Anemometers (pictured) are used to measure wind speeds. Wind striking the cups c
ID: 1310289 • Letter: A
Question
Anemometers (pictured) are used to measure wind speeds. Wind striking the cups causes the assembly to rotate; a gauge measures the frequency of this rotation and translates the data into wind speed. The length of each arm, as measured from cup to cup, is 0.445 m. The friction in the bearings is negligible, so the tangential speed of the cups matches the wind speed fairly closely. Your company makes anemometers and plans to install light bulbs on them for high wind warnmgs.The idea is to attach a coil ot wire of radius 0.132 m to one of the arms and use the EMF induced by the magnetic field of the Earth to light the bulb. The bulbs possess 234 Q of internal resistance but need 10.5 W of rms power dissipation to be visible at night. (The resistance of the wire is negligible.) Although company management wants this new feature, you have doubts about Its practicality. Assuming the magnetic field of the Earth Is 0.500 * 10"4 T and is oriented roughly horizontal with the Earth, and that 42.2 m/s (94.5 mph) is considered the threshold of dangerous wind speed, calculate the number of turns in the coll wire needed to light the bulb. Would this high wind warning feature be useful?Explanation / Answer
Here,
P = V * I and I = V/R. So, P = V^2 * R. We have P = 10.5 W AND R = 234. So, V= 49.6 V
Now, we know that EMF (volt) created in the magnet E = N A B w sine(theta). Let us assume sine theta to be 90 degrees and therefore sine(theta) = 1. Required E = 49.6 V, N is what we must calculate and A stands for the Area of the coil, B is the magnetic field strength that is give and w is the angualr velocity. Let us find A and w.
A = pi*r^2 = .132^2 * pi = 0.0547 m^2
w = v/R = 42.2/0.445m = 94.83 per second.
So N = E/ (A* B * w) = 49.6/ (94.83* .5*10^-4 * 0.0547) = 191239.9 = 191240 coils. (has to be integer)
the coil would probably not fit the meter.