Calculating Torque, Power, and Work in a Magnetic Motor - Solving Part D

[andrew410]

The rotor in a certain electric motor is a flat rectangular coil with 86 turns of wire and dimensions 2.60 cm by 3.80 cm. The rotor rotates in a uniform magnetic field of 0.950 T. When the plane of the rotor is perpendicular to the direction of the magnetic field, it carries a current of 7.3 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 3700 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?

So...I got a, b, and c but can't seem to get the right solution to part d.
Any help would be great thx! :)

 

[clive]

If you have the work done in one full revolution (c) then the power is
P=\frac{W}{T}
where
T=\frac{3700}{60} s.

 

[andrew410]

I did that before...but the answer isn't right.
For part c, I got zero joules of work. So, 0/(3700/600) = 0 watts.
But, the answer isn't right...
What am I doing wrong?

 

[clive]

Because I'm trying to figure out this problem please tell me first if these answers are correct:

a) BNIlD
b) BNIlD\omega, \omega=2\pi f
c) 4BNIlD
d) \frac{4BNIlD}{T}.

 

[andrew410]

a) I got I*B*A*(# of turns of wire), where A is the area.
b) I got (part a)*60*2*pi
c) 0
d) ?

 

[clive]

Not your answers...the answers from the book you found this problem!

 

[andrew410]

its an even problem so the answer isn't found in the book...anyways thanks for the help, I got it now. :)