A circular armature coil is used in a motor with a uniform magnetic field of 3.3

Question

A circular armature coil is used in a motor with a uniform magnetic field of 3.30 T. The coil has a diameter 75 cm, a resistance of 35 ohm and generates a back emf of 93 V_RMS when running a full speed under no load at the angular frequency omega_A At what angular frequency, omega_B relative to omega_A does the coil be need to be rotated to generate a back emf of 73 V_RMS? omega_B= omega _A If this motor is operating on 120 V_RMS now with a load (e.g running a fan) and at the angular frequency omega_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).

So, for (A): wB=73/93wA = 0.785wA

On part B just take the 120 Vrms and subtract the A emf value at full speed under no load i.e 93V and then divide by the resistance.
for (B): I=(120-93)/35 = 0.77 A

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