# Computing the force of the induced electric field on the moving charges that create the non-steady current & thus the same electric field. [closed]

Previously, I asked a question on whether charges suffer force from the magnetic field they create while constituting current.

I got responses from Timaeus as well as Sebastian Riese; the former told me use Jefimenko's equations while later confirmed that charges are indeed affected by the electric & magnetic field they create:

so the statement:charges are forced by the electric field that got created due to the non-steady current those charges produce true? – user36790

Yup, but they are also affected by the magnetic field, one example are plasma pinches, where the current carrying plasma tube is compressed due to the magnetic field generated by the current itself. – Sebastian Riese

I'm having some problem in getting how to compute the force on the charges from the time-varying magnetic field as well as electric field that those charges create.

In order to find the force exerted on a charge by those magnetic forces as well as the electric field; we have to use Jefimenko's equation.

But exactly what $r$ in the equation should I use to find the force on the charge that created the field?

IMO, $r$ is zero since it is the distance between the charge that created the field from itself is zero. That means the force from the induced electric field is indeterminate.

But that can't be.

So, what $r$ should I use to compute the force on the charge that created that field?

One statement from Riese :

[...] interaction of a single particle with the electro-magnetic field it generates – this is indeed non-trivial as it leads to inconsistencies of electrodynamics

I think Riese is telling about the same thing that when the force is being computed for a single charge, there can be indeterminate results.

However, it is well known that charges are indeed affected by the changing magnetic flux which they create at non-steady state, that is self-induction.

How can I show mathematically that at non-steady state, the electric field & the magnetic field produced by the moving-electrons which create the field force those same charges without getting blown up?

Also, at steady state, there is a magnetic field associated with the time-independent current; why can't it exert force on the charges constituting the time-independent current?

Could anyone please explain these to me ?

• You haven't paid any attention to Jefimenko: "neither Maxwell's equations nor their solutions indicate an existence of causal links between electric and magnetic fields. Therefore, we must conclude that an electromagnetic field is a dual entity always having an electric and a magnetic component simultaneously created by their common sources: time-variable electric charges and currents." The field is the electromagnetic field. So stop talking about electric fields and magnetic fields. When you move past one electron such that it is "current", you do not create a magnetic field round it. – John Duffield Dec 22 '15 at 22:03
• @JohnDuffield It's a red herring to complain about the mere words involved. A given electromagnetic Faraday tensor field can be broken into electric and magnetic parts for any frame, and there isn't anything wrong about that. There is an error when the OP thinks it is the magnetic field in the frame of the thin wire that causes self induction. It's the electric field in the frame of the thin wire that causes the EMF and that field is caused at each moment and place on the wire by the time varying current on the past light cone of that moment and place on the thin wire. – Timaeus Dec 28 '15 at 18:27
• @Timaeus : it's no red herring. A frame is not something that actually exists, nor does a light cone. But the electron's electromagnetic field does, and it has a 'screw' nature. Maxwell spoke of this: "a motion of translation along an axis cannot produce a rotation about that axis unless it meets with some special mechanism, like that of a screw". – John Duffield Dec 29 '15 at 14:08

You need to decide whether you are interested in force on point particles or on extended particles or on a charged fluid. Force formulae and resulting equations are different for each.

• If you consider point particle with charge $q_k$, position $\mathbf r_k$ and velocity $\mathbf v_k$, the standard way to calculate force is

$$\mathbf F_k = q_k \mathbf E_\text{ext}(\mathbf r_k,t) + q_k \mathbf v_k\times \mathbf B_\text{ext}(\mathbf r_k,t)$$

where $\mathbf E_\text{ext}(\mathbf x,t),\mathbf B_\text{ext}(\mathbf x,t)$ are external electric and magnetic field (not including the field due to particle). This formula is actually used to define electric and magnetic field; the fields at the point $\mathbf r_k$ are such that the above formula gives accurately force on the particle of small charge $q_k$ when it is placed at the point $\mathbf r_k$ with velocity $\mathbf v_k$.

To use this formula for given $\mathbf r_k$,$\mathbf v_k$ we need to know the fields $\mathbf E_\text{ext}(\mathbf x, t)$, $\mathbf B_\text{ext}(\mathbf x, t)$. It is practically impossible to find out these fields exactly. Instead quasi-static field or Liénard-Wiechert retarded field of prescribed sources are often used as approximations.

For example, when electron moves inside a charged capacitor, the above formula is often used in textbooks to calculate EM force acting on the electron and with that it is possible to calculate its trajectory. In this calculation, the external EM field is due to capacitor only and is often approximated as electrostatic field of charge on the capacitor plates (although charge density on the plates changes during the fly-by of the electron due to interaction of plates with the electron, effect of this changing on the EM field of the capacitor can be neglected).

• If you consider matter as made of extended charged particles or charged fluid, the common formula for EM force acting on the fluid in volume $V$ is not the above one, but it involves integration. The simplest formula is

$$\mathbf F = \int_V \rho\mathbf E(\mathbf x, t) + \mathbf j \times \mathbf B(\mathbf x, t) dV$$

where $\rho,\mathbf j$ are total charge and total current density and $\mathbf E(\mathbf x, t),\mathbf B(\mathbf x, t)$ are total electric and magnetic field. Although this formula is generally assumed to give correctly total force when integrated over the whole body, the integrand

$$\rho\mathbf E(\mathbf x, t) + \mathbf j \times \mathbf B(\mathbf x, t)$$

does not necessarily give correct force per unit volume as a function of position . In practice special medium-specific formulae are used that express force density in terms of other quantities. You can find good introduction to this in the book by Panofsky & Phillips, Classical Electricity and Magnetism, sec. 6.6.

For example, these authors give formula for electrostatic force per unit volume of dielectric whose dielectric constant $\kappa$ is a function of mass density $\rho_m$: $$\rho\mathbf E - \frac{\epsilon_0}{2}E^2 \nabla \kappa + \frac{\epsilon_0}{2}\nabla \left(E^2\frac{d\kappa}{d\rho_m}\right).$$ (formula No. 6-68 in the book).

• As you can see if I want to compute the 'self-force' on the charge, $r$ becomes zero, which makes the force from the field infinite. Probably, that was pointed by Sebastian Riese as interaction of a single particle with the electro-magnetic field it generates – this is indeed non-trivial as it leads to inconsistencies of electrodynamics. – user36790 Dec 23 '15 at 12:38
• You pointed out not including the field due to particle in the answer. But could you tell why during self-induction, the flux of the changing magnetic field exert force (back-emf) on the charges that created that field? Doesn't that mean the charges are being forced by the very magnetic field they created? Am I missing something by saying during self-induction, charges are forced by their very-own created magnetic field? But also, we know the field at the charge which created it is infinite, isn't it? What am I mistaking in considering self-induction as case of self-force? Please help. – user36790 Dec 23 '15 at 19:05
• Self-induction is due to self-action, but not self-action of particles on themselves, but only action of one part of circuit on another. Say you have $10^{23}$ electrons in circuit conducting electric current, then one electron experiences non-stationary EM forces due to $10^{23} - 1$ other electrons. That is enough to explain why self-induction happens. – Ján Lalinský Dec 23 '15 at 20:18
• Suppose we are considering a solenoid where there is a non-steady current; there would be a self-induction & hence a back-emf. The magnetic field created by the current in solenoid is $\mu_0 nI(t)$; the magnetic flux is $\mu_0 n I(t)\pi r^2$ . The back EMF is then given by $$\varepsilon_0 = -(nl)(\mu_0\pi r^2)\;\frac{dI(t)}{dt}.$$ Now, if I don't include one charge as you said, wouldn't it affect the overall field $\mu_0\; nI(t)\;$ as the field has been calculated by Ampere's law taking in consideration all the charges. So, wouldn't, by excluding that charge, affect the magnetic field ? – user36790 Dec 24 '15 at 13:44
• The formula you give is derived assuming the electric current is distributed over surface of the solenoid. In this picture, there are no point particles, the charge is distributed on a surface with some density. In a model where current consists of discrete particles, the magnetic field is more complicated than this calculation provides, but for common-size solenoid the above formula is still a very good and often sufficient approximation. One electron has negligible effect on macroscopic EM fields. – Ján Lalinský Dec 27 '15 at 15:03

it is well known that charges are indeed affected by the changing magnetic flux which they create at non-steady state, that is self-induction.

This isn't correct. A flux is through an area not a point.

The time rate of change of the magnetic flux through a fixed loop in space is equal to the line integral of the electric field through that same loop.

The charges are affected by the field at the point where the charge is.

So a rate of change of a magnetic flux can be numerically equal to the line integral of an electric field. And they can be equal because they have a common cause. But the flux of $\partial \vec B/\partial t$ only tells you the line integral of the electric field, the EMF. It doesn't tell you the electric field at any particular point. For that you can use Jefimenko.

Except to use Jefimenko you need to know the charges and currents at earlier times.

How can I show mathematically that at non-steady state, the electric field & the magnetic field produced by the moving-electrons which create the field force those same charges without getting blown up?

You first need to pick a model. Discrete charges? A charged fluid? Extended charges? Most models don't talk about the microscopic fields, but only about the macroscopic fields.

Also, at steady state, there is a magnetic field associated with the time-independent current; why can't it exert force on the charges constituting the time-independent current?

The magnetic field does exert a force. But for a stationary wire, the magnetic force is unrelated to the EMF.

And again, your model matters. If you have a solid stationary positively charged wire to which you add some mobile negative charges with a random and equal distribution of magnetic moments for each negative charge then there are two forces. The Lorentz Force doesn't do work for the magnetic part of the force and for a thin stationary wire the magnetic part of the Lorentz force pushes a charge out of the wire and a Hall voltage keeps it inside the wire. Whereas any forces due to the magnetic moments of the charge carriers are just torques on the magnetic moment. And now the mathematical model matters too, whether your dipole is a Gilbert or an Ampere model, you have to use the corresponding forces and torques.

So in general, for a stationary wire the magnetic force tries to push the charges outside of a thin stationary wire and other non-magnetic forces then have to keep the charges inside the wire, and since both are orthogonal to the thin wire, neither affects the EMF.

For a steady current a Hall voltage develops across the ends of the wire to exactly cancel the magnetic force.

So imagine some net positive electric charge on the inner part of the wire and some net negative charge on the outer part of the wire, making an electric voltage that exactly cancels the magnetic force on each piece of current.

That is how a steady current flows around a loop in a resistance free wire. No net force on any charge other than that needed to go in a circle.

A certain acceleration is required for a steady current to go in a circle. And a Hall voltage develops until that acceleration is exactly what you need for a steady current. The same as for any steady current in a circuit.

Could you please explain me in an easier language why my above thinking that the charge is affected by its own field it created at the past, is wrong? Really, I've no idea of light cone.

Let's say you work based on reports your boss sends you. And your boss doesn't want you doing out of date things, so you always have to look at reports you are getting right now And your boss has decided that his instructions shall go at the fastest allowed speed, boss speed. And that nothing is allowed to go faster and that employees have to always go slower than boss speed. After all if you sent faster, you could outrun your boss's reports and who knows what you'd get up to!

Now you might wonder if your boss ever sends you a message about him meeting you somewhere else. And the answer is no. If he meet you somewhere else, he'd send reports in all directions at boss speed. Whereas you'd have to leave at less than boss speed. So the reports of that meeting always get someplace before you do. And remember you always have to follow the newly arrived reports, not the old ones.

So you never receive a report arriving at boss speed about you doing something somewhere else. And it is just because you go slower than the reports. Nothing deeper than that.

And a report about your here and now can't really affect your here and now becasue only the past can affect the now.

So it might seem like your own field can't affect you.

But that's the real issue. There are two things. You, the charge. And the field. And the point is that you interact. You affect the field and the field affects you. The charge affects the field and the field affects the charge.

Specifically you have a position and a velocity and you affect how the field there changes based on your velocity. And your velocity there changes based on the field.

And your only option is for both to change in a mutually consistent way that conserves energy and momentum. The total energy and momentum of the charge and the field.

• Really sorry! I didn't notice you answered this:/ Okay! as that of your reply to my second question, I'm talking about discrete charges and charged fluid. Also, I've seen you often mention microscopic field; what is it actually? What do you want to mean by 'microscopic'? – user36790 Jan 5 '16 at 3:19
• Also, since, cause precedes effect as is evident from Jefimenko's equations, can't I say a charge can be affected by its own field also? Suppose, our charge is at position 1 at time $t-r/c$; when it reaches at position 2 at $t$, at distance $r$ from position 1, wouldn't it feel its own electric field- the retarded field due to the same charge at position 1 at time $t-r/c\;?$ Does it happen so? – user36790 Jan 5 '16 at 5:28
• @user36790 There is absolutely no way to get such a statement out of Jefimenko. Firstly the equations clearly talk about how changes and currents cause fields and have nothing whatsoever to do with how fields affect charges. Plus if you move at sublight speeds your past light cone is not going to intersect your past self. – Timaeus Jan 5 '16 at 6:07
• @user36790 Correct. The field at the location of the particle depends on the other charges in the past, on any incoming vacuum solution (or boundary conditions) and the own particle's own kinematics right here and right now. – Timaeus Jan 9 '16 at 14:27
• @user36790 I would say it doesn't feel a self force when it has a uniform velocity. But I guess I'd have to check every person's version of self forces to know for sure that everyone says the same. – Timaeus Jan 11 '16 at 15:15