Answer to Question 2

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A coil, which is part of an electric motor armature, has dimensions 4.5 6.0 cm with 500 turns.  It is placed in a magnetic field of flux density 140 mT. A steady current of 2.0 A flows through it. Calculate:
(a) The maximum torque;
(b) The power if the motor is spinning at 3000 revolutions per minute.

(c) The voltage of the source, if the motor is 80 % efficient.

(a) t = 140 10-3 T 27 10-4 m2 2.0 A 500 = 2.625 10-3 N m = 0.38 N m (2 s.f.)

 

 

(b)  f = 3000 rpm 60 s = 50 Hz.

 

      P = 0.378 N m 2 p 50 Hz  = 118.75 W = 120 W (2 s.f.)

 

 

(c)  Total power = 118.75 W 0.80 = 148.44 W

 

       V = 148.44 W 2.0 A = 74.22 V = 74 V (2 s.f.)