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

In summary, the given electric field can be written as a complex exponential with a constant phase. Taking the curl of the electric field and using the vector calculus identity, we can derive the magnetic field as a complex exponential with a constant phase as well and a coefficient of k cross E divided by the angular frequency. This is consistent with the given relationship between the curl of the electric field and the time derivative of the magnetic field.
  • #1
Blue Kangaroo
41
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:
Physics news on Phys.org
  • #2
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
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.
 
  • Like
Likes Charles Link

What is Faraday's Law?

Faraday's Law, named after physicist Michael Faraday, describes the relationship between a changing magnetic field and an induced electric current. It states that an electric current is induced in a conductor when it is placed in a changing magnetic field.

How does Faraday's Law relate to optics?

In optics, Faraday's Law is used to explain the phenomenon of electromagnetic induction, which is the process of generating an electric current by changing the magnetic field. This is the principle behind devices such as generators and transformers, which are essential components in many optical instruments.

What are the applications of Faraday's Law in optics?

Faraday's Law has numerous applications in optics, including the production of polarized light, the operation of optical isolators, and the generation of electromagnetic waves in fiber optics. It also plays a crucial role in the functioning of devices such as lasers, magneto-optical modulators, and optical switches.

What are the limitations of Faraday's Law?

Faraday's Law is limited in its applicability to linear materials and does not account for nonlinear phenomena. It also assumes that the conductor is in an ideal environment with no external interference. Additionally, it does not take into account the effects of temperature and other environmental factors.

How is Faraday's Law related to other laws of electromagnetism?

Faraday's Law is closely related to other laws of electromagnetism, such as Ampere's Law and Gauss's Law. Together, these laws form the basis of Maxwell's equations, which describe the behavior of electromagnetic fields. Faraday's Law is also a fundamental principle in the study of optics and electromagnetic waves.

Similar threads

  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
776
  • Introductory Physics Homework Help
Replies
3
Views
647
  • Introductory Physics Homework Help
Replies
5
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
135
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
921
  • Introductory Physics Homework Help
Replies
1
Views
160
  • Introductory Physics Homework Help
Replies
4
Views
1K
Back
Top