Homopolar Motor: Energy Input, Output, & Magnetic Field

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SUMMARY

The discussion focuses on the energy dynamics of a homopolar motor, specifically addressing the relationship between input and output power. When operating at 10 volts and 1000 amps, the output power will always be less than the input due to inherent losses such as friction and resistive heating. The equation for calculating resistive heating losses is given as P=I²R, where P represents lost power, I is the current, and R is the resistance. The conversation emphasizes that while ideal conditions suggest power output equals power input, real-world applications reveal unavoidable energy losses.

PREREQUISITES
  • Understanding of electrical power concepts, including voltage and current.
  • Familiarity with homopolar motors and generators.
  • Knowledge of resistive heating and its impact on electrical systems.
  • Basic principles of energy conservation and efficiency in electrical machines.
NEXT STEPS
  • Study the principles of homopolar motor design and operation.
  • Learn about the effects of friction and air resistance on motor efficiency.
  • Explore the equation P=I²R in detail and its applications in electrical engineering.
  • Investigate methods to minimize energy losses in electrical systems.
USEFUL FOR

Electrical engineers, students studying electromagnetism, and anyone interested in the efficiency and performance of homopolar motors and generators.

John Shillington
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I was curious in regards to a homopolar motor where the energy imparted into the rotor comes from if the energy input is say 10v at 1000 amps and there is 10v 1000amps output. Aswell if the effect of the magnetic field produces a different power on the rim of the disk vs the input from the central rod.
 
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The input power comes from whatever we are using to drive a current through the loop of the motor.

How are we getting an output from the motor? Using it to drive a generator? In any case the output power will be less than the input; if we are passing 1000 amps at 10 volts through the loop of the motor, then its usable output must be less than 10kW.

(And as a practical note, a sustained 1000A current is a big scary dangerous thing - I hope that you are doing a thought experiment here, not planning to actually build something).
 
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Thanks for your reply.

The question had more to do with if any of the power through the motor gets absorbed and turned into kinetic energy and if say there where 2 current and voltmeters on where the input is passed verse to where the output is taken from the rim of tge disk if the current and or voltage when being read on the output would be altered at due to said absorption or some other effect.

Im assuming from your answer the power on the side where current flows out of the disk stays just 10kw. Can you clarify that please?
 
John Shillington said:
Thanks for your reply.

The question had more to do with if any of the power through the motor gets absorbed and turned into kinetic energy and if say there where 2 current and voltmeters on where the input is passed verse to where the output is taken from the rim of the disk if the current and or voltage when being read on the output would be altered at due to said absorption or some other effect.

Im assuming from your answer the power on the side where current flows out of the disk stays just 10kw. Can you clarify that please?
Yeah, for an ideal motor/generator, the power output equals the power input. In the real world, there are always at least small losses for bearing friction and wire losses, etc., so the power output is always less than the power input.

For example, in my company's previous building, we used a motor/generator to convert 120Vrms @ 60Hz to 240Vrms @ 50Hz for testing the European versions of our products. The medium power motor/generator unit was about 4m^3 in volume, and made an impressive loud humming noise and was always pretty warm. Do you think that the fact that it made noise and was warm might detract from an ideal 100% efficiency? :smile:
 
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I understand that but specifically for a homopolar generator what power is consumed by the rotor? An equation would be appreciated.
 
John Shillington said:
what power is consumed by the rotor? An equation would be appreciated.
There will be friction in the bearings, resistive heating from the current through the wire loop, and (if we aren’t enclosing the whole thing in a vacuum chamber) air resistance. The only one that will have a straightforward equation will be the resistive heating: ##P=I^2R## is the lost power.
 

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