Asked by Anonymous
A simple generator has a 780-loop square coil 21.0 cm on a side. How fast must it turn in a 0.650 T field to produce a 150 V peak output?
I got 6.71 rev/s but that is wrong
I got 6.71 rev/s but that is wrong
Answers
Answered by
drwls
In angular velocity (rad/s) units, call the coil rotation rate w. The peak induced voltage is
Vmax = N*B*A*w = 150 V
where N is the number of turns (780),
A = (0.21)^2 = 0.0441 m^2, and
B = 0.650 T
Thus
w = 6.71 rad/s
Your numerical answer is correct but the units are wrong.
You have to divide 6.71 by 2 pi to get the answer rev/s, which would be 1.07. Perhaps they want rpm. That would be 64 rpm
Vmax = N*B*A*w = 150 V
where N is the number of turns (780),
A = (0.21)^2 = 0.0441 m^2, and
B = 0.650 T
Thus
w = 6.71 rad/s
Your numerical answer is correct but the units are wrong.
You have to divide 6.71 by 2 pi to get the answer rev/s, which would be 1.07. Perhaps they want rpm. That would be 64 rpm
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