# Problem related to application of Maxwell's equation for point charge moving uniformly

Maxwell's 4th equation which describes magnetic field, has two terms:

$$\oint \mathbf{B}\cdot d\mathbf{l}=\mu I+\mu \varepsilon \frac{\mathrm{d}\Phi}{\mathrm{d}t}$$

Now, I wanted to derive the magnetic field caused by a uniformly moving point particle and was able to derive Biot Savart law using just the second term of the above equation, which gives:

$$\oint \mathbf{B}\cdot d\mathbf{l}=\mu \varepsilon \frac{d\Phi }{dt} = \frac{\mu q \mathbf{v} \times \mathbf{r} }{2{\pi}\mathbf{r}^3 }$$

Here, $q$ and $\bf v$ are the charge and velocity of the particle.

Now I realize that for a point particle, the magnetic field will mainly be caused only by the changing electric field. But that is not entirely true. When the particle perpendicularly passes a plane, it acts as a temporary current of magnitude ${\mu}.I$. passing across the plane. So, for all these planes where the particle is passing, there should be temporarily 'extra' magnetic field which is not predicted by Biot-Savart law. Hence, when a point charge passes a plane, the magnetic field induced on the plane will be given by:

$$\oint \mathbf {B}\cdot d\mathbf{l}= \frac{\mu.q.\mathbf{v} }{2\pi\mathbf{r}} +\frac{\mu.q. \mathbf{v}}{4{\pi}\mathbf{r}^2 }$$

This added term is not insignificantly small to not to be detected in experiments, so I assume that the first term in equation (1) plays no role in describing magnetism caused by a point charge. So, what does equation (1) means by current 'I'? Continuous current traditionally seen in electric wires? But there exist no continuous currents as at small enough scales we can always view current as caused by distinct separable electrons. So is this why Maxwell's equations are called approximate laws?

Thank you. If I did not make my question clear, tell me and I will explain.

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Edit: Solved the problem. Maxwell's equation is correct after all!

In case of point charge, the current caused by the charge plays no role in the magnetic field around it because the charge is a point, and it would pass a plane in 0 time since point is very small.

In case the charge is bigger than point, the current as well as rate of change of electric flux will together cause magnetism such that the result is again equal to biot-savart law for complicated reasons.

• I was going to fix up some of your formatting, but I'm not clear on something: what do things like $\vec{r^2}$ and $\vec{r^3}$ mean? $\vec{r^2} = \vec{r}\cdot\vec{r} = \lVert\vec{r}\rVert^2$ (a scalar), or $\vec{r^2} = \lVert\vec{r}\rVert\vec{r}$ (a vector), or something else? And $\vec{r^3} = \lVert\vec{r}\rVert^2\vec{r}$ (a vector), or $\vec{r^3} = \lVert\vec{r}\rVert^3$ (a scalar), or $\vec{r^3} = (\vec{r}\otimes\vec{r})\cdot\vec{r}$ (another vector), or something else? – David Z Dec 8 '15 at 9:25
• Ok. Vector.vector= scalar. I Forgot that. Sorry and thanks. – Prem kumar Dec 8 '15 at 9:59
• @Raja - Are you trying to divide by 3-vectors? Regardless, I would recommend Jackson's E&M book, specifically pages 248-258 of the third edition (i.e., blue cover). There he goes through a very detailed explanation of why and how one can go from a truly microscopic description of Maxwell's equations to the macroscopic version most people are used to and use... – honeste_vivere Dec 8 '15 at 12:57

The derivation of the electromagnetic field generated by a moving point charge is fully calculated and discussed as Lienard-Wiechert potential. Essentially, one has to take into account the retarded effect of field propagation as stated in special relativity: since the particle is moving, it takes a finite amount of time for the field to propagate and one cannot just assume true the expression for the electric field generated by a static point particle and take derivatives thereof (as you have done above).

There will by all means be two contributions playing the game: the moving particle, as a current, generates a magnetic field (which propagates in space and time) on top of an electric field (because it is still a charge, after all). Those two contributions can be described by means of a retarded potential as pair $(\varphi(x,t), \mathbf{A}(x,t))$, with the resulting electric and magnetic fields to be derived accordingly.

Manifestly covariant formalism will make things pretty easy to calculate, and a general walkthrough can be found in standard literature as, for example, the following$^1$.

So is this why Maxwell's equations are called approximate laws?

Maxwell's equations are not approximate laws. Covariantly following all the steps after having chosen an appropriate inertial reference frame precisely works out what the result must be.

• Note, that there is a simpler solution than the Lienard-Wiechert potentials in this case: Lorentz transformation of the field in the rest frame of the particle. – Sebastian Riese Dec 8 '15 at 12:58
• Oh yes, of course, that would be far easier. – gented Dec 8 '15 at 13:10

I want to address the biggest problem with your work. Namely, magnetic fields are not caused by changing magnetic fields.

This is a surprisingly common misconception that seems to be entirely due to lazy notation for flux. Magnetic flux could be written as $\Phi_B$ and then you can get $$\int\vec E\cdot d\vec \ell=-\frac{d}{dt}\Phi_B.$$

Electric flux could be written as $\Phi_E$ and then you can get $$\int\vec B\cdot d\vec \ell=\mu_0I_\text{enc}+\mu_0\epsilon_0\frac{d}{dt}\Phi_E.$$

And now it is clear from the first equation that changing magnetic fields are associated with circulating electric fields. And it is equally clear from the second equation that changing electric fields are associated with circulating magnetic fields.

Absolutely everything else you wrote is also completely and totally wrong.

• Thanks for picking that basic error, but actually the error was not due a misconception of mine. I wrote- magnetic field is caused by changing 'magnetic' field. I 'meant' magnetic field is caused by changing 'electric' field, and was surprised to find the mistake today. The rest of the question still makes sense to me. What I actually ask is that suppose a charged particle perpendicularly approaches a closed circular loop. Then the magnetic field on any point of the loop will equal as given by biot-savart's law if i just use the second term of Maxwell's 4th equation. – Prem kumar Dec 15 '15 at 2:43