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It is sometimes said that a point charge is equivalent to an electric current. If it were a steady current, I should be able to find it from Ampere’s law or Biot-Savart’s law. Even if the current is time dependent I have to first determine $\vec J(\vec r,t)$ and then use $$\vec \nabla\times \vec B=\mu_0\vec J+\mu_0\epsilon_0\frac{\partial \vec E}{\partial t}$$ However, I find it difficult to think of a moving charge in a straight line (or any other open trajectory) not only as steady current but also I think that the concept of current itself is not well-defined in this case. By current we mean putting a surface perpendicular to the trajectory and calculate the current by counting number of charges flowing through it in a given time and then by dividing that by the time of observation. In this case $q$ is fixed, but time of observation $t$ is not, the current not a very meaningful object here. Therefore, I think the magnetic field produced by a moving charge cannot be calculated from Biot-Savart’s law. I think the magnetic field has to calculated as follows. A moving charge has a time-dependent charge density, which creates a time dependent electric field and the time-variation of the electric field is what creates this magnetic field from the equation $$\vec \nabla\times \vec B=\mu_0\vec J+\mu_0\epsilon_0\frac{\partial \vec E}{\partial t}$$ through the second term and I think $\vec J$ is irrelevant in this case. Am I correct?

I think I can use $\nabla\cdot\vec E=\frac{\rho(\vec r,t)}{\epsilon_0}$ to find out $\vec E(\vec r,t)$ and then put in $$\vec \nabla\times \vec B=\mu_0\epsilon_0\frac{\partial \vec E}{\partial t}$$ to determine $\vec B$? But I find it equally difficult to solve this. I cannot make out what should be the boundary conditions to solve this differential equations. For a charge moving in a straight line with uniform velocity $\vec v$ I think I can take the charge density to be $\rho(\vec r,t)=q\delta^{(3)}(\vec r-(\vec r_0+\vec vt))$ at time $t$, where $\vec r_0$ is the position of the charge at $t=0$.

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  • $\begingroup$ You can make your life easier and take the origin of coordinates at $\vec r_0$. $\endgroup$ – Sofia Feb 13 '15 at 14:47
  • $\begingroup$ You work too hard. Can you get the "Field Theory" of Landau and Lifschitz? You will find there the calculus of the fields $\vec E$ and $\vec H$ produced by a moving charge, in the chapter V, paragraph 38 (Field of a charge in uniform movement). $\endgroup$ – Sofia Feb 13 '15 at 16:07
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However, I find it difficult to think of a moving charge in a straight line (or any other open trajectory) not only as steady current but also I think that the concept of current itself is not well-defined in this case.

The current density is well defined in this case, and is rather straightforward. For any moving charge density we have $J=\rho v$. For a point charge it is particularly simple. $$J=q \delta(r-r_q(t)) v_q(t)$$

From there you can solve Maxwell’s equations in the standard manner. The general solution is called the Lienard Wiechert potentials and is the full relativistic fields created by an arbitrarily moving point charge.

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The definition of electric current came from observations:

Electric current is the rate of charge flow past a given point in an electric circuit, measured in Coulombs/second which is named Amperes.

From the microscopic view of current

driftvelocitycurrent

There is no problem with a single ion of velocity v measured passing a given x point in vacuum, a value can be defined.

The electron taken as an elementary point particle presents a problem until remembering it is a quantum mechanical entity, the uncertainty principle will give an area to enter the equation, and should also be used for ions. Where the charge particle passes, the effects of current can be seen.

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  • $\begingroup$ We could've also proceeded from $I=Q/t$ to $I=Q/(d/v)$ to $I d = Q v$. To me that seems the straightest way to observe that current in a length of wire is the same thing as that charge moving with some velocity. $\endgroup$ – Džuris Aug 8 at 23:43
  • $\begingroup$ @Džuris except that a length of wire has zillions of charges in the lattice and one electron cannot really observed, just the statistical effect of many can. $\endgroup$ – anna v Aug 9 at 3:12

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