Anemometers (pictured) are used to measure wind speeds. Wind striking the cups c
ID: 1451669 • 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 0.455 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 warnings.The idea is to attach a coil of 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 of internal resistance but need 14.5 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 40.0 m/s (89.5 mph) is considered the threshold of dangerous wind speed, calculate the number of turns in the coil wire needed to light the bulb.
Would this high wind warning feature be useful?
1. No, the coil would be too large to fit on the anemometer
2. Yes, this device is practical from a design standpoint.
Explanation / Answer
d = 0.455 m
r = 0.132m
n = ?
R = 234 ohm
P = 14.5 W
B = 0.5*10^(-4) T
v = 40m/s
Now, angular velocity of the anemometer, w = v/d = 40/0.455 = 87.912 rad/s
Time taken for one quarter rotation = pi/2*w = 0.01785s
Flux through the coil at the time when B is perpendicular to the coil = B*A = 0.5*10^(-4)*pi*0.132^2
Flux rate in quarter rotation = BA/0.01785 = 1.5317*10^-4
emf = n*flux rate = n*1.5317*10^-4
but P = VI = 14.5 W = V^2/R
V = sqroot(14.5*234) = n*1.5317*10^-4
n = 380272.98 turns
The coil would be too large to fit anemometer