# Tag Info

0

No, because the coefficients of any Fourier expansion are usually independent of one other (unless some further conditions hold). The decomposition into real and imaginary part (or equivalently sine and cosine) is a special case of a more general Fourier decomposition $$\psi(\textbf{x},t)=\int ... 2 \nabla\cdot\mathbf B=0 does indicate that there are no magnetic monopoles, so there isn't a "starting" or "ending" point for field lines is mostly correct. So this must mean that magnetic field lines either form closed loops extend to infinity intersect the domain boundary (wall, stellar surface, etc) So the "starting & ending points" issue ... 8 One must be very careful in making the step from \nabla\cdot\mathbf{B}=0 to a statement such as "magnetic field lines do not start or end". Consider the field in the region of an X-point type magnetic null (in two dimensions). Take a 'volume' (i.e. an area) centred on the null point, and look at the field lines through the bounding curve. No matter how ... 0 The electric field around high voltage transmission lines (or "high tension" lines) is extremely high, and can be close to the breakdown threshold of air. That's why the highest voltage lines use multiple (often three) parallel conductors, to increase the effective conductor radius and reduce the peak electric field. Now, introduce a human into that field, ... 8 I spent a long time researching this question for Carver Mead (mentioned by Art Brown) in 2008, because we were both curious what Feynman meant. Carver thought Feynman's "better way of presenting electrodynamics" would be something along the lines of his own "Collective Electrodynamics," but that turned out to be only partly true, as I discovered in four ... 1 Your sign is wrong when computing$$\frac{\partial{(B^2)}}{\partial{(\partial_{y} A_x)}}.$$The only term in B^2 that contains \partial_{y} A_x  is (B_z)^2 = (\partial_{x}A_y - \partial_{y} A_x)^2 , and clearly by the chain rule,$$ \frac{\partial{(B_z)^2}}{\partial(\partial_{y}A_x)} = -2B_z which disagrees with what you have by a sign. Fixing ...

0

Where do Maxwell's equations come from? Oliver Heaviside. They aren't Maxwell's equations. What Anna referred to about Coulomb, Ørsted, Gauss, Biot, Savart, Ampère, and Faraday is all well and good, but see where it says this in the Wikipedia article: "The powerful and most widely familiar form of Maxwell's equations, whose formulation is due to Oliver ...

3

The answer lies in another wiki article, In electromagnetism, one of the fundamental fields of physics, the introduction of Maxwell's equations (mainly in "A Dynamical Theory of the Electromagnetic Field") was one of the most important aggregations of empirical facts in the history of physics. It took place in the nineteenth century, starting from basic ...

0

Michael Faraday by L. Pearce Williams. Lots of biographical information but also detailed explications of his work referencing his published papers and diary. No math though, since Faraday wasn't a mathematician, but he was a great experimentalist!

0

There are two solutions to this: Ampère can only be used in case of a current uniform in time, thus you wont get this problem. Actually alhough Ampère can only be used with a uniform current, I have just recently read an article (unfortunately only available in Hungarian) that the Biot-Savart law is also true if $\frac{\text{d}\vec{\mathbf{D}}}{\text{d}t}$ ...

Top 50 recent answers are included