Why is EMF (ElectroMotive Force) equal to a voltage?

  • #1
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Hello.

Let's say a current i(t) flows through an inductor of solenoid type. Time-varying i(t) generates a magnetic field B(t) inside the magnetic core of the inductor. As B(t) is time-varying, EMF, which is the line-integral of the induced electric field E(t) along a coil (which carries i(t)) around the core, is made (In fact, I'm not convinced yet that E(t) is really to be called "induced" one, as there is no clue that E(t) is "caused" by B(t) in the Maxwell equation. E(t) and B(t) here are may be better called "dual". But for convience, I'll keep call it "induced").

My question is why EMF has to be voltage applied on the inductor? If EMF-field E(t) is the only electric field present on the inductor coil, then EMF is deserved to be called a voltage. But...Is EMF-field only electric field existing on the coil? The current density is expressed as J = σE' where σ is the conductivity of the conductor. We know that voltage on the inductor and current flowing on it are not in-phase (there is actuall 90 degree phase differece between the volage and current on the inductor), so E' is not necessarilly equal to EMF-fiedl E(t), I think. (If E' = E, then voltage and current are in-phase, but it is not true in the inductor).

I thought EMF-field E is so much larger than E' in J = σE in the inductor so that E' can be ignored in the voltage calculation. But if this is true, the current should flows the same direction of E' but this is not always true.

Thanks for reading this post I hope there would be some comments on this.
 

Answers and Replies

  • #2
jim hardy
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I'm too simple minded to follow all that .

Voltage across an inductor that's carrying current
can be thought of as the sum: (current X resistance) + (rate of change of current X inductance).
Former is ohmic, latter electromagnetic.

My question is why EMF has to be voltage applied on the inductor?
Who said it does?
 
  • #3
tech99
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Hello.

Let's say a current i(t) flows through an inductor of solenoid type. Time-varying i(t) generates a magnetic field B(t) inside the magnetic core of the inductor. As B(t) is time-varying, EMF, which is the line-integral of the induced electric field E(t) along a coil (which carries i(t)) around the core, is made (In fact, I'm not convinced yet that E(t) is really to be called "induced" one, as there is no clue that E(t) is "caused" by B(t) in the Maxwell equation. E(t) and B(t) here are may be better called "dual". But for convience, I'll keep call it "induced").

My question is why EMF has to be voltage applied on the inductor? If EMF-field E(t) is the only electric field present on the inductor coil, then EMF is deserved to be called a voltage. But...Is EMF-field only electric field existing on the coil? The current density is expressed as J = σE' where σ is the conductivity of the conductor. We know that voltage on the inductor and current flowing on it are not in-phase (there is actuall 90 degree phase differece between the volage and current on the inductor), so E' is not necessarilly equal to EMF-fiedl E(t), I think. (If E' = E, then voltage and current are in-phase, but it is not true in the inductor).

I thought EMF-field E is so much larger than E' in J = σE in the inductor so that E' can be ignored in the voltage calculation. But if this is true, the current should flows the same direction of E' but this is not always true.

Thanks for reading this post I hope there would be some comments on this.
An inductor is analogous to a mass, and the voltage is equivalent to an accelerating force.
According to Newton, every force (or action) has equal and opposite reaction. These two forces are the battery voltage and the opposing EMF of the inductor, and they are always equal.
 
  • #4
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Hello.

I've summarized an inductor in my own material which is now uploaded here.

On the inductor coil, there is not only EMF-field but also an electrostatic field. In principle, the line integral of a sum of these field along the coil gives a potential difference or voltage on the inductor. However, the line integral of the electrostatic field along the coil is very close to zero as one-round line integral of this field is zero. As a result, EMF-field is a practically only contribution to the voltage on the inductor.

Please give me any comments if you don't like my physics interpretation.
 

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  • #5
jim hardy
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That particular author appears to define EMF as magnetically induced
and Est as produced by old fashioned ohmic drop
be aware that is his choice , it's his book after all

upload_2017-3-1_9-42-28.png


His Eemf arises from the QVcrossB term in Lorentz force
and his Est term from the QE term in Lorentz.


[PLAIN said:
https://www.britannica.com/science/Lorentz-force][/PLAIN] [Broken]
Lorentz force, the force exerted on a charged particle q moving with velocity v through an electric E and magnetic field B. The entire electromagnetic force F on the charged particle is called the Lorentz force (after the Dutch physicist Hendrik A. Lorentz) and is given by F = qE + qv × B.

The first term is contributed by the electric field. The second term is the magnetic force and has a direction perpendicular to both the velocity and the magnetic field.

Work is F X D and that's voltage. F has two terms in it

Author's choice.
That's my take on it.
 
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  • #6
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I agree of course, EMF it is an induced voltage in a coil measured in V.
It could be split in a effective or main EMF-in a transformer magnetic core-and a leakage
one [in vicinity air or oil or else] and it is a scalar.
E=ρ*J is a vector represented an electric field intensity [or electric field strength]
The module is measured in V/m.
If E=gradient(V) then V=integral[E*dl] and then along a conductor:
E=ρ*J then V=integral( ρ*J*dl]=ρ*length*I/sq=R*I [sq=cross section area-J=I/sq].
That means the electric field intensity is constant along a wire and this voltage it is not negligible [always]:smile:.
 
  • #7
jim hardy
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That means the electric field intensity is constant along a wire and this voltage it is not negligible [always]:smile:.

I was thinking about this during my morning walk.
Thank you Babadaq for your as always erudite contribution.

As a practical guy i came at it from an electrical machinery background.
In an electric motor, QVcrossB is what transforms electrical to mechanical energy. It usually dominates.
Ohmic drop becomes significant when the motor is heavily loaded enough that its armature resistance contributes noticeably to voltage across the winding.

Thank you for presenting the math.

old jim
 
  • #8
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Thank you for your kind words Jim.
 
  • #9
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Thanks for giving many comments on my question.

Especially, I appreciate Jim's consistent comments on me:)

The last comments you and Badadag made are something I need to take some time to consume.

I'll give anything on this post If I come up with some idea regarding this.
 
  • #10
jim hardy
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Thanks for the feedback. We like to feel we might have helped a little bit.

old jim
 

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