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Finding the magnetic field of a wave from the E field

  1. Dec 13, 2016 #1
    1. The problem statement, all variables and given/known data
    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.
    2. Relevant equations
    $$\nabla \times E = - \frac{\delta B}{\delta t}$$

    3. 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?
     
  2. jcsd
  3. Dec 13, 2016 #2

    nrqed

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    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.
     
  4. Dec 20, 2016 #3
    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
     
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