Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: chris on 28/05/2017 23:02:32
-
Phillip would like to know:
Would it be possible to create, generate, and control a static electromagnetic field?
What does everyone think?
-
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.
-
That's more or less what you are doing every time you set yourself up to get a static shock from a doorknob.
-
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.
-
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.
-
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?
-
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.
-
Sorry Pete; I didn't follow that.
-
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.
-
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 - 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!
-
@chiralSPO Excellent explanation; now I understand what @PmbPhy is saying too. :D
-
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?
-
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.
-
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.
-
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!
-
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.
-
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.
-
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.
-
The electrons can be in a circular orbit due to a magnetic field which has to be continuously supplied.
-
I forgot to mention that a rotating magnet can generate a static electric field.
-
I'm still quite lost, I'm not sure which bits PmbPhy is retracting and which not (if any).
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.
This seems to me a bit like arguing that there is a difference, in kind, between the metre used to measure the position of a point in empty space, and the metre used to measure the height of a reservoir that can generate power in a hydro plant. Distance is distance.
FWIW, and digressing a bit from electrostatic fields, my usage of the two terms as a radio engineer is that EMF is conventionally the source voltage of a Thevenin source, and the PD is the terminal voltage. My point is that these are both voltages, with no difference in kind, and the terms merely distinguish between two different locations in the equivalent circuit. (Both would be the same magnitude of course when no current flows, as with an electrostatic field.)
Re: a charged particle, if I charged a speck of dust using a van der graaf, and then placed it between two parallel plates connected to a battery, would it not be attracted to one of the plates?
-
This seems to me a bit like arguing that there is a difference, in kind, between the metre used to measure the position of a point in empty space, and the metre used to measure the height of a reservoir that can generate power in a hydro plant. Distance is distance.
Your analogy is flawed. In almost no difference in the examples you just gave. To specify a point in space you need a point of reference and a displacement vector from that reference point to the point you wish to specify. The reference point is arbitrary. The magnitude of the displacement vector is the distance. The displacement vector is called the position vector. The height of a reservoir is the magnitude of the distance from the reference point to the height of the reservoir. The reference point is still arbitrary. Its usually chosen to be the point of lowest place that the water can go to. However you can also dig a well in which the water can fall further and then use the bottom of the well as the reference point.
While distance is distance, EMF is physically different than difference in potential. That's because they are defined differently. In fact its impossible to have a difference in electric potential in the absence of an electric field but its quite possible to generate an EMF without an electric field. In fact a dynamo works by moving a conductor through a magnetic field which does work on the charged particles in the conductor which can then be used to generate energy in a circuit. But there's no need for an electric field in this case and therefore no electric potential.
FWIW, and digressing a bit from electrostatic fields, my usage of the two terms as a radio engineer
I don't see what being a radio engineer has to do with anything.
The definition of difference in potential and EMF as defined in physics are as follows:
An electric potential is the amount of work needed to move a unit positive charge from a reference point to a specific point inside the field without producing any acceleration.
An electromagnetic force (EMF) is defined as that which can produce energy to generate an electric current or in another context a line integral containing a magnetic term.
There is, in fact, no requirement for an electric field or a battery to even be present to generate an EMF.
See the definition of EMF in any one of the classic texts in EM such as Jackson. Other great texts on the subject are Zangwill or Grtiffiths. All too easily people confuse the terms merely because they have the same units.
-
I posted the following link above: https://en.wikipedia.org/wiki/Electromotive_force#Electromotive_force_and_voltage_difference
I should point out that there are misleading comments in that page. E.g. it says
The electric charge that has been separated creates an electric potential difference that can be measured with a voltmeter between the terminals of the device.
In the region surrounding a charged object there is an electric field and as such differences in potentials. But you cannot measure those differences with a volt meter.
-
In fact a dynamo works by moving a conductor through a magnetic field which does work on the charged particles in the conductor which can then be used to generate energy in a circuit. But there's no need for an electric field in this case and therefore no electric potential.
If I move a wire through a magnetic field there will be a voltage difference generated between the ends. If I were to place a speck of charged dust nearby, would it not be attracted to one end of the wire because there is now an electric field surrounding the wire?
I don't see what being a radio engineer has to do with anything.
The relevance is that radio engineers all understand the terms EMF and PD in the context of a Thevenin source, as defined above.
An electromagnetic force (EMF) is defined as that which can produce energy to generate an electric current
Ok, so how is that different from PD? Using my definitions of EMF and PD, I could construct a Thevenin source with an infinite resistance, such that the EMF can deliver energy but the PD cannot, would that not then satisfy your distinction between the two terms?
-
If I move a wire through a magnetic field there will be a voltage difference generated between the ends. If I were to place a speck of charged dust nearby, would it not be attracted to one end of the wire because there is now an electric field surrounding the wire
That applies only to a static situation and is indeed a difference in potential. But its not am EMF because (1) it does not conform to the definition of EMF and (2) it cannot supply
I don't see what being a radio engineer has to do with anything.
The relevance is that radio engineers all understand the terms EMF and PD in the context of a Thevenin source, as defined above.
[/quote]
Irrelevant since your responses suggest you are confusing an EMF with a legitimate PD.
I will not continue with this since there are numerous authoritative sources available online with clealy explain the differences. If you refuse to look them up then it seems to imply that you don't want to learn but would rather argue. E.g. The line integral defining a motional EMF is the integral of vxB.
You're confusing the induced electric field for an open circuit with an EMF?
Since there's ample information online which I already pointed out and you chose to ignore. i.e. Jackson, Zangwill and Grtiffiths I see no reason to repeat myself.
-
How about this scenario:
- Take a 1.5V battery, which produces a 1.5V EMF
- Connect it to a parallel-plate capacitor, which now has a 1.5V Potential Difference between the plates
- There is a static electromagnetic field between the two plates, with the field direction being pretty much a straight line between the plates
- There is a slight fringing field extending beyond the edge of the plates, with the field direction bulging out and curving back in to the other plate
- You can manipulate the electromagnetic field of the capacitor in various ways, eg:
- You can increase or decrease the separation of the plates; early experiments with electrostatics mounted the parallel plates on a screw thread so you could increase or decrease the separation
- You could change the area of the plates; analogue radios had a mechanical method of changing the amount of overlapping area of the plates
- You could insert materials with different e values
What happens to the voltage depends whether you have the battery still connected or not; the battery will tend to maintain a constant 1.5V across its terminals, and so a current will flow to keep the voltage across the capacitor plates constant while it is connected.
However, if you disconnect the battery before manipulating the capacitor, the charge on the plates is constant, and the voltage on the plates will vary as you change the physical structure of the capacitor.
You can image the electric field strength by using commercial mathematical simulation packages.
- Or you can use a very high impedance voltmeter, with higher impedance that air leakage
- Or, in principle, you could use a device that counts the individual electrons flowing from the more positive terminal to the more negative terminal. This could use totally insulated probes and draw no current from the EMF or PD.
With an old-fashioned "pile" battery, consisting of alternating copper and zinc disks of equal size, the internal EMF may cancel the straight-line electric field between the top and bottom plate.
But if you make the top and bottom plates bigger - in fact, like a capacitor, you will have a static electric field between these plates, plus a fringing field.