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I am given $\boldsymbol A(r,\phi) = \frac{i}{e}a(r)U(\phi)\nabla U^{\dagger}(\phi)$, where $U(\phi)=e^{in\phi}$.

I want to obtain the magnetic field $\boldsymbol B$ from this vector potential, so thus I must use the equation:

$$\boldsymbol B = \nabla \times\boldsymbol A$$

Since $\boldsymbol A$ does not depend on a variable $z$, I will be left with

$$\boldsymbol B = \frac{1}{r}\left(\frac{\partial}{\partial r}(rA_{\phi}) - \frac{\partial}{\partial \phi}(A_{r})\right)\boldsymbol z$$

How do I get the values for $A_{\phi}$ and $A_{r}$?

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closed as off-topic by Emilio Pisanty, Kyle Kanos, Jon Custer, heather, ZeroTheHero Jul 8 '18 at 2:52

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    $\begingroup$ Just calculate A, using the gradient in cylindrical coordinates. $\endgroup$ – Javier Jun 19 '18 at 15:41
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The potential provided is,

$$\vec A = \frac{i}{e}a(r)e^{in\phi}\nabla e^{-in\phi} = \frac{i}{e}a(r)e^{in\phi} \left( -\frac{in}{r}e^{-in\phi}\hat\phi\right) = \frac{n}{e}\frac{a(r)}{r} \hat\phi.$$

using the definition of the gradient of a scalar in cylindrical coordinates. Thus, there is only one non-zero component, $A_\phi$. The magnetic field follows from $\vec B = \nabla \times \vec A$.

Can you take it from here?

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  • $\begingroup$ Thank you for pointing out the gradient in the function that gives a direction to the vector potential. Missed that part! $\endgroup$ – user570877 Jun 19 '18 at 15:54

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