4

Let a bunch of charge move with a constant velocity $\mathbf v\;.$ Since, the charges are moving, they would create magnetic field $\bf B$ as it is current that produces magnetic field.

Now, would $\bf B$ affect these moving charges which themselves create $\mathbf B\;?$

My thoughts:

For electrostatics, electric field is given by the formula $$\mathbf E = \frac{q}{r^2}\hat {\mathbf {r}};$$ since $r= 0$ for the same charge that created the field, that would imply the electric field is exerting an infinite force on the charge which is not possible.

Now, if the bunch of charge is moving to constitute a steady-state current $I$ through a small-section $d\mathbf l,$ then magnetic field due to them is given by Biot-Savart Law:

$$d\mathbf B= \frac{I\;d\mathbf l\times \hat{\mathbf{r}}}{r^2}\;;$$ force on each charge moving through $d\mathbf l$ is given by $$\mathbf F= \mathbf v\times\frac{I\;d\mathbf l\times \hat{\mathbf{r}}}{r^2} \;;$$ since $d\mathbf l$ & $\hat{\mathbf r}$ are in the same direction, the numerator is zero & in the denominator $r$ is zero which would again make the force infinite. This is not possible.

However, when the current is non-steady, the moving charges produce time-dependent magnetic field which results in self-induction.

The self-induced emf caused due to change in flux of the magnetic field created by the charges does work on the charges; each charge feels force $\mathbf F= q\; \mathbf v\times \mathbf B $ where $\mathbf B$ is created by themselves.

So, that means, charges are moving & they are creating a magnetic field which again acts on the charges which created the field.

At steady state, charges are not affected by their own magnetic field whereas on the other hand, at non-steady state, the magnetic field, that those moving charges produced, imparts force on each charge leading to the production of self-emf.

But why is it so?

Fields are independent entities; they can share momentum & energy with charges as said by Timaeus.

Charges are indeed affected by their field when they read. But up to my reading, I've not found, at non-steady state of current, charges radiate. Still they are affected by the changing flux of their own created magnetic field. This produces self-emf; which means the magnetic field impart force on those moving charges which create the field.

However, why is it that moving charges are not affected by their own magnetic field when the current is at steady state but at non-steady state, charges are affected by the force from the magnetic field they create?

If charges are indeed affected by the magnetic field they create, is there any way to mathematically prove that during self-induction, the moving charges are forced by their own created magnetic field?

I'm quite confused; am I missing something here? Please help.

  • 2
    Coil does not only contain one charge carrier, so I do not see why self induction should be contradictory. But there are effects of self-interaction, especially the Abraham-Lorentz force. That is a particle does interact with the electromagnetic waves it emits (only in this fashion can momentum be conserved). – Sebastian Riese Dec 14 '15 at 19:57
  • @Sebastian Riese: I've not not said 'one charge'; from the very beginning line, I've been saying 'bunch of charge' –  Dec 15 '15 at 02:48
  • I think the flaw is that the Bio Savart law is not enough in this case, one needs the whole panoply of the of Maxwell's equations for this complexity. One can derive the Bio Savart law from Maxwell's equations but one cannot derive Maxwell's equations from the BS law. J equations seem to be derivable from the framework of maxwell's equations too http://oldstupac2.blogspot.gr/2007/04/bizarre-and-intriguing-story-of-oleg.html – anna v Dec 15 '15 at 04:41
  • @anna v: This is what I'm not understanding; which law will be applicable when:/ –  Dec 15 '15 at 04:43
  • Maxwell's equations can go to regions of phase space where the phenomenologically derived "law" like coulomb and bio savart cannot hold because of the complexity of the problem. After all many charges means a many body problem, and the classical laws were derived from simple observations assuming two body interactions. Even three body interactions cannot be computed analytically in gravitation, which is simpler than electromagnetism. J. equations can be derived from Maxwell equations and are applicable for the complexity of many moving charges, within the classical electromagnetic domain. – anna v Dec 15 '15 at 04:56
  • 2
    @Benito Ciaro : no, light doesn't propagate because the electric wave generates the magnetic wave which generates the electric wave etc. That's a popscience myth I'm afraid. See this answer. E and B are the spatial and time derivatives of four-potential. It's an electromagnetic wave. – John Duffield Dec 15 '15 at 13:43
  • What is the thing you are calling self induction? There is a thing called self inductance of a loop. It's about how a current in a loop is related to the magnetic field in that same loop and so the dynamics of the current is related to the dynamics of the magnetic field and hence to the EMF around the loop. If the loop is moving the EMF is non trivially related to actual forces. – Timaeus Dec 20 '15 at 04:27
  • @Sebastian Riese: I've been thinking for quite a few days about whether the electric field produced due to the changing current exerts force on the charges that create it; I always get confused in this. I tried & still trying to contemplate what your statement above Coil does not only contain one charge carrier, so I do not see why self induction should be contradictory; what did you really want to say? could you please elaborate a bit? –  Dec 21 '15 at 14:44
  • @user36790 I simply assumed your question to be solely about self-interaction (that is the interaction of a single particle with the electro-magnetic field it generates – this is indeed non-trivial as it leads to inconsistencies of electrodynamics). The title of the question can be interpreted in this way. And otherwise, I really don't get the point of the question. Why shouldn't they? Another example, where the field of the moving charges affects themselves, is, by the way, the skin effect, where high frequency currents only travel on the outer layer of a conductor. – Sebastian Riese Dec 21 '15 at 14:50
  • @Sebastian Riese: Okay, so the statement: charges are forced by the electric field that got created due to the non-steady current those charges produce true? –  Dec 21 '15 at 14:55
  • @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 Dec 21 '15 at 15:02
  • @Sebastian Riese: I was dying for this single yup:P Okay, that statement being taken true, I've a bit query on this: what $r$ should I use in Jefimenko's equation for computing the force due to electric field (& also the magnetic field); suppose I want to compute the force of the electric field (as well as magnetic field) created by a differential charge density $\rho$ on themselves; shouldn't then $r$ be $0$? If $r$ is zero, then the force would be indeterminate :( –  Dec 21 '15 at 15:09

3 Answers3

8

Neither Coulomb nor Biot-Savart are correct equations for the electromagnetic field except in statics. There are time dependent generalizations, such as Jefimenko's equations.

$$\vec E(\vec r,t)=\frac{1}{4\pi\epsilon_0}\int\left[\frac{\rho(\vec r',t_r)}{|\vec r -\vec r'|}+\frac{\partial \rho(\vec r',t_r)}{c\partial t}\right]\frac{\vec r -\vec r'}{|\vec r -\vec r'|^2}\; \mathrm{d}^3\vec{r}' -\frac{1}{4\pi\epsilon_0c^2}\int\frac{1}{|\vec r-\vec r'|}\frac{\partial \vec J(\vec r',t_r)}{\partial t}\mathbb{d}^3\vec r'$$ and $$\vec B(\vec r,t)=\frac{\mu_0}{4\pi}\int\left[\frac{\vec J(\vec r',t_r)}{|\vec r -\vec r'|^3}+\frac{1}{|\vec r -\vec r'|^2}\frac{\partial \vec J(\vec r',t_r)}{c\partial t}\right]\times(\vec r -\vec r')\mathbb{d}^3\vec r'$$ where $t_r$ is actually a function of $\vec r'$, specifically $t_r=t-\frac{|\vec r-\vec r'|}{c}.$

These reduce to Coulomb and Biot-Savart only when those time derivatives are exactly zero, which is statics. So Jefimenko is an example of proper time dependent laws for the electromagnetic field. Note that both the electric and the magnetic part of the electromagnetic field have parts that depend on the time variation of current. Faraday's law relates the two together.

And the next fact is also key: the EMF around a stationary wire is 100% due to the electric field. And if you have a universe with neutral objects everywhere then Jefimenko predicts the electric field is solely and 100% caused by time variation of current. And I do mean cause, as in cause and effect. The equations are causal since the charge (and change in charge) and the current (and change in current) are things in the past ($t_r\leq t$) that affect the present.

So lets review. Coulomb and Biot-Savart are incorrect when not in statics. Self induction of a stationary wire is caused by electric fields making a nonzero EMF. These electric fields are caused by time varying currents. And yes, these time varying currents also cause (and yes I do mean cause) magnetic fields.

And when people say that changing electric fields cause magnetic fields and vice versa, they are not this kind of causality where one causes the other, they are just equality since they related two things that happen at the same time. Causality relates the present (the effects) to the past (the causes). To quote 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 (emphasis added).

Now let's get to self fields. It is easy to imagine that each charge only reacts to the fields of the other charges. But do not over simplify when considering radiation reaction and other such complications.

In reality you have charges and fields. And you have to specify both. Each has energy, each has momentum, and they exchange energy and momentum with each other. Fields are not just mathematical fictions to compute forces between particles, they are real things with their own degrees of freedom, their own energy, momentum, and even stress.

Then you can find out the energy and momentum of the fields and the charges and then find out how they mutually exchange energy and momentum amongst themselves.

But as everyone said, there can be self-force only if the charges are radiating.

That is not what everyone says. Self force is much more general than radiation reaction forces. The most common case of harmonic motion had the particle gain as much energy as it loses from the inductive fields (Schott fields, that fall off faster than the radiation fields but still store energy) and only on average loses energy to the radiative fields.

Lots of people study this. Stephen Lyle wrote an entire book on just a uniformly accelerating charged particle and an entire book on self forces. Herbert Spohn wrote an entire book on the mutual coupled dynamics of electromagnetic fields and charges. And Fritz Rohrlich wrote a classic book on Classical Charged Particles. And those are just books, there are probably at least a dozen articles published every year for at least the last hundred years on the topic.

So. There are lots of self forces. Debate continues to this day about whether it is changing forces, jerk (changing acceleration), or acceleration itself that causes self forces. Or whether it's something completely different.

Keep in mind that a textbook case of a classical sized wire is usually discussing macroscopic fields, where you average over a large enough number of atoms that the fields and charge and current become smoother fields that don't jump around due to thermal effects or based on where in the lattice of the solid you are. And when someone does care about those effects, they use statistical physics and/or quantum physics. Not just pure Maxwell.

Timaeus
  • 25,523
  • Even if charges are moving with uniform velocity, they would be affected by their own field? –  Dec 15 '15 at 02:53
  • Okay, I'm wrong. But could you say why? I've presented my thought; where did I mistake? Could you point that? Also, why would Coulomb's law & Biot-Savart Law be wrong? Please, please reply /\ –  Dec 15 '15 at 03:25
  • "You need to get over the fairy tale that fields are just mathematical fictions to compute forces between particles and accept that they are real things", and is this Jefimenko the new bible for physics? I thought we are discussing within the current pov of physics, not new theories.Physics is consistent and allows for alternate formulations even of Maxwell's equations, which seems to be the case here. To set up doctrines on reality of mathematics , I do not think it is wise. this is interesting http://oldstupac2.blogspot.gr/2007/04/bizarre-and-intriguing-story-of-oleg.html – anna v Dec 15 '15 at 04:28
  • @annav It is better to use Jefimenko than to use Coulomb and Biot-Savart outside of statics. It is misleading to students to have them think that Coulomb and Biot-Savart are general laws for nonstatic situations. And I emphasized the even better approach to treat the fields and the charges as fundamental entities. Because electrons and photons are fundamentally equally real. I don't think math is real, I just think it isn't right to make students think electrons are more real than photons and that fields are just an afterthought without their own degrees of freedom. – Timaeus Dec 15 '15 at 04:33
  • Well, I agree with most of what you state here, but then one should separate quantum from classical. The question is about classical, and yes one can derive the coulomb and bio savart from ME, and the reverse is not true. ME needed Maxwell to think outside the box. Seems to me that J equations do not need the quantum frame but are also derivable from ME. – anna v Dec 15 '15 at 04:47
  • @annav Coulomb only holds in electrostatics. Biot-Savart only holds in Magnetostatics. And since they are wrong, neither is derivable from Maxwell. Both are just Jefimenko applied to a special case where the charge density and current density don't change in time. Yes, Jefimenko just finds a particular solution to Maxwell, but it works when Coulomb and Biot-Savart fail. I only brought up quantum language to point out how silly it is to expect electromagnetic fields to not exist as a separate thing, that would be just like expecting photons to not exist as a separate thing. Use fields & charges – Timaeus Dec 15 '15 at 04:55
  • well, physics is consistent. The 1/r Coulomb law behavior can be derived from ME with some assumptions https://en.wikipedia.org/wiki/Gauss's_law#Deriving_Coulomb.27s_law_from_Gauss.27s_law – anna v Dec 15 '15 at 05:02
  • 1
    @annav Coulomb does NOT work when the charge density changes in time, which is exactly a case when the field doesn't fall off like $1/r^2$ which is exactly why you need Jefimenko (or Maxwell) rather than Coulomb. And similarly when current changes in time you also get electromagnetic fields that fall off other than $1/r^2$ which again, is exactly a case where Coulomb and Biot-Savart fail utterly. They only work in statics. Use Maxwell (or Jefimenko if you must have unique fixed fields for a given charge and current and their time derivatives) rather than Coulomb and Biot-Savart! – Timaeus Dec 15 '15 at 05:09
  • @Timaeus: Oh! I got the point. When the current is non-steady, Biot-Savart Law doesn't work. During self-induction, the current is changing; so I cannot apply BS law. –  Dec 15 '15 at 05:13
  • @Timaeus Of course I agree. I guess it was in my comment to user directly where I said that for more than "two body" problems coulomb etc laws cannot work. Above I was talking of the limit in the obsrevations which were used to posit the pre maxwell laws. – anna v Dec 15 '15 at 06:12
  • @user36790 Why you are asking? See my answer and think of it longer than only a moment. – HolgerFiedler Dec 15 '15 at 07:12
  • I've been waiting for four days; probably you were busy. I am in a dire need to clear this confusion; I got from your answer that I've to use Jefimenko's equations at non-steady state & also got the point that fields are also realistic - they have momentum & energy as well. for that I thank you a lot. But really what I didn't get is whether charges suffer force from their magnetic field during self-induction. To me, it seems that they are forced by the changing flux of their own magnetic field during self-induction & that implies the magnetic field exert force on the charges [...] –  Dec 20 '15 at 03:14
  • which create the field. But as everyone said, there can be self-force only if the charges are radiating. I don't think during the self-induction, that is during the non-steady state, the charges are radiating; at least no book write this. But if they are not affected by their own magnetic field, how can self-induction be even possible? So, I would really appreciate you if you could tell whether during self-induction, the magnetic field exerts force (opposing the current) on those charges that create the field. Please help. –  Dec 20 '15 at 03:18
  • @user36790 I wasn't busy, I was done answering. The question of charges, currents ,fields, forces and the mutual dynamics of them all is not trivial and not a common textbook problem. And I don't think you've defined your terms and I don't think you are using standard terminology and I never thought your question was well posed. I'd even suggest asking about a specific experiment in a separate question. I did edit my answer, but it probably won't make you happy. And finally EMFs are less related to forces than you might think. The F in EMF is just a historical leftover not actually a force. – Timaeus Dec 20 '15 at 04:24
  • @Timaeus: Yes, I know EMF is not force; it is the work done by the force per unit charge (of an induced field, here). But who is providing the force? The changing flux of the field created by those moving charges provide the force. Okay. I'll make a separate question on self-induction. I always learn new things from your answer; thanks for all that. I'll notify you about that & hope you would help. –  Dec 20 '15 at 04:43
  • @user36790 No, it is not the work done. It is the line integral around an instantaneous closed loop of the force per unit charge. If the loop is moving than the direction through the instantaneous loop is not in the same direction as the motion of the object, so it isn't the work. Again, elementary textbooks don't always give correct definitions. They want to avoid complications. So they might define voltage in statics, and an EMF as work per charge when a circuit isn't moving. And they might keep using those words in situations beyond where they were introduced with narrow definitions. – Timaeus Dec 20 '15 at 04:50
  • @Timaeus: Yes, I meant to say that. I got the derivation in the same language at the beginning of induction by Purcell. –  Dec 20 '15 at 04:51
  • @Timaeus: Oh! I didn't see you edited your answer. Sorry, I'm reading it now. –  Dec 20 '15 at 04:57
  • @user36790 As for who is providing the force, it is the regular ordinary Lorentz Force: the electric part for the flux of $\partial \vec B/\partial t$ and the magnetic part for the motion of the loop (if the loop is thin and the charges are confined to stay inside the loop). – Timaeus Dec 20 '15 at 05:32
  • Yes, Lorentz force. But who is producing the $B$? The charges which are constituting the non-steady current. That means the charges are forced by the $B$ they created, isn't it? –  Dec 20 '15 at 19:11
  • @user36790 I did try to tell you right at the beginning that everything you wrote is wrong, and I did mean absolutely everything. For instance currents are not the only thing that cause magnetic fields. Changing currents also cause magnetic fields, and changing currents make electric fields too. This is relevant becasue if the loop isn't moving the EMF is 100% entirely from the Lorentz force due to the electric field. If you use the Lorentz force you don't really expect magnetic fields to do work, right? – Timaeus Dec 20 '15 at 19:34
  • @user36790 Even for moving wires the EMF only equals the change in flux when the wire is thin and the charges are confined to stay in the wire. When currents change this produces electric and magnetic fields, the electric fields have nonzero line integrals around closed curves, and the magnetic fields have changing fluxes and the two are related and have the same cause, changing current. And for a stationary loop, the EMF is from the electric field, and electric fields can be caused by changing currents. So currents make magnetic fields and changing currents make electric fields. – Timaeus Dec 20 '15 at 19:37
  • This isn't an answer, it's a tirade, and it's wrong. See Jefimenko: "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". An electron has an electromagnetic field. If you move past it you don't change its field into a magnetic field because it is now a current in your frame of reference. – John Duffield Dec 22 '15 at 21:52
  • @JohnDuffield I've edited the answer, but I never understood which part you claim is wrong. If there is a single line of text in the answer you think is wrong please do point it out. I only emphasize the electric field in the EMF since the OP seemed to think magnetic forces were involved in the EMF for self induction in a wire with nonsteadt current. If the wire is at rest in a frame, then in that frame it is the electric force (not the magnetic force) that is responsible for the EMF in that frame. – Timaeus Dec 28 '15 at 00:08
  • @Timaeus: BTW, in the Jefimenko's equation, retarded time is given by $t_r=t-\frac{|\vec r-\vec r'|}{c}.$ But what happens when $\vec r'= \vec r;?$ How should I deal with it? Shouldn't I take the value of $\vec r' = \vec r$ when computing the integral? –  Jan 11 '16 at 13:30
  • @user36790 For discrete charges, to what end? If your goal is to find an acceleration by finding a force, note that the self field has a term $\partial \vec J/\partial t(\vec r',t_r)$ which itself already depends on the acceleration. So it's a catch 22. If it's for continuous charges, you can remove that point from the integral to make it an improper integral. Which is how you handle functions that blow up at a point and aren't Dirac deltas. – Timaeus Jan 11 '16 at 15:30
  • Got it; hope that the improper integral converges. Also, check this; you can see the first term of the integral of Jefimenko doesn't blow up at $r=0$. Can also the other term not blow up at $r= 0;?$ just like the first term? –  Jan 11 '16 at 19:46
  • @user36790 It's a catch 22, if the force blows up then $\partial \vec J/\partial t$ blows up, which could cause the field to blow up. And even if it doesn't, its not a straight forward computation when you need the acceleration right now to find the field right now. Some people use an iterative approach to hope to find a consistent acceleration though the acceleration is already fixed by the past unless there is an impulse. – Timaeus Jan 12 '16 at 16:27
-2

Have you ever thought yourself that the magnetic dipole moments are the reason for any magnetic field? if they are aligned, they form an external field. The alignment happens by acceleration because the magnetic dipole moment is bonded to the intrinsic spin (it's really a rotation according to a really performed experiment from Einstein and de Haas). And that the statement of moving charges is a misinterpretation, the charges have to be accelerated (like in a coil). The experiment with magnetic field between two wires was performed first with switch on of the currents (an acceleration of charges).

HolgerFiedler
  • 10,334
  • 1
    This is just wrong. Where did you get that idea? Take a pure electric monopole at rest (which admittedly does not exist) and then do a Lorentz boost. A magnetic field will arise, that is not related to any dipole that is aligned. – Sebastian Riese Dec 16 '15 at 16:10
  • @SebastianRiese Two questions. Did you or someone else perform such an experiment? And if yes, are you sure that at the moment you measure the magnetic field, you not induce this field? Additional questions: Has you measured a magnetic field with a measurement instrument witch don't has its own magnetic field? BTW, I'm familiar with the common theory. – HolgerFiedler Dec 16 '15 at 16:53
  • You are well aware, that there are numerous experimental verifications of special relativity to high accuracy? The things I stated follow quite directly from special relativity. Refuting them means refuting special relativity, which is quite a far stretch. – Sebastian Riese Dec 16 '15 at 16:56
  • @SebastianRiese To state something, what is obvious, is OK. I give a deeper explanation. How about chat room? – HolgerFiedler Dec 16 '15 at 17:00
-4

Let a bunch of charge move with a constant velocity $\mathbf v\;.$ Since, the charges are moving, they would create magnetic field $\bf B$ as it is current that produces magnetic field.

We're all familiar with the current-in-the-wire as per the picture below, but moving charges don't really "create" a magnetic field. See Jefimenko for a bit about that: "...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..."

enter image description here

You can work this out by imagining that you're a positron, and I set you down in front of a nailed-down electron. This electron has an electromagnetic field. It doesn't have an electric field or a magnetic field, it has an electromagnetic field. As does the positron. Because of this, when I let you go, you move linearly towards the electron. Then because you only see a linear motion, you think in terms of the radial electric field as per Andrew Duffy's Physics 106 course:

enter image description here

But what happens when I throw you past the electron? You still move towards it, but you also experience a rotational motion, which you attribute to the electron's magnetic field. But note that you didn't create the electron's magnetic field just because you moved. All that happened is that your relative motion revealed the rotational magnetic aspect of the electromagnetic interaction.

I always thought charge cannot be affected by its own field, be it electric or magnetic.

It isn't. The electron has an electromagnetic field, and it's affected by another charged particle, or by a photon. Not by itself.

Electric field is given by the formula $\mathbf E = \frac{q}{r^2}\hat {\mathbf {r}};$ since $r= 0$ for the same charge that created the field, that would imply the electric field is exerting an infinite force on the charge which is not possible.

Agreed. This is what Frank Close referred to in his book The Infinity Puzzle. I don't think it's much of a puzzle myself. It's quantum field theory, not quantum point-particle theory. It's the wave nature of matter, not the point-particle nature of matter.

Now, if the bunch of charge is moving to constitute current $I$ through a small-section $d\mathbf l,$ then magnetic field due to them is given by Biot-Savart Law: $d\mathbf B= \frac{I\;d\mathbf l\times \hat{\mathbf{r}}}{r^2}\;;$ force on each charge moving through $d\mathbf l$ is given by $$\mathbf F= \mathbf v\times\frac{I\;d\mathbf l\times \hat{\mathbf{r}}}{r^2} \;;$$ since $d\mathbf l$ & $\hat{\mathbf r}$ are in the same direction, the numerator is zero & in the denominator $r$ is zero which would again make the force infinite. This is not possible.

Agreed. Point particles make the maths easy, but it's only an approximation, and making the maths easy shouldn't blind people to the hard scientific evidence of things like electron diffraction and the Einstein-de Haas effect. But it does.

Conclusion: Charges cannot be affected by their own electric field & magnetic field.

Agreed.

But then, how could there be any self-induction if moving charges cannot be affected by their own magnetic fields?

Self-inductance is "the induction of a voltage in a current-carrying wire when the current in the wire itself is changing". And that current is changing because there's some kind of interaction going on with other charged particles. Like a whole bunch of charged particles in the guise of my hand flipping a switch.

The self-induced emf caused due to change in flux of the magnetic field created by the charges does work on the charges; each charge feels force $\mathbf F= q\; mathbf v\times \mathbf B $ where $\mathbf B$ is created by themselves. So, that means, charges are moving & they are creating a magnetic field which again acts on the charges which created the field.

Just simplify that current to one electron, then imagine it's not moving and it's you moving instead. Nobody created a magnetic field for the electron just because they moved. Because it's got an electromagnetic field. And electromagnetic field interactions result in linear and/or rotational motion. When we only see the former we call it an electric field, when we only see the latter we call it a magnetic field, but the fields concerned are electromagnetic fields. And the simplest example of electromagnetic field interactions involves positronium. There's a linear motion such that the electron and the positron are moving towards each other and will annihilate, and there's a rotational motion too:

enter image description here

CCASA image by Manticorp/Rubber Duck, see Wikipedia

But how can it be? That would yield an infinite force (even if not infinite, it would have to be zero since $d\mathbf l$ & $\hat{\mathbf r}$ are parallel).

There are no infinite forces, the electron has its Compton wavelength, and it has a "spinor" nature.

So, the million dollar question is: Do moving charges get affected by the magnetic field they create while moving to constitute current? How can it be if the formula is telling otherwise?

No they don't, because your motion relative to a charged particle doesn't create a magnetic field for it. It just results in rotational force in addition to the linear force because of the "spinor" nature of charged particles.

enter image description here

John Duffield
  • 11,097