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In the physical reality, where is the electric field and magnetic field? Do they go up to infinity from the x-axis?

The picture uses vectors to depict the magnitude and direction of the field but we know that a vector can describe a property at just one point. What about other points?

In other words, I am asking what happens if I keep a super tiny compass at the points A, B, C in the above diagram.

*looking for good reference to understand electromagnetic waves (with derivations) with lot of illustrations

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    $\begingroup$ a couple of related things to keep in mind: 1) $\mathbf E$, $\mathbf B$ are vector fields; for any point in space you choose, they give you a magnitude and direction. 2) the arrows do not represent the physical extent of the fields; each arrow represents the field at one specific point (presumably at the ''tail'' of the vectors). Every point in space would have one such vector associated, and if they were all drawn would overlap and make a messy picture. $\endgroup$ – anon01 Feb 16 '17 at 23:52
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Your plane wave picture shows the electric and magnetic fields of the wave at one instant in time. As at any x and t the plane wave fields are constant over the whole y-z plane you will measure the same magnetic field there if you compass needle is fast enough. You should not forget that the magnetic field (like the electric) oscillates also in time according to $B=B_0\exp{(i\omega t)}$, where $\omega$ is the angular frequency of the wave. Therefore, at any point, the direction of the magnetic field also changes with this frequency which is usually so fast that it cannot be measured with a tiny magnetic needle.

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  • $\begingroup$ Of course, the fields are changing but I have no clue how the waves behave in reality. My textbook gives the derivation from Maxwell's equations without any diagram. I have no clue what the hell is going on when Faraday's law & Ampere's Law. I blamed it on my failure to understand how an electromagnetic wave really looks like. $\endgroup$ – Yashas Nov 10 '16 at 15:17
  • $\begingroup$ The field is constant all over the y-z plane? That is incomprehensible. $\endgroup$ – Yashas Nov 10 '16 at 15:18
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    $\begingroup$ @Yashas Samaga - Yes, in a plane wave the fields at any time $t$ are constant in any plane vertical to the propagation direction. This is the reason why it is called a plane wave. Such a plane wave is an approximation of a real wave in a certain region of space. Like, e.g., the spherical wave emitted by a dipole antenna at a large distance. $\endgroup$ – freecharly Nov 10 '16 at 15:25
  • $\begingroup$ all over the y-z plane? The field exists at the other end of the universe? I feel so ashamed to ask these questions. $\endgroup$ – Yashas Nov 10 '16 at 15:38
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    $\begingroup$ @YashasSamaga A plane wave is an idealization... a simple solution to the wave equation. But it is impossible to construct a plane wave with finite equipment in a finite time. It's already too late. Not only is a plane wave infinite in extent in the $xz$ plane, but it also has infinite extent into the past and the future. Plane waves don't exist in nature. $\endgroup$ – garyp Nov 10 '16 at 15:40
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In the idealized picture you're showing, you have a plane wave (and you're viewing one point on it). That means that the fields extend infinitely in $y $ and $z $. The wave will keep going in $x $ until it hits something.

If you have an actual beam of light, it does not extend infinitely in $y,z $, but if it is wide enough, you can approximate it as a plane wave when considering points near the center of the beam. At the edges, things won't be so simple. Real light beams have divergence and other properties that you don't have to worry about with plane waves.

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The field you are describing is one possible solution to Maxwell's equations - a plane wave. It's a highly useful solution because all nonevanescent (propagating) fields in freespace / homogeneous mediums can be built from a superposition of plane waves.

But real solutions don't have to (and indeed never do) look like this particular solution. Real waves are superpositions of different plane waves; in particular, a superposition involving a spread of directions of constituent plane waves decreases swiftly with transverse distance from the center of the disturbance. Indeed, once we include a spread of directions and frequencies in the plane wave superposition, the disturbance at any time can have a truly compact support i.e. is precisely zero at any point that cannot be reached from the source travelling at less than light speed for the time since the disturbance began, and thus comply with special relativity.

Another, much more realistic propagating solution to Maxwell's equations is a spherical wave.

$$\mathbf{E}\left(r,\,\theta,\,\phi,\,t\right)=A\frac{\sin\theta}{r}\left[\cos\left(kr-\omega t\right)-\frac{\sin\left(kr-\omega t\right)}{kr}\right]\hat{\boldsymbol{\phi}}$$ $$$$ $$\mathbf{B}\left(r,\,\theta,\,\phi,\,t\right)=\frac{2A\cos\theta}{r^{2}\omega}\left[\sin\left(kr-\omega t\right)+\frac{\cos\left(kr-\omega t\right)}{kr}\right]\hat{\mathbf{r}}+\frac{A\sin\theta}{r^{3}\omega}\left[\left(\frac{1}{k}-kr^{2}\right)\cos\left(kr-\omega t\right)+r\sin\left(kr-\omega t\right)\right]\hat{\boldsymbol{\theta}} $$

where $(r,\,\theta,\,\phi)$ are the spherical polar co-ordinates.

As $r\to\infty$ such a wave looks exactly like a plane wave with amplitude varying like $1/r$ over regions subtending angles at the source.

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