# Finding electric field of the electromagnetic wave for given magnetic field [closed]

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.

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.
• 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 $\vec C=\vec 0$. Nov 21 at 16:21
• 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. Nov 21 at 16:21
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 $$\vec{k} \cdot \vec{E}_0 = 0$$.