A circular coil with area A and N turns is free to rotate about a diameter that

Question

A circular coil with area A and N turns is free to rotate about a diameter that coincides with the x-axis. Current I is circulating in the coil. There is a uniform magnetic field B  in the positivey-direction. The magnitude of the magnetic field isB. (Figure 1) ( http://postimg.org/image/iekxvpxkh/ )

A

Calculate the magnitude of the torque   when the coil is oriented as shown in part (a) of the figure.

Part B

What is the direction of the torque  when the coil is oriented as shown in part (a) of the figure?

Part C

Calculate the value of the potential energy U, when the coil is oriented as shown in part (a) of the figure.

Explanation / Answer

A) Torque = N*I*A*B*sin(theta) (here theta is the angle between normal to the plane and magnetic filed)

= N*I*A*B*sin(90)

= N*I*A*B

B) - i^

C) potential energy = M*B*cos(theta) (where M = N*I*A)

= N*I*A*B*cos(90)

= 0

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