Problem: A magnetic field of magnitude 4.6 Tesla passes through a square loop with side 2.2 meters, with the field perpendicular to the plane of the loop. The loop rotates at 5.2 Hz. The loop is connected so that the voltage is across a resistance of 3 'Ohms. What is the average power required to turn the loop, assuming that power generation is 53% efficient?
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Solution: Note that 5.2 Hz means 5.2 cycles per second. This implies a time of 1/ 5.2 seconds = .1923 sec for each cycle. One cycle corresponds to 360 degrees; a 90 degree turn corresponds to a time of ( .1923 sec)/4 = .04807 sec. The area of the loop is ( 2.2 m) ^ 2 = 4.84 m ^ 2, so the flux when the plane of the loop is prependicular to the field is ( 4.6 T)( 4.84 m ^ 2) = 22.26 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 22.26 T m ^ 2 to zero is 22.26 T m ^ 2. This occurs in .04807 seconds, so the average rate is ( 22.26 T m ^ 2)/( .04807 sec) = 926.1 T m ^ 2 / sec = 926.1 Volts. Since this is the average during each quarter of a cycle, it is equal to the overall average. Now this voltage will result in a current through the 3 `Ohms of ( 926.1 Volts)/( 3 `Ohms) = 308.7 Amps. This current through 926.1 Volts delivers ( 308.7 Amps)( 926.1 Volts) = 285800 watts. This is the power required if the efficiency is 100%; the 53% efficiency requires more power; the requirement is for 285800 watts / 53% = 540000 watts.
Generalized Response: Rotating at frequency f, a loop of area A in a magnetic field B perpendicular to the axis of the loop will generate an average voltage of V = `d`phi / `dt = 4f * B * A. Connected to a circuit with net resistance R, this will result in a current I = V / R = 4f * B * A / R and power requirement P = I V = 4f * B * A * (4f * B * A / R) = 16 (f * B * A) ^ 2 / R. If the efficiency of the generator is e, then the power required is P = 16 * e (f * B * A) ^2 / R.
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