1
$\begingroup$

We are given the following information:

The magnetic field of an electromagnetic wave travelling through vacuum is given by $$ \vec B = B_0e^{i(ky-\omega t)}\hat i + B_0e^{i(kx-\omega t)}\hat j. $$

The question is to use Maxwell's equations to find the electric field of the electromagnetic wave.

I know that $B_0 = E_0/c$ and $\vec{B_0} = \frac{\vec{k} \times \vec{E_0}}{\omega}$, but I am confused about the direction of the magnetic field and how to find the direction of the electric field.

$\endgroup$

2 Answers 2

2
$\begingroup$

Use the Ampére equation and solve for the electric field: $$\vec\nabla\times\vec{B} =\mu_0\varepsilon_0\frac{\partial\vec{E}}{\partial t} \Leftrightarrow \frac{\partial\vec{E}}{\partial t} =c^2\vec\nabla\times\vec{B}.$$ (Since we're in a vacuum, the additional term $\mu_0\vec{j}$ vanishes.) $c^2\vec\nabla\times\vec{B}$ can be calculated using your given equation (and will be proportional to $\vec e_z$ or $\hat{k}$ as you probably call it, fitting the fact that the amplitude of the electric and magnetic field are perpendicular) and then be easily integrated due to time appearing in an exponential function. Concerning the "constants" of integration, a vector field $\vec C(x,y,z)$, you have to use other Maxwell equations just like jensen paull said.

$\endgroup$
7
  • $\begingroup$ You can assume that C is always 0 for electromagnetic waves in vacuüm right? Could you tell me how you exactly show what the constant of integration is using the other Maxwell's equations? Because you can't use the curl of E to find the constant $\endgroup$
    – Stallmp
    Nov 21 at 16:07
  • $\begingroup$ Yes, $\vec C=\vec 0$. You can show it as follows: Put $\vec E$ into the Gauss equation $\vec\nabla\cdot\vec E=\vec 0$ to get $\vec\nabla\cdot\vec C=\vec 0$ and put $\vec E$ into the Faraday-Maxwell equation $\vec\nabla\times\vec E=-\frac{\partial\vec B}{\partial t}$ to get $\vec\nabla\times\vec C=\vec 0$. Combining both results yields $\vec C=\vec 0$. $\endgroup$ Nov 21 at 16:21
  • $\begingroup$ This is because of the Helmholtz decomposition (en.wikipedia.org/wiki/Helmholtz_decomposition) of a vector field into a gradient/curl-free part and a rotation/divergence-free part. This means in particular, that if $\vec\nabla\cdot\vec C=\vec 0$, then $\vec C$ is the rotation of a vector field and that if $\vec\nabla\times\vec C=\vec 0$, then $\vec C$ is the gradient of a vector field. $\endgroup$ Nov 21 at 16:21
  • 1
    $\begingroup$ A uniform electric field has no curl or divergence. Perhaps unrealistic; but so is a plane wave. $\endgroup$
    – ProfRob
    Nov 21 at 16:42
  • 1
    $\begingroup$ Yes, that is possible and only eliminated by physical interpretation. $\endgroup$ Nov 21 at 18:20
1
$\begingroup$

From the relation $\vec{B}_0 = (\vec{k} \times \vec{E}_0)/\omega$, you can actually solve for $\vec{E}_0$. To do this, take the cross product of $\vec{k}$ with both sides: $$ \vec{k} \times \vec{B}_0 = \frac{1}{\omega} \vec{k} \times (\vec{k} \times \vec{E}_0) $$ Then apply the $BAC$-$CAB$ rule to the right-hand side, and use the fact that $\vec{k} \cdot \vec{E}_0 = 0$.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.