Problem: A magnetic field of magnitude 3.1 Tesla passes through a square loop with side 1.7 meters, with the field perpendicular to the plane of the loop. The loop rotates at 9.5 Hz. What is the average magnitude of the voltage produced?
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Solution: Note that 9.5 Hz means 9.5 cycles per second. This implies a time of 1/ 9.5 seconds = .1052 sec for each cycle. One cycle corresponds to 360 degrees; a 90 degree turn corresponds to a time of ( .1052 sec)/4 = .02630 sec. The area of the loop is ( 1.7 m) ^ 2 = 2.89 m ^ 2, so the flux when the plane of the loop is prependicular to the field is ( 3.1 T)( 2.89 m ^ 2) = 8.959 T m ^ 2. After being turned 90 degrees, the loop will be parallel to the field, and will intercept none of the field. The flux will then be zero. The magnitude of the flux change from 8.959 T m ^ 2 to zero is 8.959 T m ^ 2. This occurs in .02630 seconds, so the average rate is ( 8.959 T m ^ 2)/( .02630 sec) = 680.9 T m ^ 2 / sec = 680.9 Volts. Since this is the average during each quarter of a cycle, it is equal to the overall average.
Generalized Response: If a loop rotates at frequency f in a magnetic field, then the time required for a complete rotation is 1/f (think of 10 cycles per second; the time per cycle is obviously 1/10 sec). In a quarter of a rotation the loop will rotate through 90 degrees. This will require 1/4 of the time of a cycle. Since a complete rotation takes time 1/f, the time to complete 1/4 of a cycle is `dt = 1/4 (1/f) = 1 / (4f).
If the loop rotates 90 degrees, starting from a position where the field is perpendicular to the loop, the magnitude of the flux change will be |`d `phi | = B * A. Since this takes place in time `dt = 1 / (4f), the rate of flux change will be
average rate of flux change = | `d`phi | / `dt = B * A / ( 1 / (4f) ) = 4f * B * A.
This is in fact the average rate, since the magnitude of the flux change will be B * A through every subsequent 90 degree rotation.
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