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
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
To solve this problem, you need to use Faraday's law of electromagnetic induction, which states that the induced emf (electromotive force) can be calculated by the equation:
emf = N * A * B * ω
Where:
emf is the induced electromotive force (in volts),
N is the number of turns in the coil,
A is the area of the coil (in square meters),
B is the magnetic field strength (in teslas),
and ω is the angular velocity (in radians per second).
Given:
N = 780 loops,
A = (21.0 cm)^2 = (0.21 m)^2 = 0.0441 m^2,
B = 0.650 T,
and emf = 150 V.
To find ω, rearrange the equation:
ω = emf / (N * A * B)
Substituting the given values:
ω = 150 V / (780 * 0.0441 m^2 * 0.650 T)
Calculating this expression:
ω ≈ 2.42 rad/s
Therefore, the coil must turn at approximately 2.42 radians per second (rad/s) to produce a peak output of 150 volts.