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Set 54 Problem number 12
A square loop with side 3.4 meters is connected so that any voltage produced in
the loop is in series with a resistance of .3 `Ohms.
A magnetic field of magnitude 6.1 Tesla passes through the loop, with the field
initially perpendicular to the plane of the loop.
- If the loop rotates at 6.2 Hz what is the average power required to turn the loop,
assuming that power generation is 95% efficient? Ignore the effects of
self-induction in the loop, but explain what self-induction is and what effect it might
have on your answer.
If the field is perpendicular to the axis of rotation, then at two points in
each cycle the field will be perpendicular to the plane of the loop.
6.2 Hz means 6.2 cycles per second. This implies a time of 1/ 6.2 seconds = .1612
sec for each cycle.
- One cycle corresponds to 360 degrees; a 90 degree turn corresponds to a time of
( .1612 sec)/4 = .0403 sec.
- The area of the loop is ( 3.4 m) ^ 2 = 11.56 m ^ 2, so the flux when the plane of
the loop is prependicular to the field is ( 6.1 T)( 11.56 m ^ 2) = 70.51 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 70.51 T m ^ 2 to zero is 70.51 T m ^ 2.
- This flux change occurs in .0403 seconds, so the average rate is ( 70.51 T m ^
2)/( .0403 sec) = 3498 T m ^ 2 / sec = 3498 Volts.
- Since this is the average during each quarter of a cycle, it is equal to the
overall average.
This voltage will result in a current through the .3 `Ohms of
- current = ( 3498 Volts)/( .3 `Ohms) = 11660 Amps.
This current through 3498 volts delivers
- power in circuit = ( 11660 Amps)( 3498 Volts) = 4.078E+07 watts.
This is the power required if the efficiency is 100%.
The 95% efficiency requires more power. The power required is
- power required at 95% efficiency = 4.078E+07 watts / 95% = 4.292E+07 watts.
Self-induction comes about because the current in the loop is changing. A
current in the loop results in an additional magnetic field in the vicinity of the loop,
including inside the loop. As the current changes we therefore have a change in the
magnetic field inside the loop, which results in a change in the voltage around the loop.
- Lenz' Law tells us that this voltage must oppose any change in the current.
- Since the voltage changes with the rotation of the loop, the current must also
change.
- The voltage from self-induction must therefore oppose the voltage created in
response to the main field, which results in less voltage than that computed.
- The lower the resistance of the circuit the more current tends to flow, and the
more 'opposing' voltage is therefore created by self-induction within the loop.
- For the small currents in the present problem, self-induction is a negligible
effect.
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
- voltage = 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
- power in circuit: 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
- power required at efficiency = ( 16 (f * B * A) ^2 / R ) / e = 16 f^2 B^2
A^2 / ( e * R) .
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