Magnetic field of a moving charge and Maxwell's equations

[hokhani]

How can we calculate the magnetic field of moving charge by the Maxwell equation $\nabla \times H=J+\frac {\partial D} {\partial t}$? I mean which term, $J$ or $\frac {\partial D} {\partial t}$, should be taken into account in calculations? The first, second, or both? Can we deal with the moving charge as a current wire with the magnetic field $B=\frac{\mu_0 I}{2\pi r}$? If yes, what about the term $\frac {\partial D} {\partial t}$?

 

[Ibix]

You can ignore terms that are zero, as with any equation. Do you think either of those terms will be zero?

 

[Charles Link]

The complete Maxwell equation is $\nabla \times B=\mu_o J_{total}+\mu_o \epsilon_o \dot{E}$. $\\$ If the $\dot{E}$ is ignored, the result is the curl equation $\nabla \times B=\mu_o J_{total}$. $\\$ This (inhomogeneous) curl differential (simplified) Maxwell's equation has an integral solution which is Biot -Savart's law: $\vec{B}(x)=\frac{\mu_o}{4 \pi} \int \frac{ \vec{J}_{total}(x') \times (x-x')}{ |x-x'|^3} \, d^3x'$. $\\$ If you do the same thing with the $H$, (in this case the $B$ without the $\mu_o$), there will sometimes be a homogeneous solution to $\nabla \times H=0$ that needs to be included, but not in this case. $\\$ [The homogeneous solution to $H$ arises in problems involving magnetic materials with magnetic poles, but is not applicable here. In addition, the equation $\nabla \times H=\mu_o J_{free}+\frac{\partial{D}}{\partial{t}}$ is a form of $\nabla \times B=\mu_o J_{total}+\mu_o \epsilon_o \dot{E}$ , where $J_{total}=J_{free}+J_m+J_p$. Starting with $B=\mu_o H+M$, and using $\nabla \times M= \mu_o J_m$ and $J_p=\dot{P}$, this form of Maxwell's equation with $\nabla \times H$ arises.(The $\nabla \times M$ cancels the $\mu_oJ_m$ leaving $\nabla \times H$ and $J_{free}$). It really is unnecessary to employ the $\nabla \times H$ form of the equation, unless you have magnetic materials present, in which case the $\nabla \times H$ equation can be quite useful, e.g. in working with transformers, (and deriving the MMF equation), where the $J_{free}$ is the current in the windings ]. $\\$ Oftentimes in the textbooks they present Ampere's law (using Stokes' theorem): $\int \nabla \times B \cdot dA= \oint B \cdot dl=\mu_o I$ as being an integral form of Maxwell's equation, but Biot-Savart's law is also an integral solution of the Maxwell $\nabla \times \vec{B}$ equation. $\\$ If the $\dot{E}$ needs to be taken into account, the result requires a Lienard-Wiechart type solution.

 

[Cryo]

There are many levels to this problem. Firstly, is the charge accelerating? If not, then don't waste your time solving differential equations, simply find the field in rest frame of the charge:

$\mathbf{E}=\frac{e\mathbf{r}}{4\pi\epsilon_0 r^3}$
$\mathbf{B}=\mathbf{0}$

Then boost into the reference frame where the charge is moving https://en.wikipedia.org/wiki/Classical_electromagnetism_and_special_relativity#The_E_and_B_fields

If the charge is accelerating, life can get progressively more difficult. If it is not accelerting too much, and not moving too fast, you can use [SIZE=4][B][URL='https://www.physicsforums.com/members/charles-link.583509/']Charles Link[/URL][/B][/SIZE] approach. Beyond this, you need to start looking into Lienard-Wiechert potential https://en.wikipedia.org/wiki/Liénard–Wiechert_potential.

Going further still, if acceleration is so strong that the radiation produced by the accelerating charge starts carrying momentum comparable to the momentum of the charge, you need serious electrodynamics texts, such as Jackson's "Classical Electrodynamics". As far as I remember, this problem does not have a general solution, only some special cases are covered.

 

[Delta2]

How can we calculate the magnetic field of moving charge by the Maxwell equation $\nabla \times H=J+\frac {\partial D} {\partial t}$? I mean which term, $J$ or $\frac {\partial D} {\partial t}$, should be taken into account in calculations? The first, second, or both? Can we deal with the moving charge as a current wire with the magnetic field $B=\frac{\mu_0 I}{2\pi r}$? If yes, what about the term $\frac {\partial D} {\partial t}$?

As other have noted and I also recommend, read a book or Wikipedia's entry on Lienard-Wiechert potential, there you will find out what current density and what charge density we have for the case of a moving point charge. The current is not exactly the same as a long straight wire with current I, it involves dirac delta functions and the velocity of the point charge...
My only other comment would be that @Charles Link solution using the Biot-Savart law is valid as an approximation only when the acceleration of the point charge is relatively small, so we can consider the quasi-static approximation.

 

[hokhani]

There are many levels to this problem. Firstly, is the charge accelerating? If not, then don't waste your time solving differential equations, simply find the field in rest frame of the charge:

$\mathbf{E}=\frac{e\mathbf{r}}{4\pi\epsilon_0 r^3}$
$\mathbf{B}=\mathbf{0}$

Thank you. By this approach, could you please prove that the force exerted on a moving charge, in a constant magnetic field B, is $F=qv \times B$?

 

[hokhani]

As other have noted and I also recommend, read a book or Wikipedia's entry on Lienard-Wiechert potential, there you will find out what current density and what charge density we have for the case of a moving point charge. The current is not exactly the same as a long straight wire with current I, it involves dirac delta functions and the velocity of the point charge...
My only other comment would be that @Charles Link solution using the Biot-Savart law is valid as an approximation only when the acceleration of the point charge is relatively small, so we can consider the quasi-static approximation.

Thanks. Do you mean that the magnetic field of a current carrying wire is quite different from the magnetic filed of a moving charge?

 

[hokhani]

The complete Maxwell equation is $\nabla \times B=\mu_o J_{total}+\mu_o \epsilon_o \dot{E}$. $\\$ If the $\dot{E}$ is ignored, the result is the curl equation $\nabla \times B=\mu_o J_{total}$. $\\$ This (inhomogeneous) curl differential (simplified) Maxwell's equation has an integral solution which is Biot -Savart's law: $\vec{B}(x)=\frac{\mu_o}{4 \pi} \int \frac{ \vec{J}_{total}(x') \times (x-x')}{ |x-x'|^3} \, d^3x'$. $\\$ If you do the same thing with the $H$, (in this case the $B$ without the $\mu_o$), there will sometimes be a homogeneous solution to $\nabla \times H=0$ that needs to be included, but not in this case. $\\$ [The homogeneous solution to $H$ arises in problems involving magnetic materials with magnetic poles, but is not applicable here. In addition, the equation $\nabla \times H=\mu_o J_{free}+\frac{\partial{D}}{\partial{t}}$ is a form of $\nabla \times B=\mu_o J_{total}+\mu_o \epsilon_o \dot{E}$ , where $J_{total}=J_{free}+J_m+J_p$. Starting with $B=\mu_o H+M$, and using $\nabla \times M= \mu_o J_m$ and $J_p=\dot{P}$, this form of Maxwell's equation with $\nabla \times H$ arises.(The $\nabla \times M$ cancels the $\mu_oJ_m$ leaving $\nabla \times H$ and $J_{free}$). It really is unnecessary to employ the $\nabla \times H$ form of the equation, unless you have magnetic materials present, in which case the $\nabla \times H$ equation can be quite useful, e.g. in working with transformers, (and deriving the MMF equation), where the $J_{free}$ is the current in the windings ]. $\\$ Oftentimes in the textbooks they present Ampere's law (using Stokes' theorem): $\int \nabla \times B \cdot dA= \oint B \cdot dl=\mu_o I$ as being an integral form of Maxwell's equation, but Biot-Savart's law is also an integral solution of the Maxwell $\nabla \times \vec{B}$ equation. $\\$ If the $\dot{E}$ needs to be taken into account, the result requires a Lienard-Wiechart type solution.

Thanks. Consider the calculation of the magnetic field at a point in the vacuum, around the moving charge with constant velocity. There are two types of points where we should calculate the magnetic field. One, is the instant location of the charge where we have both terms $\mu_o J_{free}$ and $\frac{\partial{D}}{\partial{t}}$ present. The other point is out of the location of the charge where we have only the second term, $\frac{\partial{D}}{\partial{t}}$. Is the magnetic field different at the two points?

 

[Cryo]

Thanks. Do you mean that the magnetic field of a current carrying wire is quite different from the magnetic filed of a moving charge?

Of course. Magnetic field from a single moving charge will have a lot of complicated features. When you sum fields from many moving charges, to get the field from a current-carrying wire, you loose a lot of these features.

For example. The field from a single moving charge will decay as $1/r^2$ with the distance ($r$) between the observer and the charge. The field from the (straight) current-carrying wire will decay as $1/\rho$ with the (shortest) distance ($\rho$) between the observer and the wire.

 

[Cryo]

Thank you. By this approach, could you please prove that the force exerted on a moving charge, in a constant magnetic field B, is $F=qv \times B$?

I was answering the question about calculating magnetic field, not the force.

You can try to use this approach to get the force (i.e. use Coulomb force to get Lorentz force via boost), but it will be messy. You will not be able to isolate Coulomb force from Lorentz force completely.

If I wanted to derive the equation of motion for the charged particle in electromagnetic field (i.e. get the force), I would start with the Lagrangian for the field + the charge and then extract equations by extremizing the action. I am not sure this is the route you would want to take.

 

[Charles Link]

Thanks. Consider the calculation of the magnetic field at a point in the vacuum, around the moving charge with constant velocity. There are two types of points where we should calculate the magnetic field. One, is the instant location of the charge where we have both terms $\mu_o J_{free}$ and $\frac{\partial{D}}{\partial{t}}$ present. The other point is out of the location of the charge where we have only the second term, $\frac{\partial{D}}{\partial{t}}$. Is the magnetic field different at the two points?

I think you need to think a little about what the equation $\nabla \times H=J_{free}+\frac{\partial{D}}{\partial{t}}$ or even the equation $\nabla \times B=\mu_o J_{total}+\mu_o \epsilon_o \frac{\partial{E}}{\partial{t}}$ represent. Just because $J$ is absent in a region does not mean that the $J$ outside that region doesn't cause a $B$ in that region. For a single point charge, in general $J$ is absent almost everywhere, but a non-zero $E$ and $B$ still exist everywhere. $\\$ Meanwhile in the region where the $J$ of a single charge is present, the equations (e.g. Biot-Savart) break down at that point because they predict $\pm \infty$ for $E$ and $B$ even though for real particles, $E$ and $B$ are never infinite. $\\$ For many problems that are considered (e.g. currents in conductors), $J$ represents a macroscopic average, and any possible point-like behavior of the charged particles is ignored. In any case, $J$ can be zero in a region, but to solve for $B$ using $\nabla \times B=\mu_o J +...$ you do need to find the entire spatial solution for $B$. You can't just say since $J=0$ there, that we are solving $\nabla \times B=0$ at that point.

 

[vanhees71]

Well, first of all there's no electric and magnetic field but only one electromagnetic field. For a moving point charge (neglecting the not completely solved problem of radiation reaction) you need to write down the Lienard-Wiechert potentials or equivalently the Jefimenko equations for the field components.

If it's a uniformly moving charge, there's a tremendous shortcut. It's just a Lorentz boost of a charge at rest. The pontial in the rest frame is given by
$$A^{\mu} = \begin{pmatrix} \phi(\vec{x}) \\ 0 \\ 0 \\0 \end{pmatrix}.$$
Here (in Heaviside-Lorentz units)
$$\phi(\vec{x})=\frac{q}{4 \pi |\vec{x}|}.$$
This can be written in manifestly covariant form
$$A^{\mu}=u^{\mu} \frac{q}{4 \pi \sqrt{(u \cdot x)^2-x \cdot x},$$
where $u$ is the four-velocity of the particle, which in the rest frame of the particle is $u=(1,0,0,0)$ and in an arbitrary inertial frame $u=(\gamma,\gamma \vec{v}/c)$ with the usual (constant) three-velocity of the particle.