I confuse the four kinds of fundamental interactions, so I think the electric force and magnetic force should not be classified as a big class called electromagnetism.

Here is my evidence:

  1. The Gauss law of electric force is related to the surface integration but the Ampere's law corresponds with path integration.

  2. The electric field can be caused by a single static charge while the magnetic force is caused by a moving charge or two moving infinitesimal current.

  3. The electric field line is never closed, but the magnetic field line (except those to infinity) is a closed curve.

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    $\begingroup$ Special relativity mixes electric and magnetic fields, and having one kind of field together with relativity inevitably leads to having the other, see e.g. this question. $\endgroup$
    – ACuriousMind
    Commented May 21, 2016 at 14:11
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    $\begingroup$ Your evidence only suggests that the magnetic field is not the same as the electric field. It doesnt suggest, on the other hand, that these fields are related. In fact, they are. $\endgroup$ Commented May 21, 2016 at 14:12
  • $\begingroup$ when you study the electric field of a dipole en.wikipedia.org/wiki/Dipole you will see that it has the same geometry as the magnetic dipole. It is just that in nature as we have found it there are no magnetic monopoles like the single charges which are charge monopoles. Otherwise there is symmetry $\endgroup$
    – anna v
    Commented May 21, 2016 at 15:18
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    $\begingroup$ It seems that you haven't seen Maxwell's correction term to Ampere's law. Ampere's law with only the current as source is not consistent with the conservation of electric charge. For Ampere's law to be consistent with conservation of electric charge, one must include a source term proportional to the rate of change of the electric field. So it is not possible to formulate satisfactory theories of the electric and magnetic field separately. They will always be coupled. $\endgroup$ Commented May 21, 2016 at 15:22
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    $\begingroup$ Electricity and Magnetism are very much closely related natural entity. Charge (with electric field)+motion---->Magnetism(Magnetic field). As well, Magnet (With magnetic field)+motion---->Electric field(As in electromagnetic-induction in generator). P.S. 1. read this like a chemical equation, not like a mathematical expression. 2. I've wrote it in an oversimplified manner. $\endgroup$
    – user115953
    Commented Jul 21, 2016 at 19:09

7 Answers 7


Consider this: A charged particle at rest creates an electric field, but no magnetic field. Now if you walk past the charge, it will be in motion from your point of view, that is, in your frame of reference. So your magnetometer will detect a magnetic field.

But the charge is just sitting on the table. Nothing about the charge has changed.

Evidently the space around the charge is filled with something that at times appears to be a pure electric field, and at other times appears to have a magnetic field. We conclude that the field is something other than an electric field or a magnetic field. It is another type of field which combines the two into one entity.

  • $\begingroup$ Yes, this disproves my second evidence and third evidence. How about the first evidence? is there any quantification between magnetic field and the electric field? $\endgroup$
    – Zack Ni
    Commented May 21, 2016 at 23:21
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    $\begingroup$ @ZackNi Ampere's law must include a term proportional to the time rate of change of the electric field, or it will be inconsistent with conservation of electric charge. Students are usually first shown the special static cases, in which case magnetostatics and electrostatics are independent. But in the full theory en.wikipedia.org/wiki/… the electric and magnetic fields are definitely coupled. $\endgroup$ Commented May 22, 2016 at 21:13
  • $\begingroup$ I don't really see how 1) supports your argument. The two field components, E and B, have different properties. I don't see the logical connection to your suggested conclusion. $\endgroup$
    – garyp
    Commented May 23, 2016 at 0:30
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    $\begingroup$ @ZackNi: There's actually a very elegant way of combining Gauss's Law and Ampere's Law into one equation, in the context of special relativity: $\sum_\beta \partial F^{\alpha \beta}/\partial x^\beta = \mu_0 J^\alpha$, where $\alpha$ and $\beta$ are indices in space and time (just like regular 3-D vectors have $x$, $y$ and $z$ components, only with a $t$ component as well.) Chapter 12 of Griffiths's book on E&M has a pretty good treatment of the basics of it; see also the Wiki article for a summary. $\endgroup$ Commented May 23, 2016 at 13:04
  • $\begingroup$ @MichaelSeifert Ah I still remember my first lesson on EM. The prof posted that on the board and announced that (sans indices) we were done. If memory serves, to use that equation, one needs to apply the Minkowski metric. $\endgroup$
    – Aron
    Commented May 24, 2016 at 9:46

The arguments from special relativity given in the other answers is correct. What is charge according to one observer is current according to another observer that is in relative motion to the first. But this is, from a historical perspective, somewhat backwards. This consideration is what led Einstein to develop special relativity -- the paper is called On the Electrodynamics of Moving Bodies. But long before Einstein, it was known that electricity and magnetism were not independent.

Have you seen Faraday's law of induction? If a magnet is moved through a coil of wire, there will be an induced electric field in the wire and a measurable current produced as a result. This is a hint that electricity and magnetism are related.

It also seems that you haven't seen Maxwell's correction term to Ampere's law. Ampere's law with only the current as source is not consistent with the conservation of electric charge. For Ampere's law to be consistent with conservation of electric charge, one must include a source term proportional to the rate of change [time derivative] of the electric field. Again the electric and magnetic fields are interrelated.

It is not possible to formulate satisfactory, independent theories of the electric and magnetic field. They will always be coupled. From a modern point of view, it has to be like this because there is no way to be consistent with Einstein otherwise, but it was because of Maxwell's theory of electromagnetism that special relativity was developed, not because of special relativity that it was realized that electricity and magnetism are related.

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    $\begingroup$ But, time is reversible, so it doesn't matter in which order things happened :) $\endgroup$
    – user95006
    Commented May 21, 2016 at 22:14
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    $\begingroup$ Actually time is not reversible on fundamental levels, some transitions in particle physics violate this symetry. And certainly it isn't in macroscopic world. $\endgroup$
    – Blazej
    Commented May 22, 2016 at 21:08
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    $\begingroup$ I would add to Robins answer, why "It is not possible to formulate satisfactory, independent theories of the electric and magnetic field." The context is that charged particles have not only an electric field. Electrons as well as other particles have also a magnetic dipole moment. Usually this moments are distributed randomly and the net field is zero. But due to the related properties intrinsic spin and magnetic dipole moment (they always or parallel or anti-parallel) it is enougth to rotate a body and this will induce a magnetic field inside the body. $\endgroup$ Commented May 23, 2016 at 4:28
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    $\begingroup$ The simplification that a particle in rest does not have a magnetic field and the big number of upvotes for garyps answer shows that that the unknown before 1907 magnetic dipole moment and the unknow before 1923 intrinsic spin is still not teached as the reason for magnetic induction of accelerated particles. $\endgroup$ Commented May 23, 2016 at 4:34
  • $\begingroup$ People should note that what @HolgerFiedler is saying is irrelevant at best and false at worst. Even in vacuum the $\mathbf E$ and $\mathbf B$ fields are coupled. Even in a theory of scalar particles, i.e., completely discarding the spin, there will be magnetic fields from particles in motion. You can just look up the Liénard–Wiechert potentials in Jackson, or any electromagnetic wave mode in a plasma. $\endgroup$ Commented May 23, 2016 at 11:41

You are right, the electric field and the magnetic field are distinct fields that have different properties. The reason why they are still classified as the cause for the "electromagnetic force" are the following:

In higher theories, like the field theory, the electric and the magnetic field are caused by the same gauge principles. There is just "one" interaction between a particle and the electromagnetic field, and this electric field and the magnetic field are two fields to describe this interaction.

Furthermore: If you change your frame of reference, for example by running by the experiment you are looking at at the moment, then electric field and magnetic field will mix up. What part of a force acting on a particle is magnetic and what is electric is a property that depends on your relative speed towards the particle. Here you see again, that electric field and magnetic field are just two ways to describe the "electromagnetic interaction". How this interaction splits up to the E-field and the B-field depends on your frame of reference.

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    $\begingroup$ I don't understand the down vote here. $\endgroup$
    – garyp
    Commented May 21, 2016 at 14:20
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    $\begingroup$ I highly disagree with statement that electric and magnetic field are distinct. Split of electromagnetic field into two parts (electric and magnetic) depends on the point of view and can't be done in Lorentz-invariant way. Electromagnetic field really is fundamentally one object. However, I need to agree that it can be "more magnetic" or "more electric" in some situations, due to existence of Lorentz invariant such as $B^2-E^2$/ $\endgroup$
    – Blazej
    Commented May 21, 2016 at 14:33
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    $\begingroup$ @Blazej The manner in which they interact with charged particles is experimentally distinct. You use different equipment to measure them. And the post goes on to say that they are fundamentally unified. $\endgroup$
    – garyp
    Commented May 21, 2016 at 15:22
  • $\begingroup$ I admit that my downvote here was not a good decision. I don't fully agree with this post but clearly it is not of poor quality and doesn't contain anything blatantly wrong. $\endgroup$
    – Blazej
    Commented May 21, 2016 at 19:25
  • $\begingroup$ I'm going to edit this answer, taking the things you mentioned into account. $\endgroup$ Commented May 22, 2016 at 0:42

Several answers have given a physical explanation as to why electric and magnetic forces are tightly coupled, and why you can't develop independent theories of "just electric" and "just magnetic" fields.

Your subquestions (especially #1) make me think you're looking for some kind of symmetry. It turns out, there's a really nice one!

All the asymmetry between electric and magnetic fields can be entirely chalked up to just one fact:

There are electric, but not magnetic, monopoles.

If you allow the existence of magnetic charge, Maxwell's equations become completely symmetric: $$ \begin{array}{ccc} & \textrm{Without monopoles} & \textrm{With monopoles} \\ \textrm{Gauss's Law} & \nabla \cdot E = \frac{\rho}{\epsilon_0} & \nabla \cdot E = \frac{\rho}{\epsilon_0} \\ \textrm{Gauss's Law (Magnetism)} & \nabla \cdot B = 0 & \nabla \cdot B = \mu_0 \rho_m \\ \textrm{Faraday's Law} & - \nabla \times E = \frac{\partial B}{\partial t} & - \nabla \times E = \frac{\partial B}{\partial t} + \mu_0 J_m \\ \textrm{Ampere's Law} & \nabla \times B = \mu_0 \epsilon_0 \frac{\partial E}{\partial t} + \mu_0 J & \nabla \times B = \mu_0 \epsilon_0 \frac{\partial E}{\partial t} + \mu_0 J \\ \end{array} $$

If the symmetry isn't clear, recall that $E$ and $cB$ have the same units, and that $\mu_0 \epsilon_0 = 1/c^2$, and rearrange accordingly. $$ \begin{array}{cc} \nabla \cdot E = \frac{\rho}{\epsilon_0} & \nabla \cdot (cB) = \frac{\rho_m / c}{\epsilon_0} \\ - \nabla \times E = \frac{1}{c} \left( \frac{\partial (cB)}{\partial t} + \frac{J_m / c}{\epsilon_0} \right) & \nabla \times (cB) = \frac{1}{c} \left( \frac{\partial E}{\partial t} + \frac{J}{\epsilon_0} \right) \end{array} $$

In fact, if you "rotate" between $(E, B)$, $(\rho, \rho_m)$, and $(J, J_m)$, you get a set of fields still satisfying these extended Maxwell's equations. So if you rotated by $\pi/2$, you could turn any field with no magnetic monopoles into a field with no electric monopoles.

Unfortunately, I don't know enough about the physics to explain why those additional terms are they way they are, so hopefully someone who understands better will. I only know about them from an antenna design perspective, where magnetic charge is a useful conceptual tool that simplifies computation.

EDIT: There's an answer to a different question that gives an even cleaner form for Maxwell's laws (no $\epsilon_0$s or $\mu_0$s!)

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    $\begingroup$ Dear Henry, this symmetry is more clearly seen if one writes down Maxwell equations in fully relativististic notation, involving electromagnetic field tensor $F$ and its dual. One has $\nabla_{\mu}F^{\mu \nu} = 4 \pi J_e^{\nu}$ and $\nabla_{\mu}*F^{\mu \nu} = 0$ but one could easily have $\nabla_{\mu}*F^{\mu \nu} = 4 \pi J_m^{\nu}$. However if you do this, concept of potential for electromagnetic field loses any meaning. $\endgroup$
    – Blazej
    Commented May 22, 2016 at 21:07
  • $\begingroup$ Oh cool, that is a lot cleaner! Why does this kill the idea of a potential though? There's a potential for relativistic E&M, and there's one (well, two) for magnetic charge, what about the combination kills the idea? $\endgroup$ Commented May 22, 2016 at 21:19
  • $\begingroup$ I will stick to language of vector calculus as it is probably more well known. In ordinary electromagnetism divergence of B field vanishes. This implies that there exists a vector field (which we call A) such that its curl is equal to magnetic field, $\nabla \times A=B$. Then Faraday Law implies that there exists scalar field $\Phi$ such that $E=-\nabla \Phi - \frac{\partial A}{\partial t}$. In other words, existence of potentials relies on Maxwell equations in the form without magnetic charges. $\endgroup$
    – Blazej
    Commented May 22, 2016 at 21:37
  • $\begingroup$ The trick I've usually seen to get around that is: remove all the magnetic charges, find the standard potentials. Then remove all the electric charges and find the other potentials. You can't use superposition to add them into one potential, but the two things you get have an $E$ and $B$ like relationship. Maybe that's not a true potential though. $\endgroup$ Commented May 22, 2016 at 21:52
  • $\begingroup$ Ok, this would work. $\endgroup$
    – Blazej
    Commented May 23, 2016 at 15:22

Regarding 1) observe that there is a pattern in common - namely that there is some region (volume for Gauss and a surface for Ampere) and integral of the source on this region is equal to the integral of the field on the boundary. This is a striking similarity.

2) currents are nothing else than moving charges. So both fields are generated by charges. These are two sides of the same coin. Both are related to motion of charges but motion in different direction. Static charge is "moving in time, but not in space" and it produces only electric field. Moving charges move both in time and space and they produce both electric and magnetic field. This is all nicely unified in relativity. There charge density and current are incorporated in one object - current 4-vector. 4-vectors have one time component and 3 spatial components. Charge density is the time component of current (it is the current moving in time) while what we ordinarily think of as current (in space) is spatial component of this 4-vector.

3) This is an interesting argument, I need to think about it.

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    $\begingroup$ As to 3), I don't believe that our current theory requires closed magnetic field lines. That is, I don't believe we have a good understanding for the experimental lack of magnetic monopoles. I may be wrong, and I invite clarification. $\endgroup$
    – garyp
    Commented May 21, 2016 at 14:22
  • $\begingroup$ @garyp hopefully not, because pulsars seem to have lots of open field lines... $\endgroup$ Commented May 21, 2016 at 23:32

The classical electromagnetic effect is perfectly consistent with the lone electrostatic effect but with special relativity taken into consideration. The simplest hypothetical experiment would be two identical parallel infinite lines of charge (with charge per unit length of $ \lambda \ $ and some non-zero mass per unit length of $\rho \ $ separated by some distance $ R \ $. If the lineal mass density is small enough that gravitational forces can be neglected in comparison to the electrostatic forces, the static non-relativistic repulsive (outward) acceleration (at the instance of time that the lines of charge are separated by distance $ R \ $) for each infinite parallel line of charge would be:

$$ a = \frac{F}{m} = \frac{ \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda^2}{R} }{\rho} $$

If the lines of charge are moving together past the observer at some velocity, $ v \ $, the non-relativistic electrostatic force would appear to be unchanged and that would be the acceleration an observer traveling along with the lines of charge would observe.

Now, if special relativity is considered, the in-motion observer's clock would be ticking at a relative rate (ticks per unit time or 1/time) of $ \sqrt{1 - v^2/c^2} $ from the point-of-view of the stationary observer because of time dilation. Since acceleration is proportional to (1/time)2, the at-rest observer would observe an acceleration scaled by the square of that rate, or by $ {1 - v^2/c^2} \ $, compared to what the moving observer sees. Then the observed outward acceleration of the two infinite lines as viewed by the stationary observer would be:

$$ a = \left(1 - v^2 / c^2 \right) \frac{ \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda^2}{R} }{\rho} $$


$$ a = \frac{F}{m} = \frac{ \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda^2}{R} - \frac{v^2}{c^2} \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda^2}{R} }{\rho} = \frac{ F_e - F_m }{\rho} $$

The first term in the numerator, $ F_e \ $, is the electrostatic force (per unit length) outward and is reduced by the second term, $ F_m \ $, which with a little manipulation, can be shown to be the classical magnetic force between two lines of charge (or conductors). The electric current, $ i_0 \ $, in each conductor is

$$ i_0 = v \lambda \ $$

and $ \frac{1}{\epsilon_0 c^2} $ is the magnetic permeability

$$ \mu_0 = \frac{1}{\epsilon_0 c^2} $$

because $ c^2 = \frac{1}{ \mu_0 \epsilon_0 } $

so you get for the 2nd force term:

$$ F_m = \frac{v^2}{c^2} \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda^2}{R} = \frac{\mu_0}{4 \pi} \frac{2 i_0^2}{R} $$

which is precisely what the classical E&M textbooks say is the magnetic force (per unit length) between two parallel conductors, separated by $ R \ $, with identical current $ i_0 \ $.

  • $\begingroup$ "If the lines of charge are moving together past the observer at some velocity, " How is that possible? Two lines were moving in opposite directions. How could they move together and that too past an observer $\endgroup$ Commented Jul 27, 2017 at 3:36
  • $\begingroup$ well, @AnubhavGoel, it's a thought experiment. i dunno where you get an infinite amount of charge to make an infinite line of charge, let alone two infinite lines of charge. but the thought experiment is to let them both move together in the same direction (so the return current path would have to be behind you by a couple of lightyears). an observer moving with the lines of charge observes them as stationary. but for the other observer (that observes them moving), that observer will see the moving observer's clock ticking more slowly. $\endgroup$ Commented Jul 27, 2017 at 5:02
  • $\begingroup$ Ok.. You mean by cgarge moving that current is flowing i. e. charge is most moving along its axis? $\endgroup$ Commented Jul 27, 2017 at 6:25
  • $\begingroup$ i mean that when the infinite lines of charge move, that constitutes a current. $\endgroup$ Commented Jul 27, 2017 at 7:00
  • $\begingroup$ Along or perpendicular to direction of line? $\endgroup$ Commented Jul 27, 2017 at 7:20

This answer has been hinted at in the others, but it's worth stating their collective knowledge as a succinct one liner that every physicist should know:

Electric and Magnetic force only make sense in the light of special relativity if they are unified because if they were thought of as separate entities, then relatively moving observers would reach different conclusions about the causes of any electromagnetic interaction observed by both observers.

Otherwise put, as stated in garyp's answer, a wholly electric field to one observer is measured as an electric and magnetic field to another, and contrariwise.

But let's simply focus on one, fundamental and highly "visceral" (in the sense of being easily observed in a highschool physics laboratory) result. The qualitative form of the Lorentz force law, leaving Maxwell's equation's aside altogether. That is, the force on a charged particle is a homogeneous, linear function of the particle's velocity. From this statement and the postulate of the constancy of the speed of light alone, it follows that the electromagnetic field must be a rank 2 two tensor, antisymmetric in its covariant form. So electricity and magnetism get welded into one entity that mixes them up as it is Lorentz transformed, and we know this from a simple highschool lab experiment even before we try to guess at one symbol of Maxwell's equations. See my answer here for how this conclusion is reached. One can even argue the antisymmetry from the highschool lab alone without relativity: the fact that charged particles move in circular / helical paths in magnetic fields (i.e. you can guess at a glance that force is at right-angles to the velocity without doing any detailed measurement simply by looking at a cathode ray tube[see footnote 1]) and that a stationary charge is uninfluenced by a static magnetic field but feels an electric field shows that the matrix of the transformation $v^\mu\mapsto F^\mu{}_\nu\,v^\mu$ is antisymmetric when the raised index is lowered ($F^\mu{}_\nu\mapsto g_{\mu\,\sigma}\, F^\sigma{}_\nu$), although this last statement assumes that you already know that the notion of velocity is broadened to a four-veclocity.

Of course, there is also the great body of knowledge before special relativity that betokened unification, as outlined in Robin Eckman's answer. One cannot forget to cite the great Faraday in any discussion like this one.

[1]: You can still get hold of spherical cathode ray tubes like this wonderful Teltron one, they sometimes come up on Ebay and other second hand sites for modest prices. The lovely thing about these big spherical ones is that they are designed for demonstrating the measurement of electron charge to mass ratio from the curvature of electron paths, and you can curl the electron beam nearly into a complete circle with a strong enough magnet, making for a thorough compelling demonstration of the qualitative aspects of the Lorentz law I discuss. I find it's quite something to stare at the circular paths and to ponder that I can actually see the form of the Faraday tensor in a piece of laboratory apparatus. You can also get the Teltron tubes filled with argon as mine is, meaning you'll get really funky purple electron paths which are a little unusual these days.

  • $\begingroup$ Just a simple objection: even though the forces are completely different and there are no magnetic monopoles, through ad hoc manipulations one can transform one force into an[y ]other in another system, but one force becomes or looks like the other in another system. The crucial point is that , here, in one system there is one force and in the other there are two. Therefore the two forces cannot be one ane the same seen by different observers in different frames $\endgroup$
    – user104372
    Commented May 23, 2016 at 5:44
  • $\begingroup$ @user11374 Let me think on that one when my brain is less fried than it is right now and get back to you! $\endgroup$ Commented May 23, 2016 at 6:01

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