How Electromotive Force is Produced by Electric Generator?

AI Thread Summary
Electromotive force (EMF) is generated by the Lorentz force, which causes electrons to move in a specific direction due to the influence of a magnetic field. This movement results in electric current, as charged particles are displaced to areas of higher potential energy. The concentration of energy in an electric field is key to understanding EMF, which can be calculated by dividing energy by charge. The discussion emphasizes that the Lorentz force is not a potential force, and the correct expression for EMF in Faraday's Law involves both electric and magnetic fields. Understanding these principles is crucial for grasping how electric generators operate.
mo0nfang
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at 5:42 - What is Electromotive Force? Why does it make electrons move (make electric current) in one direction?
 
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That force is called the Lorentz force, which google.
 
Electromagnetic Induction @1:40 <--- Why protons group themselves at the top and electrons at the bottom?
 
mo0nfang said:
Electromagnetic Induction @1:40 <--- Why protons group themselves at the top and electrons at the bottom?
In that video, the electrons (-) are moving to the top. It's because of the direction of the magnetic field. It could have been either way.
 
mo0nfang said:
Electromagnetic Induction @1:40 <--- Why protons group themselves at the top and electrons at the bottom?
Again, it is the Lorentz force, of which you are well aware.
 
mo0nfang said:
What is Electromotive Force?
How concentrated energy is in an electric field. Lorentz force moves charged particles to positions of higher potential energy.
Divide the energy by the amount of charge and you get electromotive force.
 
To argue with a "potential" is misleading in this case since the Lorentz force is not a potential force but reads
$$\vec{F}=q \left (\vec{E}+\frac{\vec{v}}{c} \times \vec{B} \right),$$
and thus the correct electromotive force in Faraday's Law of induction in integral form is
$$\mathcal{E}(t)=\int_{C} \mathrm{d} \vec{r} \cdot \left [\vec{E}(t,\vec{x})+\frac{\vec{v}(t,\vec{x})}{c} \times \vec{B}(t,\vec{x}) \right ]=-\frac{1}{c} \dot{\Phi}_B,$$
where $$\vec{v}(t,\vec{x})$$ is the velocity field of the path ##C##.
 
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