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I did an exercise which probably is quite popular,
in which you draw an electromagnetic wave and prove that it should propagate at the speed of light $1 \over \sqrt {\mu_0\epsilon_0}$ using Farday's law and Ampere's law.

Basically if this is the wave:


Electromagnetic wave

Let's say the E-field (red) is in the X direction, the B-Field (blue) is in the Y direction, and the velocity of the wave is in the Z direction.

You take for example for ampere's law a surface in the ZY plane with a length L equal to the amplitude of the wave, and a width equal to $\lambda\over 4$ You do a similar thing with Faraday's law and you get the speed of light, assuming you know that the E-field and B-field propagate in this manner.

I got the right answer but I wondered about this: Let's say I only had the E-field and I know the wave propagates at the speed of light, I assume this is enough information to draw the B-field at each point.

But how will I know the direction? Both Faraday's law and Ampere's law say you need a closed loop integral and the rules I've been taught say you go over the loop in a clockwise direction for example and take the normal to the surface according to the right hand rule etc.

But clockwise and counter-clockwise direction don't really give me much information in this case, so how can I determine the direction of the B-field if I only have the E-field?

fiftyeight
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2 Answers2

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If you're careful about how you define the surface, then you will get the correct direction out of Maxwell's equations. In vector calculus (and generally in math and physics), a surface has an orientation, which also specifies the orientation of the loop that forms its boundary. So you can't just pick a direction to go around the loop at random. The direction in which you go around the loop is related to the orientation of the normal vector to the surface.

Of course, you don't actually need to do a surface integral to figure this out. The electric and magnetic fields in an EM wave (at any given position and moment in time) are related by the equation

$$\mathbf{B} = \frac{1}{c}\hat{\mathbf{k}}\times\mathbf{E}$$

where $\hat{\mathbf{k}}$ is a unit vector that points in the direction of propagation of the wave. If you don't know which direction the wave is moving, you can't tell which way the magnetic field points.

David Z
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  • I understand that you have to choose the direction around the lop according to the direction of the normal to the surface, but I jut don't understand what is the meaning of this loop in this context, because in other exercises I did the loop represented a wire, or in the case of the B field, a circular B-field around the wire – fiftyeight Feb 14 '12 at 17:21
  • The loop is the boundary of the surface, that's all. Sometimes you choose a surface that happens to be bounded by a wire, but you don't have to. – David Z Feb 14 '12 at 19:39
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$ {\bf E} \times {\bf B}$ (which is the Poynting vector multiplied by $\mu_0$) should be in the direction of wave motion.

I you want an aide-memoire, if you imagine taking a screwdriver and turning a (standard) screw in the direction of ${\bf E}$ towards ${\bf B}$, then the wave should be travelling in the direction that the screw is driven (either in or out).

ProfRob
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