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  30 \degree with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?

Answers (1)
S Sayak

The magnitude of torque experienced by a current-carrying coil in a magnetic field is given by

\tau =nBIAsin\theta

 where n = number of turns, I is the current in the coil, A is the area of the coil and \theta is the angle between the magnetic field and the vector normal to the plane of the coil.

In the given question n = 20, B=0.8 T, A=0.1\times0.1=0.01 m2, I=12 A, \theta=30o

\tau =20\times 0.8\times 12\times 0.01\times sin30^{o}

=0.96 Nm

The coil therefore experiences a torque of magnitude 0.96 Nm. 

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