Problem: A magnetic field of magnitude 3 Tesla passes through a square loop with side .6000 meters, with the field at an angle of 3.4 degrees with a perpendicular to the plane of the loop. If the loop is suddenly turned , in .003000 seconds, to make an angle of 6.2 degrees, then what average voltage will be produced? If the voltage is connected to a resistance of 7.5 `Ohms, then how much average power is required?
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Solution: The area is .3600 m ^ 2, so at 3.4 degrees the flux will be ( .3600 m ^ 2)( 3 T)(cos( 3.4 deg)) = 1.078 T m ^ 2. At 6.2 deg, a similar calculation yields flux 1.073 T m ^ 2. The change in flux is thus .004900 T m ^ 2. Occuring in .003000 sec, the average rate is ( .004900 T m ^ 2) / ( .003000 sec) = 1.633 T m ^ 2 / sec 1.633 Volts. Through 7.5 `Ohms this will result in a current of .2177 amps, and a power of ( .2177 amps)( 1.633 Volts) = .3555 watts.
Generalized Response: The flux at angle `theta from perpendicular will be B * A cos(`theta). The flux change between angles `theta1 and `theta2 will therefore be `d `phi = B * A * ( cos(`theta2) - cos(`theta1) ).
If the flux change takes place in time interval `dt, then the voltage produced will be
V = `d `phi / `dt = B * A ( cos(`theta2) - cos(`theta1) ) / `dt.
The resulting current will be I = V / R, and the resulting power will be P = I V = V^2 / R; the above expression for V may be substituted.
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Figure description: