[chris (OP)]

Phillip would like to know:

Would it be possible to create, generate, and control a static electromagnetic field?

What does everyone think?

[PmbPhy]

Phillip would like to know:

Would it be possible to create, generate, and control a static electromagnetic field?

What does everyone think?

In principle, yes. All one has to do is distribute charges and currents in a manner to generate the electric field and magnetic fields desired. However there are certain configurations of fields which cannot be generated by any distribution of charge and currents.

[Kryptid]

That's more or less what you are doing every time you set yourself up to get a static shock from a doorknob.

[evan_au]

Your average AA battery from the corner store generates a static electromagnetic field while it is sitting on the shelf.
There is no current, so the magnetic component is zero. And the electric field is about 1.5 Volts.

[PmbPhy]

Your average AA battery from the corner store generates a static electromagnetic field while it is sitting on the shelf.
There is no current, so the magnetic component is zero. And the electric field is about 1.5 Volts.

I assume that you're speaking of the voltage of the battery. That's an EMF and as such there is no electric field associated with it. The units of the electric field is not volts, its volts per meter.

[evan_au]

That's an EMF and as such there is no electric field associated with it. The units of the electric field is not volts, its volts per meter.

Could you clarify that for me, please, Pete?
The length of an AA battery is around 4cm. That makes it around 1.5V/4cm = 40V/m; is it now an electric field?

[PmbPhy]

Could you clarify that for me, please, Pete?
The length of an AA battery is around 4cm. That makes it around 1.5V/4cm = 40V/m; is it now an electric field?

No. Its only when the voltage is a difference in potential that there is an electric field associated with it. An electromotive force has the same units as a difference in potential but has a different physical meaning.

Suppose I took a charged pith ball and walked it around a closed circuit. If there were forces to overcome in the process then I'd have to apply a force to do it. The force I applied would not be conservative in general and would not necessarily be done by an electric field But work would be done in the process. For example: The work might  be done by an magnetic field, which exerts a force on a moving charge. In fact that's how a dynamo works. See: https://en.wikipedia.org/wiki/Electromotive_force#Electromotive_force_and_voltage_difference

This shouldn't be confused with the fact that a circuit which generates an EMF can create an electric field.

Note also that the if there was an electric field inside the battery then it has to be created by a non-zero charge density which is zero inside a battery.

[chris (OP)]

Sorry Pete; I didn't follow that.

[chiralSPO]

I believe that what Pete is saying is that a battery does not necessarily have an electric field about it. An electrochemical (galvanic) battery is not a capacitor, which has an EMF that is determined purely by a macroscopic electric field. Batteries instead have an EMF that is determined by microscopic electric fields (atomic scale), and so when completely isolated from a circuit, a battery is unlikely to have a significant dipole moment (uneven distribution of charge within the battery)

Imagine a simple electrochemical cell, containing a piece zinc metal at one end of a tube, and a piece of copper at the other end. The copper end of the battery contains an electrolyte solution composed of copper chloride dissolved in water (Cu²⁺ and Cl^– ions), while the zinc side is dissolved sodium chloride (Na⁺ and Cl^– ions), and a thin membrane separates the two electrolytes, allowing chloride ions to cross, but not copper. When the two electrodes are not connected to anything, no electrochemical reaction occurs, and the tube shouldn't have any significant electric field about it. (you could measure the amplitude and direction of the field anywhere around the cell, and would probably not observe any--if you touch a meter to the two ends of the cell it would detect the EMF, but the potential difference between points in space each a micron away from the electrode is probably pretty close to zero).

Once a circuit is completed, electrons flow from the zinc electrode to the copper electrode, which dumps the electrons into the empty orbitals of the Cu²⁺ ions, reducing them to copper metal, and chloride ions move to the zinc end of the cell. So negative charge (moving either as electron or chloride) has moved all around the circuit, but the cell itself doesn't have any change in charge distribution on a macroscopic scale, only changes in the composition of matter within.

[PmbPhy]

Sorry Pete; I didn't follow that.

Sorry. It's a difficult subject to explain. I did my best. I wrote to the author of a well known EM text whom I keep in contact with and asked him to help me explain it better. His texts are famous for explaining things well. Meanwhile chiralSPO did better than I.

chiralSPO  - Great explanation. Please note that a capacitor does not have an EMF, it has an actual difference in potential across it due to the separation of charge across the capacitor plates. When applying Kirchhoff's around a circuit a capacitor is treated as a difference in potential, not an EMF.

[chiralSPO]

chiralSPO  - Great explanation. Please note that a capacitor does not have an EMF, it has an actual difference in potential across it due to the separation of charge across the capacitor plates. When applying Kirchhoff's around a circuit a capacitor is treated as a difference in potential, not an EMF.

Thanks! Good to know!

[chris (OP)]

@chiralSPO Excellent explanation; now I understand what @PmbPhy is saying too.

[vhfpmr]

I believe that what Pete is saying is that a battery does not necessarily have an electric field about it. An electrochemical (galvanic) battery is not a capacitor, which has an EMF that is determined purely by a macroscopic electric field. Batteries instead have an EMF that is determined by microscopic electric fields (atomic scale), and so when completely isolated from a circuit, a battery is unlikely to have a significant dipole moment (uneven distribution of charge within the battery)

Imagine a simple electrochemical cell, containing a piece zinc metal at one end of a tube, and a piece of copper at the other end. The copper end of the battery contains an electrolyte solution composed of copper chloride dissolved in water (Cu²⁺ and Cl^– ions), while the zinc side is dissolved sodium chloride (Na⁺ and Cl^– ions), and a thin membrane separates the two electrolytes, allowing chloride ions to cross, but not copper. When the two electrodes are not connected to anything, no electrochemical reaction occurs, and the tube shouldn't have any significant electric field about it. (you could measure the amplitude and direction of the field anywhere around the cell, and would probably not observe any--if you touch a meter to the two ends of the cell it would detect the EMF, but the potential difference between points in space each a micron away from the electrode is probably pretty close to zero).

Once a circuit is completed, electrons flow from the zinc electrode to the copper electrode, which dumps the electrons into the empty orbitals of the Cu²⁺ ions, reducing them to copper metal, and chloride ions move to the zinc end of the cell. So negative charge (moving either as electron or chloride) has moved all around the circuit, but the cell itself doesn't have any change in charge distribution on a macroscopic scale, only changes in the composition of matter within.

You've lost me.

What distinction are you drawing between EMF and PD?

If there's a voltage between the terminals of a cell, then there's an electric field in the space surrounding the cell, which is what I assume evan_au was getting at. The electric field looks similar to the magnetic field surrounding a bar magnet:

https://qph.ec.quoracdn.net/main-qimg-dd8c242f454fa5fc88a4efb47f5dfb55

Or are you arguing that a voltage only exists when a current is flowing? In which case how would the current ever start flowing in the first place?

[PmbPhy]

You've lost me.

What distinction are you drawing between EMF and PD?

Well, first off it is not I who is drawing the distinction but physics. To learn what an Electromotive Force is please see: https://en.wikipedia.org/wiki/Electromotive_force

You appear to be confusing two physical concepts because they have the same units. Let me give you an analogy to help you understand. There is a position 4-vector defined as R = (x, y, z, ct) where (x, y, z) is a point in space and t is the time that an event occurs at that point. t has the units of time. ct has the units of length just like x, y and z but they have quite a different physical meaning.

An electromotive force or EMF is defined by something called a line integral whereas a difference in potential is defined by something called a gradient. In words an EMF is the work done per unit charge by exerting a force on a charged particle around a circuit. An illustrative example, one in which there is no electric field whatsoever present, is a dynamo. This is a simple device in which a conductor is moved through a magnetic field. The magnetic field exerts a force on the charges in the conductor. This will allow the "device" to create a steady flow in a resistor so long as work is being done on the conductor to keep it moving.

There's a difference in potential between two points in a static electric field but such a field cannot produce a steady current in a resistor whereas an EMF can. A battery is a chemical based device which can produce an EMF. A battery is called a seat of EMF. However the EMF is not created by charges but by chemical actions.

If there's a voltage between the terminals of a cell, then there's an electric field in the space surrounding the cell, which is what I assume evan_au was getting at.

That is incorrect. Again you're confusing an EMF with a difference in potential. All you have to do in order to see the difference place charge on something like a rubber rod by rubbing it with silk or wool. It will then become charged. Then suspend it by a thread. When you've done that place a battery next to the charged rod. If there was any charge on it then it would move. You'll find that it won't.

In the case of creating and EMF using a dynamo there is an electric field inside the wire but there is no electric field around the wire. If there is current flow inside the wire there will be an electric field inside. However, again, no electric field outside. The electric field inside a conductor is proportional to the product of the current and the conductivity of the conductor.

[PmbPhy]

I believe that what Pete is saying is that a battery does not necessarily have an electric field about it. An electrochemical (galvanic) battery is not a capacitor, which has an EMF that is determined purely by a macroscopic electric field.

A capacitor does not have an EMF. It has a difference in potential. Do you recall the difference between an EMF and difference in potential? Just because they have the same units does not mean they have the same physical meaning.

[evan_au]

Then suspend (the rubber rod) by a thread. When you've done that place a battery next to the charged rod. If there was any charge on it then it would move. You'll find that it won't.

Doesn't this just reflect the fact that the insulating rod is charged to thousands of volts, while your typical battery only has an EMF of 1.5V?

Put a few thousand batteries end-to-end, and you might get some motion!

[PmbPhy]

Doesn't this just reflect the fact that the insulating rod is charged to thousands of volts, while your typical battery only has an EMF of 1.5V?

No. There is nothing in the experiment I suggested that responds to that. Also, I don't know what you mean by "charged to thousands of volts". There is no unique voltage associated with a given charge.

The point here is that if there is an electric field around a battery then it should cause the charged rod to be deflected from equilibrium. The large charge on the rod assures it.

[PmbPhy]

I need to post a retraction. Evan may very well be correct and as such I may very well be wrong. I asked an expert in this field. Here is his response

There's a chemical force that is responsible for the emf of the battery.  This chemical force will drive positive charge to the +
>terminal, and/or minus charge to the - terminal.  This accumulation of charge will (of course) set up an electric field (pointing, incidentally, in the opposite direction to the chemical force, so the net force on charges inside the battery is zero, if it's not connected to a circuit, and hence no current flows).  So the answer to your question is YES: there WILL be an electric field inside the battery.  If you now connect the battery up to a circuit, it is this electric field that drives charge around the loop (only indirectly the chemical force, which after all is localized within the battery, and cannot directly push charge through the distant resistor).

I hope this retraction serves to remind people that we can all be wrong. It's not that big of a deal. What is a big deal is the willingness to be open to being wrong and the ability to admit to ones mistakes.

Please don't take this to mean that whenever someone disagrees with us/you that they can rightly say "Hey. You can be wrong!" That's not what I meant. If I was 100% certain of this subject I never would have asked an expert. But I know when I'm 100% correct on a point.

[PmbPhy]

If I seemed not 100% certain about my retraction its because I'm still studying it. I'm willing to say that in a voltaic cell there's an E-field. But in general there need not be. You might have elves moving charges around and that doesn't require an E-field but work can be done in doing so.

[Walterhurley56]

The electrons can be in a circular orbit due to a magnetic field which has to be continuously supplied.