After reading this question on this site I learned that the direction of the magnetic field is given by $\boldsymbol{B}=\frac{1}{\omega}\boldsymbol{k}\times \boldsymbol{E}$

The left diagram below is a left-handed coordinate system, while the right diagram is a right-handed coordinate system:

left and right-handed coordinate systems

The image above and the quote below are taken from this page on Wikipedia for the Right-hand-rule

Coordinates are usually right-handed. For right-handed coordinates, the right thumb points along the Z-axis in the positive direction, and the curl of the fingers represents a motion from the first or X-axis to the second or Y-axis. When viewed from the top or Z axis the system is counter-clockwise. For left-handed coordinates, the left thumb points along the Z-axis in the positive direction and the curled fingers of the left hand represent a motion from the first or X-axis to the second or Y-axis. When viewed from the top or Z axis the system is clockwise. Interchanging the labels of any two axes reverses the handedness. Reversing the direction of one axis (or of all three axes) also reverses the handedness. (If the axes do not have a positive or negative direction then handedness has no meaning.) Reversing two axes amounts to a 180° rotation around the remaining axis.

I am tasked with (what I thought was) a rather easy question:

The electric and magnetic fields of a plane electromagnetic wave propagating in a vacuum have the following form: $\boldsymbol{E} = \boldsymbol{E_0}e^{i\left(\boldsymbol{k}\cdot\boldsymbol{r}−\omega t\right)}, \boldsymbol{B} =\boldsymbol{B_0}e^{i\left(\boldsymbol{k}\cdot\boldsymbol{r}−ωt\right)}$. Given that the wave is propagating in the $+\hat{\boldsymbol{z}}$ direction and ${\boldsymbol{E_0}}$ is in the $+\hat{\boldsymbol{y}}$ direction. Find the direction of ${\boldsymbol{B_0}}$.

Now the problem here is that the direction to be determined is not along z, ie. the direction of propagation, $\hat{\boldsymbol{k}}$ is not the result of the vector product. Proceeding anyway using $\boldsymbol{B}=\frac{1}{\omega}\boldsymbol{k}\times \boldsymbol{E}$ with the right thumb pointing in the $+\hat{\boldsymbol{z}}$ direction and right index finger pointing in the $+\hat{\boldsymbol{y}}$ direction. The third (middle) finger is now in the $-\hat{\boldsymbol{x}}$ direction; just as in this left-handed coordinate system below:

left-handed coordinate system

The correct answer is that the magnetic field is in the $-\hat{\boldsymbol{x}}$ direction. I am confused for 2 reasons:

  1. I thought the set of vectors $\{\boldsymbol{k},\boldsymbol{E_0},\boldsymbol{B_0}\}$ form a right-handed set, but I have just shown they are a left-handed set.
  2. The first 2 fingers (I thought) are supposed to be reserved for the vectors $\hat{\boldsymbol{k}}$ and ${\boldsymbol{E_0}}$, their vector product should be given by the direction of the thumb. Now, I had to point my thumb in the direction of propagation, $\hat{\boldsymbol{k}}$ to determine the direction of ${\boldsymbol{B_0}}$. Is this method I used even valid?

1 Answer 1


The order in which you are referring to your fingers is incorrect. For a vector equality of the form $\textbf{C} = \textbf A \times \textbf B$, the thumb of your right hand is supposed to represent the direction of the quantity on the left-hand-side: that is, $\textbf{C}$. The index finger then represents $\textbf{A}$, while the middle finger, $\textbf{B}$.

In your case with $\textbf{B} = \frac{1}{\omega} \left(\textbf{k} \times \textbf{E} \right)$, you will need to assign the index finger of your right hand to $\textbf{k}$ and the middle finger to $\textbf{E}$. If you do it correctly, your thumb will point toward $-\hat{x}$, which is the correct direction of $\textbf{B}$.

  • $\begingroup$ Hi, thanks for your answer. "For a vector equality of the form $\textbf{C} = \textbf{B} \times \textbf{A}$, the thumb of your right hand is supposed......The index finger then represents $\textbf{A}$, while the middle finger, $\textbf{B}$" I thought that the letter which comes first, in this case, $\textbf{B}$ is assigned the index finger then middle finger for $\textbf{A}$? $\endgroup$
    – Electra
    Jul 20, 2020 at 12:05
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    $\begingroup$ There are different conventions, but I find the one I outlined to be the least confusing. Either way, it seems like you are getting something wrong with the convention you're following. For what it's worth, it's best to avoid using one's actual hand altogether in calculating cross-products. Try following the cyclic cross-product rules: i.e. C = A X B => A = B X C, etc. $\endgroup$
    – Yejus
    Jul 20, 2020 at 16:24
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    $\begingroup$ @Yejus I think you typoed in your answer! I think you meant $\mathbf{C}=\mathbf{A}\times\mathbf{B}$; the rest of the answer makes sense $\endgroup$
    – DanDan面
    Dec 14, 2020 at 10:03

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