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The magnetic vector potential is given as:

[tex]\mathbf{A}(\mathbf{r}) = \frac{\mu _0}{4\pi }\int \frac{\mathbf{J}(\mathbf{r'})}{|\mathbf{r} - \mathbf{r'}|}\d\tau '[/tex]

I am asked to show that [itex]\mathbf{\nabla } \cdot \mathbf{A} = 0[/itex] by applying the divergence to the equation above. When I've done that, I've used some product rules to simplify and end up with:

[tex]\mathbf{\nabla } \cdot \mathbf{A} = -\frac{\mu _0}{4\pi }\int \frac{\mathbf{J}(\mathbf{r'})\cdot(\mathbf{r} - \mathbf{r'})}{|\mathbf{r} - \mathbf{r'}|^3}d\tau '[/tex]

I can't see why this must be zero.

Second question: Consider a surface carrying uniform surface current density. We know that the component of [itex]\mathbf{B}[/itex] parallel to the surface but perpendicular to the direction of flow is discontinuous at the surface, and the "derivative" of [itex]\mathbf{A}[/itex] inherits this discontinuity. This is expressed (supposedly) in the equation:

[tex]\frac{\partial \mathbf{A}_{above}}{\partial n} - \frac{\partial \mathbf{A}_{below}}{\partial n} = -\mu _0\mathbf{K}[/tex]

I am asked to prove this using the equations:

[tex]\mathbf{A}_{above} = \mathbf{A} _{below}[/tex]

[tex]\mathbf{B} _{above} - \mathbf{B} _{below} = \mu _0 (\mathbf{K} \times \hat{\mathbf{n}})[/tex]

[tex]\mathbf{\nabla } \cdot \mathbf{A} = 0[/tex]

Now, I'm not entirely sure what the equation I'm trying to prove even says. Earlier in my book, we have the equation:

[tex]\frac{\partial V}{\partial n} \equiv \mathbf{\nabla }V\cdot \hat{\mathbf{n}}[/tex]

This gives a definition for the

[tex]\frac{\partial \mathbf{A}_{above}}{\partial n} = \frac{\partial \mathbf({A_x\hat{\mathbf{x}} + A_y\hat{\mathbf{y}} + A_z\hat{\mathbf{z}}})_{above}}{\partial n} = \frac{\partial A_x}{\partial n}\hat{\mathbf{x}} + \frac{\partial A_y}{\partial n}\hat{\mathbf{y}} + \frac{\partial A_z}{\partial n}\hat{\mathbf{z}}\ ?[/tex]

Also, doesn't the given equation:

[tex]\mathbf{A}_{above} = \mathbf{A} _{below}[/tex]

Imply that the right side of the equation I have to prove should be 0, and not [itex]-\mu _0\mathbf{K}[/itex]?

[tex]\mathbf{A}(\mathbf{r}) = \frac{\mu _0}{4\pi }\int \frac{\mathbf{J}(\mathbf{r'})}{|\mathbf{r} - \mathbf{r'}|}\d\tau '[/tex]

I am asked to show that [itex]\mathbf{\nabla } \cdot \mathbf{A} = 0[/itex] by applying the divergence to the equation above. When I've done that, I've used some product rules to simplify and end up with:

[tex]\mathbf{\nabla } \cdot \mathbf{A} = -\frac{\mu _0}{4\pi }\int \frac{\mathbf{J}(\mathbf{r'})\cdot(\mathbf{r} - \mathbf{r'})}{|\mathbf{r} - \mathbf{r'}|^3}d\tau '[/tex]

I can't see why this must be zero.

Second question: Consider a surface carrying uniform surface current density. We know that the component of [itex]\mathbf{B}[/itex] parallel to the surface but perpendicular to the direction of flow is discontinuous at the surface, and the "derivative" of [itex]\mathbf{A}[/itex] inherits this discontinuity. This is expressed (supposedly) in the equation:

[tex]\frac{\partial \mathbf{A}_{above}}{\partial n} - \frac{\partial \mathbf{A}_{below}}{\partial n} = -\mu _0\mathbf{K}[/tex]

I am asked to prove this using the equations:

[tex]\mathbf{A}_{above} = \mathbf{A} _{below}[/tex]

[tex]\mathbf{B} _{above} - \mathbf{B} _{below} = \mu _0 (\mathbf{K} \times \hat{\mathbf{n}})[/tex]

[tex]\mathbf{\nabla } \cdot \mathbf{A} = 0[/tex]

Now, I'm not entirely sure what the equation I'm trying to prove even says. Earlier in my book, we have the equation:

[tex]\frac{\partial V}{\partial n} \equiv \mathbf{\nabla }V\cdot \hat{\mathbf{n}}[/tex]

This gives a definition for the

**normal derivative**for a scalar field, but not for a vector field. Would it just be:[tex]\frac{\partial \mathbf{A}_{above}}{\partial n} = \frac{\partial \mathbf({A_x\hat{\mathbf{x}} + A_y\hat{\mathbf{y}} + A_z\hat{\mathbf{z}}})_{above}}{\partial n} = \frac{\partial A_x}{\partial n}\hat{\mathbf{x}} + \frac{\partial A_y}{\partial n}\hat{\mathbf{y}} + \frac{\partial A_z}{\partial n}\hat{\mathbf{z}}\ ?[/tex]

Also, doesn't the given equation:

[tex]\mathbf{A}_{above} = \mathbf{A} _{below}[/tex]

Imply that the right side of the equation I have to prove should be 0, and not [itex]-\mu _0\mathbf{K}[/itex]?

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