The axis of rotation of a square loop with side 5 meters is perpendicular a magnetic field of magnitude 7.1 tesla. What is the average magnitude of the voltage produced when the loop rotates at 5.4 Hz?
Since the field is perpendicular to the axis of the loop, at two instants in its cycle the field will be perpendicular to the loop.
5.4 Hz means 5.4 cycles per second. This implies a time of 1/ 5.4 seconds = .1851 sec for each cycle.
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 | 6.8 `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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