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Static electromagnetic fields implies: $$\frac{\partial\mathbf E}{\partial t} = 0 \quad\mbox{ and }\quad \frac{\partial\mathbf B}{\partial t} = 0$$ This means for electrostatics: $$\nabla\cdot\mathbf E = \frac{\rho}{\epsilon_0}, \quad \nabla\times\mathbf E = 0 \quad$$ And for magnetostatics: $$\nabla\cdot\mathbf B = 0, \quad \nabla\times\mathbf B = ... 1 I guess different authors use different definitions. For me, it is that the E- and B-fields do not have time derivatives, hence curl free, conservative E-fields and B-fields that can depend only on steady currents. The condition that the divergence of \partial {\bf E}/\partial t = 0 is not the same thing. The E-field could be time variable and have this ... 0 Electrostatic and magnetostatic are specific cases of the general electromagnetism. Defining a special case does not require to know a law/model that rules the phenomena. I don't need maxwell equations to define electrostatics or magnetostatics. I only need them if I want to know that my choice of special case is clever or useless. For instance, I can ... 0 It's hard to understand what you're asking, but I believe you're wondering how Maxwell modified Ampere's Law? Originally, Ampere's Law was:$$\nabla \times \mathbf{B} = \mu_0 \mathbf{j}$$Maxwell noticed that, with this alone, you could have weird things happen like energy conservation violation. To get rid of this problem he added another term to the ... 0 This looks an incredibly difficult way of solving what is quite an easy problem. If you use Faraday's law in integral form, constructing a small, rectangular loop that goes into and out of the interface, it is easy to show that the component of the E-field that is perpendicular to the normal surface vector (i.e. the E-field parallel to the interface plane) ... 1 What York is doing is not "physics" but an ancient antenna engineering trick whereby one first solves for the field of the linear electric Hertzian dipole and then to solve for the radiated field of a closed loop current one considers the latter as a formally equivalent "magnetic dipole" source resulting in the interchange of E and B everywhere. The same ... 2 If we let \mu_0=1, \epsilon_0 =1 (adopting a system of units where c=1), then Maxwell's equations become completely symmetric to the exchange of {\bf E} and {\bf B} via a rotation (see below).$$ \nabla \cdot {\bf E} = 0\ \ \ \ \ \ \nabla \cdot {\bf B} =0 \nabla \times {\bf E} = -\frac{\partial {\bf B}}{\partial t}\ \ \ \ \ \ \nabla \times ...

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Under what additional assumptions can $\vec{M}$ be made zero so that the conventional maxwell's equation is consistent with the extended maxwell's equation? The standard electromagnetic theory assumes there are no magnetic charges and no corresponding current density. The author of the text you referenced included $\vec{M}$ there for unclear reasons. ...

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If it is eventually found that magnetic monopoles exist, then M would not be always zero and Maxell's equations would become fully symmetrycal on E and B.

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A picture is worth a thousand words. Here's how it looks as a function of space, evolving in time: Here blue is real part, and purple is imaginary part of the complex exponent $\exp(i(kx-\omega t))$. If you instead just look at $\exp(-i\omega t)$, you'll get this:

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If you look at a wave at a moment in time, you can see how it varies spatially by plugging in different values of r: $e^{ikr}$. If you look at a point in space, you can see how it varies in time by fixing r and varying t: $e^{-i\omega t}$. If you want the behavior in both space and time, you end up with the expression you have - and you can see how the ...

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The covariant formulation of EM is precisely this. The formulation as a gauge theory also does this. ($c = 1$ in the following) Given the $E$- and $B$-fields as spatial three-vectors in some frame, we construct the antisymmetric field strength tensor as (roman indices are spatial indices, summation over repeated indices implied) $$F^{0i} := E^i \; ... 5 Maxwell's equations in vacuum are symmetric bar the problem with units that you have identified. In SI units$$ \nabla \cdot {\bf E} = 0\ \ \ \ \ \ \nabla \cdot {\bf B} =0 \nabla \times {\bf E} = -\frac{\partial {\bf B}}{\partial t}\ \ \ \ \ \ \nabla \times {\bf B} = \mu_0 \epsilon_0 \frac{\partial {\bf E}}{\partial t} If we let $\mu_0=1$, ...

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