# Relationship between the magnetic and electric component of an EM field

I am studying Maxwell's equations and their use to derive a wave equation to derive the behaviour of electromagnetic waves in vacuum. In the case of plane waves, EM fields can be described by:

$\vec E_z = \vec E_0e^{i(kz - wt)}$

$\vec B_z = \vec B_0e^{i(kz - wt)}$

$\nabla^2E = \mu_o\epsilon_0\frac{\partial^2 E}{\partial t^2}$

$\nabla^2B = \mu_o\epsilon_0\frac{\partial^2 B}{\partial t^2}$

where $z$ is an arbitrary direction of propagation.

The book I am using as my guide is Introduction to Electrodynamics by Griffiths. In the book, the author writes the following:

However, I do not understand what is the reasoning behind equation 9.45. How does the Maxwell equation involving the curl of the electric field generate that piece of work?

You insert into the Maxwell equation $$\vec \nabla \times \vec E=-\frac {\partial \vec B}{\partial t} \tag 1$$ the wave solutions $$\vec E=\vec E_0 exp (i\vec k \vec r - i\omega t)$$ and $$\vec B=\vec B_0 exp (i\vec k \vec r - i\omega t)$$ In the cross product of equ. (1), you can move the exponential function to the nabla. Thus you have a gradient of the wave exponential giving $$\vec \nabla exp(i\vec k \vec r)=i\vec k expi(\vec k \vec r)$$ and you obtain $$\vec k \times \vec E=\omega\vec B$$ This should answer your question.
• Thanks for the answer. However, I do not understand what you mean by "move the exponential function to the $\nabla$". Could you please clarify your explanation a bit more or point me towards a reference with a detailed account of your reasoning? – daljit97 Mar 5 '18 at 0:00
• @daljit97 - Simply put, when you have a cross product of two vectors (here $\vec \nabla \times \vec E_0 exp (i\vec k \vec r)$ ) and one vector has a numerical factor (here $exp (i\vec k \vec r)$), you can move this factor to the other vector (here in front of the $\vec \nabla$) without change of the cross product. Thus you can do this, take the gradient of the exponential and then move the remaining exponential back. – freecharly Mar 5 '18 at 0:12
• @daljit97 $\vec a \times f \vec b=\vec a f\times \vec b$ where f is a number. – freecharly Mar 5 '18 at 0:18
• @daljit97 You move in the cross prodcut the $exp(\vec k \vec r -\omega t)$ back to the $\vec E_0$ giving $\vec E_0exp(\vec k \vec r -\omega t)$ again. Then you have left $$i \vec k \times \vec E_0exp(\vec k \vec r -\omega t)=i\omega \vec B_0 exp(\vec k \vec r -\omega t)$$ and you can divide this by $i$. In the end you can also divide by the expontial wave factor $exp(\vec k \vec r -\omega t)$. – freecharly Mar 5 '18 at 13:37