Consistency of Faraday's Law and Electric Field Equation in Optics

AI Thread Summary
The discussion focuses on demonstrating the consistency of Faraday's Law with the given electric field equation. Participants suggest expressing the electric field in exponential form and taking its real part for easier manipulation. A vector calculus identity is highlighted to aid in calculating the curl of the electric field. Additionally, a potential typo in the phase factor is pointed out, and Cartesian coordinates are recommended for computing the gradient. Overall, the conversation provides guidance on how to approach the problem effectively.
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Homework Statement


Suppose that an electric field is given by E(r,t)=E0cos(k·r−ωt+φ), where k⊥E0 and φ is a constant phase. Show that B(r,t)=((k×E)/ω)B(k⋅r-ωt+φ) is consistent with ∇×E=-∂B/∂t

Homework Equations


∇×E=-∂B/∂t

The Attempt at a Solution


I know I have to take the curl of E, but I'm not sure how to go about doing it.
∂E/∂t would be ωE0sin(k⋅r-ωt+φ) and ∂E/∂r would be -kE0sin(k⋅r-ωt+φ), but I'm not sure if that helps. Also, by my calculation, -∂B/∂t= (k×E)cos(k⋅r-ωt+φ).
Any help would be greatly appreciated.
 
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Suggestion is to write the ## E ## field as ##E=\vec{ E}_o e^{i (\vec{k} \cdot \vec{x}-\omega t +\phi)} ## and take the real part afterwards. Also, there is a vector calculus identity that ## \nabla \times (\psi \vec{a})=(\nabla \psi) \times \vec{a} +\psi (\nabla \times \vec{a}) ##. The operation then becomes doing a gradient on ## e^{i(\vec{k} \cdot \vec{x}-\omega t + \phi)} ##. ## \\ ## And I think you have a typo: the second phase factor ## \phi ## should have a plus sign, like the first one. ## \\ ## Addition note: ## \vec{k} \cdot \vec{x}=k_x x+k_y y+k_z z ##. You can readily compute the gradient in these Cartesian coordinates. ## \\ ## Additional suggestion: Let ## B=B_o e^{i (\vec{k} \cdot \vec{x}- \omega t+\phi)} ##. And what you have should read ## B(r,t)=\frac{\vec{k} \times \vec{E}}{\omega} cos(\vec{k} \cdot \vec{r}-\omega t+\phi )##. The rest is straightforward.
 
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Charles Link said:
Suggestion is to write the ## E ## field as ##E=\vec{ E}_o e^{i (\vec{k} \cdot \vec{x}-\omega t +\phi)} ## and take the real part afterwards. Also, there is a vector calculus identity that ## \nabla \times (\psi \vec{a})=(\nabla \psi) \times \vec{a} +\psi (\nabla \times \vec{a}) ##. The operation then becomes doing a gradient on ## e^{i(\vec{k} \cdot \vec{x}-\omega t )} ##. ## \\ ## And I think you have a typo: the second phase factor ## \phi ## should have a plus sign, like the first one. ## \\ ## Addition note: ## \vec{k} \cdot \vec{x}=k_x x+k_y y+k_z z ##. You can readily compute the gradient in these Cartesian coordinates. ## \\ ## Additional suggestion: Let ## B=B_o e^{i (\vec{k} \cdot \vec{x}- \omega t+\phi)} ##. And what you have should read ## B(r,t)=\frac{\vec{k} \times \vec{E}}{\omega} cos(\vec{k} \cdot \vec{r}-\omega t+\phi )##. The rest is straightforward.
Thank you, I should be able to take it from here.
 
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