A simple has 200- loop square coil, 8.0cm on a side. How must it turn in a 0.40T field to produce a 12 .0v peak output
To find the number of turns, we can rearrange the equation for the peak output voltage of a coil in a magnetic field:
V = NABωsin(ωt)
Where:
V is the peak output voltage (12.0 V)
N is the number of turns in the coil (unknown)
A is the area of the coil (8.0 cm on a side = 0.08 m on a side, so A = (0.08 m)^2 = 0.0064 m^2)
B is the magnetic field strength (0.40 T)
ω is the angular frequency (unknown)
t is the time (unknown)
We can simplify the equation by assuming that ωt = 0, which means the peak output voltage occurs at that moment (ωt = 0 is equivalent to finding the maximum of the sine function). This simplification gives us:
V = NABωsin(ωt)
12.0 V = N * (0.0064 m^2) * (0.40 T) * 1
12.0 V = N * (0.0064 m^2) * (0.40 T)
12.0 V / (0.0064 m^2 * 0.40 T) = N
1875 turns = N
Therefore, the coil must have 1875 turns to produce a 12.0 V peak output in a 0.40 T magnetic field.