Question

Question 4.9: A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of 30º with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?

Class 12 - Physics - Moving Charges And Magnetism Moving Charges And Magnetism Page-169

Answer 1

torque experienced by the coil carrying current in the given magnetic field.
$\tau=BINAsin\theta$
here, $\tau$ is the torque, B is the magnetic field, I is the current passing through coil , N is the number of turns and A is cross section area.

$\tau=0.8\times12\times20\times(10\times10^{-2})^2sin30\\\\or,\tau=0.96Nm$

Answer 2

Hey!!

Please refer the attachment !

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