Tag Info

New answers tagged

2

A constant charge density does not imply a zero magnetic field. Even considering a set of isolated charges, suppose they were (mechanically) moved along a circular path. The charge density could remain the same but there would be a current flow. The curl of the magnetic field produced would be $\mu_0 \vec{J}$, where $\vec{J}$ is the current density. If the ...


2

Ampere's law (for a steady current) states that $$ \oint \vec{B} \cdot d\vec{l} = \mu_0 I$$ If we consider an infinite wire, then symmetry tells us that the B-field at the point $A$ and all other points on a circle of radius $(R+y)$ is constant in magnitude and is in the azimuthal direction. Hence the magnitude of the B-field is given by $$ 2 \pi (R+y)B = ...


1

As you can see from an example of electrodynamics book by Griffiths: For ANY wire,equation (5.35) still holds.And you can see that for an infinite wire,θ1=π/2 and θ2=-π/2.So,in your situation you can use only θ1 οr θ2 by changing the other angle to zero.It does not matter which angle you keep,mathematically,they will both give you a B of the same magnitude ...


0

You are working in Coulomb gauge, which is non-standard in radiation problems, so it's a little difficult for me to give you an in-depth analysis of your claims, but I can point out some preliminary mistakes straight away. Firstly, your assumption that the grad and curl terms fall off faster than $1/r^2$ is not a valid assumption in the time-dependent case. ...


0

Firstly, I think you've confused several points in your analysis of the situation. Working with the first picture, by standard electrostatic analysis, we see that there must be a standard Coulomb force between the plates for all times $t\ge t_2$, not just at $t=t_2$. This point is not too important, but I thought I'd mention it. Continuing to work with the ...


2

Of course it is consistent. As commented, this is exactly the equation for a conducting medium. $$ \nabla\times\mathbf B = \mu\sigma\mathbf E + \mu\epsilon\frac{\partial\mathbf E}{\partial t} $$ For instance, this is exactly the equation used to derive the wave equation in a conducting medium with no charge density: $$ \nabla^2\mathbf E = ...


1

Outside of the wire where $r>R$, the curl of the B-field is zero (no current density) and inside, it is non-zero. Using symmetry, we know that if $\vec{j}$ only depends on r, which is often the case, then the B-field must also only depend on r. $\vec{B}=\vec{B}(r)$. If we look at the curl in cylindrical coordinates, we find that $\nabla \times ...


1

The answer to As an electron drops from a higher energy level to a lower energy level, can it be modeled as a the continuous movement of a charged body, therefore causing a magnetic field to be generated around it? is "Yes, but only trivially." That is, you could probably work backwards from the far-field radiation to some imagined moving source ...


3

$\newcommand{\pdv}[2]{\frac{\partial #1}{\partial #2}} \newcommand{\div}[1]{\nabla \cdot #1} \newcommand{\curl}[1]{\nabla \times #1} $ Until Maxwell added some terms to the set of equations they were Gauss' Law, Ampere's Law, Faraday's Law, and the fact that there is no magnetic monopole. The equations then read: $$\div E = \rho / \varepsilon_0$$ $$\div B ...


0

We split $\rho=\rho_{\it ex}+\tilde\rho$ and $\jmath=\jmath_{\it ex}+\tilde\jmath$ into "explicit" and "medium" contributions. We can then define $P$ and $M$ by $\tilde\rho=-\nabla P$ and $\tilde\jmath=\nabla\times M$, and macroscopic fields by $D=E+P$ (I'm sloppy about units, there is a $4\pi$ in Gauss units) and $B=H+M$. The macroscopic equations are then ...



Top 50 recent answers are included