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Optics -- Faraday's Law

  • #1

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.
 
Last edited:

Answers and Replies

  • #2
Charles Link
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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.
 
Last edited:
  • #3
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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