The rotor in a certain electric motor is a flat, rectangular coil with 83 turns of wire and dimensions 2.45 cm by 4.40 cm. The rotor rotates in a uniform magnetic field of 0.800 T. When the plane of the rotor is perpendicular to the direction of the magnetic field, the rotor carries a current of 11.1 mA. In this orientation, the magnetic moment of the rotor is directed opposite the magnetic field. The rotor then turns through one-half revolution. This process is repeated to cause the rotor to turn steadily at an angular speed of 3.36x10³ rev/min.
(a) Find the maximum torque acting on the rotor.
(b) Find the peak power output of the motor.
(c) Determine the amount of work performed by the magnetic field on the rotor in every full revolution.
(d) What is the average power of the motor?

(a) The maximum torque acting on the rotor is  7.40 x 10⁻⁴ N.m

(b) The peak power output of the motor is 0.300W

(c) The amount of work performed by the magnetic field on the rotor in every full revolution is 2.95x10⁻³ J

(d) The average power of the motor is 0.177 W

Explanation:

(a)

        = 7.40 x 10⁻⁴ N.m

(b)

= (7.40x10⁻⁴)(3750)

         = 0.300 W

(c)

In one half revolution, the work is

    = 2NIAB = 2(7.40x10⁻⁴) = 1.48x10⁻³ J

In one full revolution, W = 2(1.48x10⁻³)

                                        = 2.95x10⁻³ J

(d)

Time for one revolution is  Δt = 60/3600 = 1/60s

= W / Δt

       = 2.95x10⁻³ / (1/60)

       = 0.177 W

an electric circuit

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