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I have solved Laplace's equation in Polar Coordinates for the scalar electric potential in a circle of radius R and have the solution $$ \phi(r,\varphi) = \phi_{0} + \sum^{\infty}_{k=1}r^{k_{\varphi}}\left(A_{k}\sin(k\varphi)+B_k\cos(k\varphi)\right) $$

Where $k_{\varphi}$ is a constant times k.

Now the boundary condition that I have is $\vec{n}\cdot\vec{J}=0$ where n is the normal to the surface, and $\vec{J}$ is the current density. This can be written as $$ \vec{n}\cdot \vec{\nabla}\phi=-\vec{n}\cdot\frac{\partial\vec{A}}{\partial t} $$ Since the system is quasi-static the magnetic vector potential, A, can be separated into a spacial and time component written as $\vec{A}(\vec{s},t)= \vec{A}(s)f(t)$ for an arbitrary function of time. The magnetic vector potential is known and is: $$ \vec{A}= \frac x{r^2}\hat{\varphi} $$ Where x is just a constant.

My attempt was to write the Boundary condition as $$ \vec{n}\cdot \left( \begin{matrix} \frac{\partial\phi}{\partial r} \\ \frac{\partial\phi}{r\partial \varphi } \end{matrix} \right) = =\vec{n}\cdot \left( \begin{matrix} 0\\ \frac{x}{r^2} \end{matrix} \right)\cdot \dot{f}(t) $$ What I am not sure about is what is the normal vector to a circle of radius R? or if my approach is correct?

Thank you.

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  • $\begingroup$ The normal vector to the circle is $(R cos \phi , R sin \phi)$ $\endgroup$
    – Semsem
    Jan 24 '14 at 4:56
  • $\begingroup$ @Semsem From the context, $\vec n$ is a unit normal vector, which is $(\cos \varphi, \sin\varphi)$. $\endgroup$ Jan 24 '14 at 5:03
  • $\begingroup$ yes it is ok !! $\endgroup$
    – Semsem
    Jan 24 '14 at 5:06
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Taking full gradient and then multiplying by $\vec n$ is doing too much. The outward normal derivative at the boundary of the circle is simply the partial derivative with respect to $r$. So, $$ \frac{\partial \phi}{\partial r} = -\vec{n}\cdot\frac{\partial\vec{A}}{\partial t} \tag{1}$$ Next, your formula for $\vec A$ says that this vector is a multiple of $\hat \varphi$, which I understand to be a unit tangent vector. If so, then the time derivative of $\vec A$ is tangent to the circle, too. Consequently, the right hand side of (1) is zero.

So, it appears that $\phi$ must be a harmonic function with zero normal derivative; the only such functions are the constant ones.


An aside: using both $\phi$ and $\varphi$ in the same computation is usually a bad idea.

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