Show that Faraday's Law holds

[Blanchdog]

Homework Statement

Suppose that an electric field is given by ##E(r, t) = E_0 \text{cos}(k \cdot r - \omega t + \phi) ##, where ##k \perp E_0## and ##\phi## is a constant phase. Show that
$$B(r, t) = \frac{k~\text{x}~E_0}{\omega} \text{cos}((k \cdot r - \omega t + \phi)$$ is consistent with Faraday's Law.

Relevant Equations

Faradays Law: $$\nabla~\text x~E = -\frac{\partial B}{\partial t}$$

I've calculated the negative time derivative of B(r, t) as: $$-\frac{\partial B}{\partial t} = k~\text x~E_0~\text{sin}(k \cdot r - \omega t + \phi)$$ The cross product can be easily expanded, I'd just rather not do the LaTeX for if I can avoid it.

The Curl of the electric field (##\nabla~\text{x}~ E##) is giving more trouble though. I should end up with a sine wave (or a cosine offset by pi/2) but as best I can tell the curl doesn't affect the stuff within the cosine at all since k and r are dotted together into a scalar. How do I show that the curl of the electric field is equal to the result of the negative time derivative of of the magnetic field above, or did I make a mistake in calculating that derivative?

 

[TSny]

as best I can tell the curl doesn't affect the stuff within the cosine at all

As a warmup exercise, calculate ##\frac{\partial}{\partial x} (\mathbf k \cdot \mathbf r)##.

 

[TSny]

I've calculated the negative time derivative of B(r, t) as: $$-\frac{\partial B}{\partial t} = k~\text x~E_0~\text{sin}(k \cdot r - \omega t + \phi)$$

Check the signs

 

[Blanchdog]

That was very helpful, I was able to solve it once I fixed the sign of the sine and saw the pattern of the derivatives I was able to do after your warm up.