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During a lecture, my lecturer writes that:

For polarisation along $x$ then $\vec E =E_0 e^{i\left(kz-\omega t\right)}\hat i$

For polarisation along $y$ then $\vec E =E_0 e^{i\left(kz-\omega t\right)}\hat j$

But then straight after he says something like "for the x polarised wave, the direction of the B-field is in the positive y-direction; and for the y polarised wave, the direction of the B-field is in the negative x-direction".

Now, I read this post from this site very carefully. For both cases above, the waves propagate in the positive z-direction. So for the x-polarised wave, using the right hand rule, the wavenumber $k$, is in the direction of my right index finger. The E-field is in the direction of my right middle finger, the only way to get my right middle finger in the positive x-direction is to rotate my right thumb so it points downward (direction of B-field). This means that by my logic the direction of the B-field is in the negative y-direction. For the y-polarised wave, I agree that the direction of the B-field is in the negative x-direction.

Who is right about the direction of the B-field in $\vec E =E_0 e^{i\left(kz-\omega t\right)}\hat i$?

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From the Maxwell equations we can derive the following: \begin{align} \vec k\times \vec E=\omega \vec B. \end{align} Assuming we have set our x-axis pointing up, z-axis pointing right, y-axis pointing out of the screen to us and assuming $\vec{k}=k\cdot \hat{z}$, the directions of your professor are correct. For your first statement, the y-axis would have to be pointing away from us into the screen, which would make the second statement (where you agree with your professor) incorrect. So for your statements, there is no possible arrangement of the axes! I always use this Trick: For $a\times b=c$: Point your flat Hand towards a, then curve your 4 Fingers towards b. The direction of c will be your Thumb when pointing out. I hope this helps. Rodrigo

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  • $\begingroup$ Many thanks for your answer, the reason I got this wrong is because I have been misled by the answer given to this question where the user states that $\textbf{C} = \textbf{B} \times \textbf{A}$ and then goes on to claim that there are 'different conventions'. I really hope there isn't different conventions, else no-one will agree on the direction of the cross product. $\endgroup$
    – Sirius Black
    Commented Dec 14, 2020 at 0:50
  • $\begingroup$ I voted up your answer but my rep is too low for it to count, sorry. $\endgroup$
    – Sirius Black
    Commented Dec 14, 2020 at 0:51
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    $\begingroup$ @SiriusBlack That's a typo. You can see from the rest of the user's explanations on that question that $\mathbf{C}=\mathbf{A}\times\mathbf{B}$ but they must have messed up the order when they typed that one line out. The different conventions refers to different mnemonics for remembering the direction of the cross product, like the right-hand rule and the Fleming left-hand rule, which are essentially the same thing. The important convention to remember is x cross y = z. $\endgroup$
    – DanDan面
    Commented Dec 14, 2020 at 10:05

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