Finding the magnetic field of a wave from the E field

Click For Summary
SUMMARY

The discussion focuses on determining the magnetic field associated with a monochromatic electromagnetic wave using Maxwell's Equations, specifically Faraday's law. The electric field is given by the equation $$E(r,\theta,\phi,t)=A\frac{\sin \theta}{r} \big(\cos (kr-\omega t)-\frac{1}{kr} \sin (kr-\omega t) \big) \hat{\phi}$$. Participants confirm that the curl of the electric field, $$\nabla \times E$$, can be computed to find the magnetic field, with the integration being straightforward as it involves basic trigonometric functions. The discussion emphasizes that the integration can be performed as an indefinite integral with respect to time.

PREREQUISITES
  • Understanding of Maxwell's Equations, particularly Faraday's law
  • Familiarity with vector calculus, specifically curl operations
  • Knowledge of electromagnetic wave properties, including monochromatic waves
  • Basic proficiency in trigonometric integration techniques
NEXT STEPS
  • Study the application of Faraday's law in electromagnetic theory
  • Learn about the properties of electromagnetic waves and their mathematical representations
  • Explore vector calculus in depth, focusing on curl and divergence
  • Investigate time-harmonic analysis in electrical engineering contexts
USEFUL FOR

Students and professionals in physics and electrical engineering, particularly those studying electromagnetic theory and wave propagation. This discussion is beneficial for anyone looking to deepen their understanding of the relationship between electric and magnetic fields in wave phenomena.

zweebna
Messages
8
Reaction score
0

Homework Statement


Many sources of electromagnetic waves (stars and light bulbs, for example) radiate in all
directions. A simple example of the electric field for a monochromatic electromagnetic wave produced by a spherical source is
$$E(r,\theta,\phi,t)=A\frac{\sin \theta}{r} \big(\cos (kr-\omega t)-\frac{1}{kr} \sin (kr-\omega t) \big) \hat{\phi}$$
where A is a constant and ##k=\omega / c##. Use one of Maxwell's Equations to determine the associated magnetic fi eld in free space (i.e. ##\rho = 0## and ##J = 0##). For simplicity of notation, let ##u = kr - \omega t##. Hint: You will not need to do any difficult integrals.

Homework Equations


$$\nabla \times E = - \frac{\delta B}{\delta t}$$

The Attempt at a Solution


I mostly just want to make sure that I'm doing this right. So to find the magnetic field, according to Faraday's law I would take the curl of the electric field and that would be equal to the negative time derivative of the B field, so I can then take a negative integral to get the B field. For ##\nabla \times E## I have
$$\nabla \times E = \frac{1}{r \sin \theta} \bigg [\frac{\delta}{\delta \theta} \big(A \frac{\sin^2 \theta}{r} ( \cos u - \frac{1}{kr} \sin u) \big) \bigg] \hat{r} + \frac{1}{r} \bigg [ - \frac{\delta}{\delta r} \big ( A \sin \theta ( \cos u - \frac{1}{kr} \sin u) \big) \bigg] \hat{\theta}$$
$$\nabla \times E = A \frac{2 \cos \theta}{r^2}( \cos u - \frac{1}{kr} \sin u) \hat{r} - A \frac{ \sin \theta}{r}(-k \sin u + \frac{\sin u - kr \cos u}{k r^2})\hat{\theta}$$

Now assuming I've done this right, then I can take the negative integral to get the B field. I'm a bit confused about this integral though. What would the limits be? Or would it just be an indefinite integral?

Also the hint is making me wary, as this integral seems like it would be kind of difficult. Is there an easier way that I'm missing?
 
Physics news on Phys.org
zweebna said:

Homework Statement


Many sources of electromagnetic waves (stars and light bulbs, for example) radiate in all
directions. A simple example of the electric field for a monochromatic electromagnetic wave produced by a spherical source is
$$E(r,\theta,\phi,t)=A\frac{\sin \theta}{r} \big(\cos (kr-\omega t)-\frac{1}{kr} \sin (kr-\omega t) \big) \hat{\phi}$$
where A is a constant and ##k=\omega / c##. Use one of Maxwell's Equations to determine the associated magnetic fi eld in free space (i.e. ##\rho = 0## and ##J = 0##). For simplicity of notation, let ##u = kr - \omega t##. Hint: You will not need to do any difficult integrals.

Homework Equations


$$\nabla \times E = - \frac{\delta B}{\delta t}$$

The Attempt at a Solution


I mostly just want to make sure that I'm doing this right. So to find the magnetic field, according to Faraday's law I would take the curl of the electric field and that would be equal to the negative time derivative of the B field, so I can then take a negative integral to get the B field. For ##\nabla \times E## I have
$$\nabla \times E = \frac{1}{r \sin \theta} \bigg [\frac{\delta}{\delta \theta} \big(A \frac{\sin^2 \theta}{r} ( \cos u - \frac{1}{kr} \sin u) \big) \bigg] \hat{r} + \frac{1}{r} \bigg [ - \frac{\delta}{\delta r} \big ( A \sin \theta ( \cos u - \frac{1}{kr} \sin u) \big) \bigg] \hat{\theta}$$
$$\nabla \times E = A \frac{2 \cos \theta}{r^2}( \cos u - \frac{1}{kr} \sin u) \hat{r} - A \frac{ \sin \theta}{r}(-k \sin u + \frac{\sin u - kr \cos u}{k r^2})\hat{\theta}$$

Now assuming I've done this right, then I can take the negative integral to get the B field. I'm a bit confused about this integral though. What would the limits be? Or would it just be an indefinite integral?

Also the hint is making me wary, as this integral seems like it would be kind of difficult. Is there an easier way that I'm missing?
I did not double check your calculation of the curl but for the integration, you may simply do an indefinite integral. It is not hard since you just integrate with respect to time, so you just need to integrate ##cos u## and ##sin u##, which is easy.
 
I haven't gone through your math, but in EE we usually attack this problem by using time harmonics.
Your maxwell equation will then become:
∇ x E = -jwB
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
689
Replies
9
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
5
Views
2K
Replies
12
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
Replies
19
Views
3K
Replies
5
Views
2K