Electric Motor and Magnetic Field

[MJC8719]

The rotor in a certain electric motor is a flat, rectangular coil with 90 turns of wire and dimensions 2.50 cm by 4.00 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, it carries a current of 9.1 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 3600 rev/min.

(a) Find the maximum torque (max) acting on the rotor.
Nm
(b) Find the peak power output (max) of the motor.
W
(c) Determine the amount of work (W) performed by the magnetic field on the rotor in every full revolution.
J
(d) What is the average power (avg) of the motor?

I cannot figure out the answer to part D.

I know for part A that the torque would equal = (90)(9.1x10-3A)(0.025mx0.04m)(0.800T) = 6.553e-4

Then for part B, it equals the answer to A x 2pi(60) = 0.247W

Then part C is given by 4(9.1 x 10-3A)(90)(0.025mx0.04m) = 0.00262J

For part D, I thought it would equal part C/(3600/60) ie you know the total work in Joules for a full revolution and now want to divide it by the number of revolutions per second (3600/60)...when I do this on my calculator, I get an answer of 4.3666e-5

The online web program is telling me that I am off by orders of magnitude...but I cannot figure out what I am doing wrong.

 

[ice109]

if c is the amount of work done on the rotor and the amount of work the rotor does per rev and it does 3600rev/min why is it not C * 3600/60, if that is correct then your answer is off by two orders of magnitude

 

[Ithryndil]

Then part C is given by 4(9.1 x 10-3A)(90)(0.025mx0.04m) = 0.00262J

Can you explain why this part is:

4NIAB

Where
N = number of loops
I = Current
A = Area of Loop
B = Magnetic Field. Thanks.

 

[Ithryndil]

I am not sure where the 4 comes from in part C. Everything else, however, makes sense.