Did I Calculate the Correct Derivative to Validate Faraday's Law?

AI Thread Summary
The discussion focuses on validating Faraday's Law through the calculation of the negative time derivative of the magnetic field B(r, t), resulting in a specific sine wave expression. The user encounters difficulties with the curl of the electric field, questioning how to demonstrate its equivalence to the negative time derivative of the magnetic field. They note that the curl seems not to affect the cosine term, leading to confusion about their calculations. A warmup exercise involving the derivative of a scalar product helps clarify the signs and patterns in their derivatives. Ultimately, correcting the sign of the sine function allows for a successful resolution of the problem.
Blanchdog
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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?
 
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Blanchdog said:
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)##.
 
Blanchdog said:
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
 
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.
 
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