# Using Maxwell's equations to find $\mathbf{B}$

The $$\mathbf{E}$$ component of an electromagnetic wave in free space is: $$\mathbf{E}(x, t) = E_0 \cos{(kx-\omega t)} \hat{\bf y}$$

How do I find the corresponding $$\mathbf{B}$$ component using one of Maxwell's equations (in differential form)?

I assume that I must use one of either $$\nabla\times\mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}$$ or $$\nabla\times\mathbf{B} = \epsilon_0\mu_0\frac{\partial\mathbf{E}}{\partial t}$$

(probably the second one?)

I also know that I am looking for an expression in the form of $$\mathbf{B}(x, t) = B_0 \cos{(kx-\omega t)} \hat{\bf z},$$ but I am not sure about the intermediate steps.

• Why not try both? Commented Sep 5, 2021 at 22:47
• I remember that from his 4 equations, in the absence of charges and currents and in vacuum, Maxwell obtained the following wave equations, of the d'Alembert type, for the electric field and magnetic field. Commented Sep 5, 2021 at 22:55

I assume that I must use one of either ... or ...
(probably the second one?)

You need to use both of them.

From your expressions for $$\mathbf{E}(x, t)$$ and $$\mathbf{B}(x, t)$$ calculate $$\nabla\times\mathbf{E}$$ and $$\frac{\partial \mathbf{B}}{\partial t}$$. Insert the results into the first of your Maxwell equations. You will get $$E_0k=B_0\omega. \tag{1}$$

Likewise calculate $$\nabla\times\mathbf{B}$$ and $$\frac{\partial\mathbf{E}}{\partial t}$$. Insert the results into the second of your Maxwell equations. You will get $$B_0 k=\epsilon_0\mu_0 E_0\omega. \tag{2}$$

From (1) and (2) and a little bit of algebra you find $$B_0=E_0\sqrt{\epsilon_0\mu_0} \tag{3}$$ and $$k=\omega\sqrt{\epsilon_0\mu_0}. \tag{4}$$

(3) is the magnetic field amplitude you were looking for. And equation (4) tells you the speed of your electromagnetic wave is $$\frac{1}{\sqrt{\epsilon_0\mu_0}}$$ which happens to be equal to the speed of light $$c=3\cdot 10^8$$ m/s.

In a region with no charges ($$\rho = 0$$) and no currents ($$\mathbf J=\mathbf 0$$) such as in a vacuum, Maxwell's equations reduce to:

\begin{align} \boldsymbol \nabla \cdot \mathbf{E} &= 0 \quad & \boldsymbol\nabla \times \mathbf{E} &= -\frac{\partial\mathbf B}{\partial t}, \\ \boldsymbol\nabla \cdot \mathbf{B} &= 0 \quad & \boldsymbol\nabla \times \mathbf{B} &= \mu_0\varepsilon_0 \frac{\partial\mathbf E}{\partial t}. \end{align}.

I suggest to see this link:

https://mathworld.wolfram.com/dAlembertsSolution.html

Taking the curl of the curl equations, you will get \begin{align} \mu_0\varepsilon_0 \frac{\partial^2 \mathbf{E}}{\partial t^2} - \nabla^2 \mathbf{E} = 0 \\ \mu_0\varepsilon_0 \frac{\partial^2 \mathbf{B}}{\partial t^2} - \nabla^2 \mathbf{B} = 0 \end{align}

The quantity $$\mu_0\varepsilon_0$$ has the dimension of (time/length)$$^2$$. Defining $$c = (\mu_0 \varepsilon_0)^{-1/2}$$, the equations above have the form of the standard wave equation's \begin{align} \frac{1}{c^2} \frac{\partial^2 \mathbf{E}}{\partial t^2} - \nabla^2 \mathbf{E} = 0 \\ \frac{1}{c^2} \frac{\partial^2 \mathbf{B}}{\partial t^2} - \nabla^2 \mathbf{B} = 0 \end{align}

I remember that the d'Alembert operator is:

$$\square = \frac{1}{c^2}\frac{\partial^2}{\partial t^2} - \nabla^2 .$$

You can integrate the first equation in time to obtain:

\begin{align} \int \nabla\times\mathbf{E} dt = \mathbf{B}(t) - \mathbf{B}(0) \end{align}

If if you have already an expression for $$\mathbf{E}(t)$$, you can just substitute it into the above equation and you'll obtain an expression for $$\mathbf{B}(t)$$.