# How to obtain surface-current density from current density?

Context

My problem is related to . I do not dispute the solution in , however, it is not helping me to understand the problem that I face. I am working through Example 10.5 in Modern Electrodynamics by Zangwill. This problem pertains to the double-curl equation.

In this example, we integrate $$\boldsymbol{\nabla}\times\left(\boldsymbol{\nabla}\times \mathbf{A}\right)$$ to find $$\mathbf{A}$$ for a long straight cylindrical wire with radius $$a$$ which carries a uniform current density $$\mathbf{j} = j\,\hat{\mathbf{z}}$$. The problem progresses until we arrive to $$\mathbf{A} = A_z(\rho)\,\hat{\mathbf{z}}$$, where is given by the case structure in the following equation in terms of an as yet to be determined constant $$C$$. \begin{align} A_z(\rho) = \begin{cases} -\frac{1}{4}\,\mu_0\,j\,\rho^2, &\rho\leq a;~\text{and} \\ -\frac{1}{4}\,\mu_0\,j\,a^2 +C \,\ln{\frac{\rho}{a}}, &\rho\geq a. \end{cases} \end{align} Zangwill states that, the matching condition $$\hat{ \mathbf{n}}_2 \times \left[ \mathbf{B}_1 - \mathbf{B}_2 \right] = \mu_0\,\mathbf{K}(\mathbf{r}_S) . \tag{10}$$ applied at $$\rho = a$$ fixes $$C$$. Since, in general, $$\mathbf{B}(\mathbf{r}) = \boldsymbol{\nabla} \times \mathbf{A}(\mathbf{r})$$, therefore, in this case, $$\mathbf{B}(\mathbf{r}) = \begin{cases} + \frac{\mu_0\,j}{2}\,\rho \,\hat{ \boldsymbol{\varphi}} , & \rho \leq a;~\text{and} \\ - \frac{ C }{ \rho} \,\hat{ \boldsymbol{\varphi} } , & \rho \geq a. \end{cases}$$ Then, since $$\mathbf{K}(\mathbf{r}_S) = K(a,\varphi,z)\,\hat{\mathbf{z}}$$, therefore Eq. 10 evaluated at $$\rho = a$$ results in $$\hat{ \boldsymbol{\rho} } \times \left[ - \frac{ C }{ a} - \frac{\mu_0\,j\,a }{2 } \right]\hat{ \boldsymbol{\varphi} } = \frac{ - 2\,C - \mu_0\,j\,a^2 }{2\, a} \,\hat{ \mathbf{z} } = \mu_0\,K(a,\varphi,z)\,\hat{\mathbf{z}} \tag{20}.$$ From Eq. 20, we find that the matching condition gives the following equation for $$C$$: $$C = - \frac{1}{2} \,\mu_0\,j\,a^2 - a \mu_0\,K(a,\varphi,z) . \tag{30}$$ Now, Zangwill claims that $$C = - \frac{1}{2} \,\mu_0\,j\,a^2 . \tag{40}$$ This means that either the surface-current density is identically zero, or that I have made a mistake along the way.

Questions

(1) Smaller question: Have I properly applied the matching conditions so to obtain the correct formula for $$C$$ (e.g., perhaps I have a sign error).

(2) Larger question: How does one obtain the surface-current density, $$\mathbf{K}$$, from the current density, $$\mathbf{j}$$?

My attempts

Option 1 (the correct option): On the one hand, I would say that the existence of a current density in the bulk of a material does not necessarily imply the existence of surface-current density.

Option 2:

Since $$I = \int d\mathbf{S}\cdot \mathbf{j}$$ therefore $$I = \int_{0}^{2\pi}\int_0^\infty \delta(\rho-a)\,\rho \,d\rho\,d\varphi \,\hat{\mathbf{z}}\cdot j\,\hat{\mathbf{z}}= 2\,\pi\, a \, j.$$ This has the outstanding issue that the units are inconsistent on the left- and right-hand side of the equation. Bearing this in mind I come to option 3.

Option 3:

$$K = \int_{0}^{2\pi}\int_0^\infty \delta(\rho-a)\,\rho \,d\rho\,d\varphi \,\hat{\mathbf{z}}\cdot j\,\hat{\mathbf{z}}= 2\,\pi\, a \, j.$$ This has its own outstanding issue. Namely, upon substitution into Eq. 30, I find that $$C = \mu_0\,j\,a^2\left[ - \frac{1}{2} - 2\,\pi\, \right],$$ which is a result that is inconsistent with Zangwill's answer (i.e., Eq. 40).

Option 4:

I adopt from joshphysics in , that
\begin{align} \mathbf j = \delta{\left(\rho -a\right)} \,\mathbf K . \end{align} However this leaves me scratching my head and asking for relief.

Bibliography

 Zangwill, Modern Electrodynamics, pp. 31-2, 310, 323-4.

• I understand from your answer that I did not make a sign error or other error. For emphasis, I repeat that the example statement does not---in any way---specify $\mathbf{K}$. Rather, the example statement only specifies $\mathbf{j}$. So do you think then that $\mathbf{K} = \mathbf{0}$? Apr 19 at 17:56