|\tau| = 50\times 2\times(0.12\times0.1)\times 0.2\times\sin 60^\circ \approx 0.20$ Nm