A circular armature coil is used in a motor with a uniform magnetic field of 3.9

Question

A circular armature coil is used in a motor with a uniform magnetic field of 3.90 T. The coil has a diameter 75 cm, a resistance of 40 , and generates a back emf of 101 VRMS when running a full speed under no load at the angular frequency A.

(A) At what angular frequency, B relative to A does the coil be need to be rotated to generate a back emf of 81 VRMS?
B =   ? A

(B) If this motor is operating on 120 VRMS now with a load (e.g running a fan) and at the angular frequency A, how much current is flowing throught the motor?

? A

Explanation / Answer

On part A you'll just want to take the rotated emf value (smaller Vrms) and divide it by the given back emf of the coil at full speed under no load at A (larger Vrms).

On part B just take the 120 Vrms and subtract the A emf value at full speed under no load and then divide by the resistance.

So, for (A): 81/101 = 0.80

wB = 0.80wA

And for (B): (120-101)/40 = 0.475 A

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