Problem: A magnetic field of magnitude 9 Tesla passes through a square loop with side 1.4 meters, with the field perpendicular to the plane of the loop. The loop is suddenly turned 90 degrees, so that its plane is parallel to the loop. If it is turned in .05500 seconds, what is the magnitude of the average rate of change of flux?
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Solution: The area of the loop is ( 1.4 m) ^ 2 = 1.96 m ^ 2, so the flux is ( 9 T)( 1.96 m ^ 2) = 17.64 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 17.64 T m ^ 2 to zero is 17.64 T m ^ 2. This occurs in .05500 seconds, so the average rate has magnitude ( 17.64 T m ^ 2)/( .05500 sec) = 320.7 T m ^ 2 / sec = 320.7 Volts.
Generalized Response: If a loop of area A is oriented perpendicular to a magnetic field B, the magnetic flux will be `phi = B * A (note: the Greek letter phi is usually used to denote magnetic flux). If the loop is suddenly turned so that it is parallel to the field, the flux will become 0 (the loop doesn't intercept any of the field when it is perpendicular to the field; also `theta = 90 deg in this orientation and `phi = B * A * cos(`theta) = B * A * 0 = 0). So the change of flux will be 0 - `phi = -`phi.
A flux change of -`phi in time interval `dt gives an average rate of flux change `d`phi / `dt = -`phi / `dt.
The rate of flux change is the voltage that results from the flux change. Any change of flux through an area results in a voltage around the boundary of that area. In particular the flux change computed here creates a voltage in the loop. This is how most commercial electrical power is generated.
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Figure description: