A circular coil having 19 turns and a radius of R = 19.2cm carries a current of

Question

A circular coil having 19 turns and a radius of R = 19.2cm carries a current of 3.30A. It is placed in a uniform magnetic field of 62.0mT. The initial angle between the plane of the coil and the magnetic field B is 58.0degrees.

What is the magnitude of the torque acting on the coil? A circular coil having 19 turns and a radius of R = 19.2cm carries a current of 3.30A. It is placed in a uniform magnetic field of 62.0mT. The initial angle between the plane of the coil and the magnetic field B is 58.0degrees.

What is the magnitude of the torque acting on the coil?

Explanation / Answer

The formula to find the torque is

= NBIAsin

= (19)(62 X 10-3)(3.3)()(.192)2(sin 58)

= .382 Nm

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