How should I think about the emf?

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1. Jan 30, 2015

ghostfolk

I was under the impression that an emf is the work done per charge by any source that is not electrostatic. However, there are such things such as emfs produced by varying magnetic fields. Being that magnetic fields never do work, what is an appropriate way of thinking of an emf?

2. Jan 30, 2015

HallsofIvy

Staff Emeritus
Where would you get the idea that "emf" ("electro-motive force) was "work"? The "emf" is the force an electric field would apply, at each point, to a unit charge.

3. Jan 30, 2015

cabraham

Actually "emf" is not a force at all. The term "electromotive force" was coined before the researchers fully understood it. The force which acts on charge carriers in the presence of E & B fields is Lorentz force.
$F=q(E + vXB)$

Claude

4. Jan 30, 2015

ghostfolk

Well isn't the emf the work per unit charge? From the emf equation looks a lot like the work equation $\mathscr E=\oint f_s \cdot d\ell$. Also, some reading online has made it appear that way.

5. Jan 31, 2015

Staff: Mentor

Actually, this is not correct. EMF is voltage, not electric field intensity. It is related to the amount of work you would have to do to move a test charge from point A to point B in an electric field. Another name for EMF is electrical potential.

Chet

6. Jan 31, 2015

ghostfolk

Now we call the electric potential emf when the source is from something other than an electrostatic field right?

7. Jan 31, 2015

Staff: Mentor

That's not my understanding. See Wiki.

Chet

8. Jan 31, 2015

ghostfolk

I was told that the electric potential and the emf are different things

9. Jan 31, 2015

Staff: Mentor

Electricity and Magnetism by Francis Weston Sears:

Any device in which a reversible transformation between electrical energy and some other form of energy can take place, is called a seat of electromotive force. The magnitude of the electromotive force of a seat may be defined quantitatively as the energy converted from electrical to nonelectrical form or vice versa (exclusive of energy converted irreversibly to heat) per unit of charge passing across a section through the seat. More briefly, electromotive force may be defined as the work per unit charge.

Since electromotive force is work per unit charge, the unit of emf in the mks system is one joule per coulomb. This is the same as the unit of potential and has been abbreviated to one volt. Hence emf's can be expressed in volts. It should be noted, however, that, although emf and potential are expressible in the same unit, they relate to different concepts.

..., the emf of the common automobile storage battery is about 6 volts or 6 joules /coulomb. This means the for every coulomb pasing across a section through the batters (or across any section of a circuit to which the battery is connected) 6 joules of chemical energy are converted to electrical form if the battery is discharging, or 6 joules of chemical energy are developed at the expense of electrical energy if the battery is being charged.

Chet

10. Jan 31, 2015

ghostfolk

Speaking of batteries, do you think you can explain to me why that in an ideal battery the emf and the potential difference are equal?

11. Jan 31, 2015

Staff: Mentor

I can't really explain. This seems to be a terminology issue rather than a physics issue.

Chet

12. Jan 31, 2015

Staff: Mentor

Huh. So that explains 'voltage drop' then. If each quantity of charge can only do so much total work in the circuit, then the work done, say by heating a resistor, reduces the amount of work available for the rest of the circuit.

I guess that would also mean that the total work done by an ideal battery is the same regardless of the circuit.

12v battery with 12 amp-hours capacity in a circuit with 12 Ohms resistance: 1 amp of current x 12 volts = 12 watts sustained for 12 hours. So 12 watts is 12 jouls/sec, which comes out to be 518,400 joules over 12 hours.

Same battery attached to a circuit with 24 Ohms of resistance: 0.5 amps x 12 volts = 6 watts, sustained for 24 hours. 6 watts is 6 joules/sec which comes out to be 518,400 joules over 24 hours.

This makes sense. If EMF is the work per charge, then moving fewer charges over time means you use less potential energy over that same time.

13. Feb 1, 2015

Staff: Mentor

Sounds like whoever said this is regarding the terminal voltage as the cell output's "potential difference" ---- this being different from the emf which includes the drop due to the cell's internal resistance (in the electrical model).

14. Feb 1, 2015

mic*

The emf produced by a battery (which is equal to the ideal voltage), ironically, is only valid for open circuits. Call it 'e'.

As soon as any current flows there is a voltage drop given across the terminals. The voltage measured at the terminals is given by V = e - Ir (where r is the internal resistance of the battery and I is current).

15. Feb 1, 2015

mic*

ghostfolk,

I'm not sure where you are at with electromagnetism so please excuse anything not at the right level,

Magnetic fields do no work because the direction of the force is always perpendicular to the motion of the charged particle. By definition work is cos (x)*F*s (where F is a force, x is the angle between the force and the displacement, and s is displacement)

In the case where F is due to a mag field, cos(x) = cos(90) = 0 so 0 work is done.

This does not mean that there is no effect. The result is a circular path with a constant speed for a charged particle moving in the constant field.

To understand the basic relationship check out right hand rule stuff.

Maxwell's equations encompass everything needed describe the realtionships quantitatively.

You may be interested in reading about JJ Thompson's charge to mass ratio experiment.

16. Feb 1, 2015

ghostfolk

The only thing I find confusing is that often the emf is defined as the work done per charge. Yet, we can say that a magnetic field produces an emf. So to me it sounds weird that we can say that magnetic field does work on the charges when it isn't. I know the work is being done by another source. Do you think you could explain this?

17. Feb 1, 2015

Staff: Mentor

A changing magnetic field induces an electric field in a conductor. It is this electric field that does the work.

The Maxwell–Faraday equation is a generalisation of Faraday's law that states that a time-varying magnetic field is always accompanied by a spatially-varying, non-conservative electric field, and vice-versa.

18. Feb 1, 2015

mic*

As Drakkith said... a CHANGING magnetic field induces an electric field.

You can then place a test charge (1C) in this electric field and calculate the potential between the test charge and the source. The potential is a value that indicates how many joules of work the field would do on the 1C test charge to move it from the source to its current location. Or you could say the work an outside force would have to do to MOVE the particle from it's location to the source of the ELECTRIC field.

As said earlier by someone clever, it's a poor name because it's not really a force. It's the work done to overcome potential energy associated with electric force. Sometimes an analogy with gravitational potential energy is used but it can lead to confusion about keeping the +/- signs right. Anyway that is a bit of a digression.

Another reason to avoid the use of 'emf' is because it can be used as an abbreviation for Electro Magnetic Field.

19. Feb 3, 2015

zoki85

I use abbreviation "EM field" to avoid confusion.

20. Feb 3, 2015

Khashishi

Electromotive force is an unnecessary and redundant term which mainly creates confusion in students. Nevertheless, it does have a definition:
$\mathcal{E} = \int_\mathbf{a}^\mathbf{b} \mathbf{E} \cdot d\mathbf{l}$
This definition differs from ghostfolk's definition in that it includes both electrostatic and electromagnetic contributions to the E field.

Note that this is not the same as the difference in electric potential ($\phi$) (except in electrostatics).
This is because the electric field depends on both the electric potential and the time-varying magnetic vector potential.
$\mathbf{E} = -\nabla \phi - \frac{\partial \mathbf{A}}{\partial t}$
In the case of electrostatics, $\frac{\partial \mathbf{A}}{\partial t}=0$ so
$\mathbf{E} = -\nabla \phi$
and
$\mathcal{E} = \phi(\mathbf{a})-\phi(\mathbf{b})$
.
But the term electromotive force is almost universally used in electromagnetics, so this equation is never useful. As far as I know, electromotive force and voltage mean the same thing, but electromotive force is used in electromagnetics and voltage is used in electrostatics for some reason. It's terrible.