A circular coil of $30$ turns and radius $8.0\, cm$ carrying a current of $6.0\, A$ is suspended vertically in a uniform horizontal magnetic field of magnitude $1.0\, T$. The field lines make an angle of $60^o$ with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning.....$Nm$
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Magnitude of torque acting on the currentcarrying coil due to the magnetic field.
$\tau=n \operatorname{IAB} \sin \theta$
$=30 \times 6 \times \pi(0.08)^{2} \times 1 \times \sin 60^{\circ}$
$=30 \times 6 \times 3.14 \times 0.08 \times 0.08 \times \frac{\sqrt{3}}{2}=3.133\,\mathrm{Nm}$
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