This is from Griffith's "Introduction to Electrodynamics" 4th edition, problem 5.33.

I need to show that

$$ \tag 1 \frac{\partial \textbf{A}_{above}}{\partial n}-\frac{\partial \textbf{A}_{below}}{\partial n}=-\mu_{0}\textbf{K} $$

where $\textbf{A}$ is the magnetic vector potential (i.e. $\textbf{B}=\nabla \times \textbf{A}$), and $\textbf{K}$ is the surface current density, and I am supposed to use the equations:

$$ \tag 2 \textbf{B}_{above}-\textbf{B}_{below}=\mu_{0}(\textbf{K}\times\hat{\textbf{n}})$$

$$\tag 3 \textbf{A}_{above}=\textbf{A}_{below}$$

$$ \tag 4 \nabla\cdot\textbf{A}=0$$

The problem is, I'm not sure what $\frac{\partial\textbf{A}}{\partial n}$ is. For a scalar, the book defines $\frac{\partial V}{\partial n}$ as: $\nabla V \cdot \hat{\textbf{n}}$

Would $\frac{\partial\textbf{A}}{\partial n}$ simply a vector with components $\nabla A_{i}\cdot\hat{\textbf{n}}$ for $i=x,y,z$?

That's what I tried, but with no luck so far.


Multiplying vectorially with $\mathbf n$ your (2), and using the identity $$\tag A \mathbf A \times ( \mathbf B \times \mathbf C) = (\mathbf A \cdot \mathbf C) \mathbf B - (\mathbf A \cdot \mathbf B) \mathbf C $$ you obtain $$ \tag B (\mathbf B_{\text{above}} - \mathbf B_{\text{below}})\times \mathbf n = - \mu_0 \mathbf K. $$

Now use in both the terms in (B) the identity $$ \tag C ( \nabla \times \mathbf A) \times \mathbf n = (\mathbf n \cdot \nabla)\mathbf A - \nabla (\mathbf A \cdot \mathbf n), $$

and remember that the normal derivative is defined (or equivalently, can be expressed) as $$ \frac{\partial \mathbf A}{\partial n} \equiv (\mathbf n \cdot \nabla) \mathbf A.$$

Can you conclude from here?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.