my questions is, what is the time dependent part of the $\vec B$ field that is not determined by Faraday's law? The problem I am working on states that we should 'set it to zero', but I don't see it.

Given $$\vec E (r, \theta, \phi, t) = A \frac{\sin \theta}{r} \cos(kr - \omega t) \hat \phi.$$

I use Faraday's law $\vec \nabla \times \vec E = - \frac{\partial \vec B}{\partial t}$ and the expression of the curl in spherical polar coordinates to find that;

$$\vec \nabla \times \vec E = \frac{2A \cos \theta}{r^2} \cos(kr - \omega t) \hat r + kA \sin \theta \sin(kr - \omega t) \hat \theta.$$

Integrating with respect to time to find $\vec B$ yields;

\begin{align} \vec B &= - \left[\frac{2A \cos \theta}{r^2} \hat r \int \cos(kr - \omega t)dt + kA \sin \theta \hat \theta \int \sin(kr - \omega t)dt\right] \\ &= \frac{2A \cos \theta}{r^2 \omega} \sin(kr - \omega t) \hat r - \frac{kA \sin \theta}{\omega} \cos(kr - \omega t) \hat \theta + C\end{align}

Assuming that what I have done is okay, what is meant by a time dependent component that is not determined by Faraday's law?

Both components of the magnetic field are time dependent and they are determined. I hope my question is clear.



closed as unclear what you're asking by Rob Jeffries, user36790, Kyle Kanos, Gert, JamalS Dec 4 '15 at 22:14

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    $\begingroup$ This is unclear. Given a time-dependent electric field, Faraday's law completely determines $\vec B$ up to a time-independent, magnetostatic component. It would help to include a reference to the original text to see exactly what they mean. $\endgroup$ – Emilio Pisanty Dec 3 '15 at 17:11
  • $\begingroup$ This is not from a text book, so I can't reference it. We are using Griffith's Intro to Electrodynamics as the course text book, but I can't find anything in it. I'll stop by my lecturer's office tomorrow and ask him what he means. $\endgroup$ – jm22b Dec 3 '15 at 18:49
  • $\begingroup$ Yes, in that case this is a question best answered by your lecturer. $\endgroup$ – Emilio Pisanty Dec 3 '15 at 18:50
  • $\begingroup$ You have certainly made a mistake, since the second term in your B-field should be inversely proportional to $r$. $\endgroup$ – Rob Jeffries Dec 3 '15 at 22:02
  • $\begingroup$ Thanks @RobJeffries I missed out the 1/r term from the curl expression of the $\hat \theta$ component. $\endgroup$ – jm22b Dec 4 '15 at 12:07

I'm not clear about your question Jacob. There seems to be some confusion somewhere. Maybe on my part. But see the Derivation from electromagnetic theory section of the Wikipedia Electromagnetic radiation article. Note this:

"Also, E and B far-fields in free space, which as wave solutions depend primarily on these two Maxwell equations, are in-phase with each other. This is guaranteed since the generic wave solution is first order in both space and time, and the curl operator on one side of these equations results in first-order spatial derivatives of the wave solution, while the time-derivative on the other side of the equations, which gives the other field, is first order in time, resulting in the same phase shift for both fields in each mathematical operation."

IMHO it's crucial to appreciate what this is saying. And to do that, we need an analogy. Imagine we have a flat calm ocean, and you're in a canoe. It's an orange canoe. Along comes a wave. This wave has no hump. As it approaches your canoe starts to slope upwards, slowly at first, then faster. Then the slope starts to reduce, and soon your canoe is horizontal on the top of the wave. Then the process is reversed and your canoe slopes downwards before flattening out. Like so:

enter image description here

The slope of your canoe denotes E, and the rate of change of slope denotes B. One is the spatial derivative, the other is the time derivative. This is why $\vec \nabla \times \vec E = - \frac{\partial \vec B}{\partial t}$. The curl of E is the rate of change of B. And since B changes, it is time-dependent. But Faraday's law doesn't determine how, all it does is tell you that for an electromagnetic wave, E and B change together. Because there's only one wave there.

See section 11.10 of Jackson's Classical Electrodynamics where he says "one should properly speak of the electromagnetic field Fμν rather than E or B separately". IMHO this should be in chapter 1, not chapter 11. Also see Wikipedia: the electric and magnetic fields are better thought of as two parts of a greater whole. I dislike the way electromagnetism is taught. Sometimes it feels like Maxwell's unification never happened.

PS : upper image courtesy of mathematica

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    $\begingroup$ This is muddled, not particularly related to the question, and unlikely to help the OP. In addition, parts of this post are only correct in the very restricted setting of plane waves, without stating this restriction, making the post misleading and detrimental to the OP. -1. $\endgroup$ – Emilio Pisanty Dec 4 '15 at 1:27
  • $\begingroup$ This isn't muddled, Emilio, this gets to the heart of it. (mod edit: nonconstructive content removed) If the OP asks for more information, I will provide it. If he revises his question I will append something to my answer. But what I won't do is send him away with a flea in his ear. $\endgroup$ – John Duffield Dec 4 '15 at 8:05
  • $\begingroup$ The curl of E gives the rate of change of B, so is it not the case that the integral wrt time of this function will tell us how B evolves? Your post seems to say this isn't the case... $\endgroup$ – jm22b Dec 4 '15 at 12:03
  • $\begingroup$ @Jacobadtr : it isn't the case. The curl of E is the rate of change of B, so the curl of E tells you how B evolves. Remember that E and B are merely two aspects of the electromagnetic field. See Jefimenko's equations. $\endgroup$ – John Duffield Dec 4 '15 at 15:01
  • $\begingroup$ I'm not sure Jacob. The "set it to zero" suggests E is time independent or static or standing. Can you give me any more information? A reference or a hyperlink? $\endgroup$ – John Duffield Dec 4 '15 at 17:36

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