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Given the far-field vector potential...

$$\vec A_{\text{ff}}(\vec r) = \left(\frac{\mu_0I}{2\pi kr}e^{-jkr}\frac{\cos\left(\frac{\pi}{2}\cos(\theta)\right)}{\sin^2(\theta)}\right)\hat z$$

I need to find the magnetic far-field; I tried by doing the following...

$$\vec H_{\text{ff}}(\vec r) = \frac{1}{\mu_0}\boldsymbol\nabla\times\vec A_{\text{ff}}(\vec r) = \frac{1}{\mu_o}\left(\boldsymbol \nabla\times\left(\hat r\cos(\theta)A_{\text{ff}} - \hat\theta \sin(\theta)A_{\text{ff}}\right)\right)$$

$$= \frac{1}{\mu_0r}\left(\frac{\partial}{\partial r}\left(-r\sin(\theta)A_{\text{ff}}\right) - cos(\theta)\frac{\partial A_{\text{ff}}}{\partial \theta}\right)\hat\phi$$

$$\implies\vec H_{\text{ff}}(\vec r) = \left(\frac{jI}{2\pi r}e^{-jkr}\frac{\cos\left(\frac{\pi}{2}\cos(\theta)\right)}{\sin(\theta)}\right)\hat\phi$$

This is the solution that I've seen but I am lost as to why the $\cos(\theta)\frac{\partial A_{\text{ff}}}{\partial\theta}$ is assumed to be $0$ when I've plotted it and it is most definitely not $0$.

Should the '$=$' at the last step be replaced with '$\approx$'?

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  • $\begingroup$ @Philip Because I wanted to take the curl in spherical coordinates, not cartesian. So I had to change the unit vector first. $\endgroup$ – Landon Sep 4 at 11:34
  • $\begingroup$ Just went through the calculations again, can't see that you've done anything wrong. I don't know enough of the physics to comment more, sadly. Good luck! $\endgroup$ – Philip Sep 4 at 15:06

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