If a magnetic field inside a betatron in cylindrical coordinates is $$\mathbf{B}=Kr^2t\ \mathbf{e}_z$$ and the accompanying electric field takes the form $$\mathbf{E}=E_\phi(r) \ \mathbf{e}_\phi$$ what would E be in terms of K and r?

I've tried using Faraday's law to do this, but I'm not sure how to correctly do the integral $$\oint _c\mathbf{E}\bullet d\mathbf{S}$$ in this situation.

I thought it would be equal to $$E_\phi \times 2\pi r$$ but I'm not sure this is the case.

Ultimately, my answer from using the intergral or differential version of Faraday's law do not match and I'm having a hard time finding out what I'm missing.


This is a simple application of the Maxwell Equation $\nabla\times\mathbf{E}=-\frac{\partial\mathbf{B}}{\partial t}$. The curl in cylindrical coordinates for the electric field in the specific form you have given reduces down to

$\nabla\times\mathbf{E}=\frac{1}{r}\frac{\partial (rE_\phi)}{\partial r}\mathbf{e}_z$

(You can look up the curl in cylindrical coordinates)

Now, the right hand side reduces to

$-\frac{\partial\mathbf{B}}{\partial t}=-\frac{\partial}{\partial t}(Kr^2t)\mathbf{e}_z=-Kr^2\mathbf{e}_z$

Setting the two equal yields

$\frac{1}{r}\frac{\partial (rE_\phi)}{\partial r}=-Kr^2\implies E_\phi(r)=-\frac{1}{4}Kr^3$

This is the answer!

  • $\begingroup$ Thank you! Would it be possible to derive the same answer using the integral version of Faraday's law? As when I attempt to, I get - 1/2 Kr^2 rather than -1/4 Kr^3. $\endgroup$ – freja Dec 20 '18 at 12:18

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