- #1
issacnewton
- 998
- 29
Hi
I am studying magnetic vector potential from Griffiths book. The eq 5.76 in his book gives
the boundary condition for the magnetic vector potential.
[tex]\frac{\partial \vec{A_2} }{\partial n}- \frac{\partial \vec{A_1} }{\partial n}=-\mu_o \vec{K} [/tex]
where n is the vector perpendicular to the boundary surface and pointing from region
1 to region 2.
There is a problem in the book asking to prove this. So here's my attempt to do it. And I couldn't do it right. The book has eq 5.74 , which is
[tex]\vec{B_2}-\vec{B_1}=\mu_o(\vec{K}\times \hat{n}) [/tex]
So I decided to use this for my purpose. Writing B in terms of the vector potential A, we have
[tex] \vec{\nabla}\times \vec{A_2}-\vec{\nabla}\times \vec{A_1}=\mu_o(\vec{K}\times \hat{n}) [/tex]
Now writing in terms of the Cartesian components explicitly and collecting x,y,z components, we have
[tex]\hat{x}\left[\frac{\partial A_{2z} }{\partial y}- \frac{\partial A_{2y} }{\partial z}-
\frac{\partial A_{1z} }{\partial y}+\frac{\partial A_{1y} }{\partial z}\right]
-\hat{y}\left[\frac{\partial A_{2z} }{\partial x}-\frac{\partial A_{2x} }{\partial z}-
\frac{\partial A_{1z} }{\partial x}+\frac{\partial A_{1x} }{\partial z}\right]
+\hat{z}\left[\frac{\partial A_{2y} }{\partial x}-\frac{\partial A_{2x} }{\partial y}
-\frac{\partial A_{1y} }{\partial x}+\frac{\partial A_{1x} }{\partial y}\right]
=\mu_o(\vec{K}\times \hat{n}) [/tex]lets call this equation 1where K is the surface current. Now I take
[tex] \vec{K} = K\hat{x} [/tex]
and
[tex] \hat{n}=\hat{z} [/tex]
so that
[tex]\mu_o(\vec{K}\times \hat{n}) = \mu_o K(-\hat{y}) [/tex]
so I take the dot products of the equation 1 with [itex]\hat{x},\hat{y},\hat{z}[/itex].
There will be three equations in all.
[tex]\frac{\partial A_{2z} }{\partial y}+\frac{\partial A_{1y} }{\partial z}=
\frac{\partial A_{2y} }{\partial z}+\frac{\partial A_{1z} }{\partial y} [/tex][tex]\frac{\partial A_{2y} }{\partial x}+\frac{\partial A_{1x} }{\partial y}=
\frac{\partial A_{2x} }{\partial y}+\frac{\partial A_{1y} }{\partial x} [/tex]
[tex]\frac{\partial A_{2z} }{\partial x}-\frac{\partial A_{2x} }{\partial z}
-\frac{\partial A_{1z} }{\partial x}+\frac{\partial A_{1x} }{\partial z}=
\mu_o K [/tex]Now after this , I am totally lost. We actually know two more things about the vector
potential.
[tex]\vec{\nabla}\cdot \vec{A} = 0 [/tex]and at the boundary, A is continuous.
[tex]\vec{A_2}=\vec{A_1} [/tex]
But I don't know how to use this information to prove the result I am seeking. Any guidance
will be appreciated.
I am studying magnetic vector potential from Griffiths book. The eq 5.76 in his book gives
the boundary condition for the magnetic vector potential.
[tex]\frac{\partial \vec{A_2} }{\partial n}- \frac{\partial \vec{A_1} }{\partial n}=-\mu_o \vec{K} [/tex]
where n is the vector perpendicular to the boundary surface and pointing from region
1 to region 2.
There is a problem in the book asking to prove this. So here's my attempt to do it. And I couldn't do it right. The book has eq 5.74 , which is
[tex]\vec{B_2}-\vec{B_1}=\mu_o(\vec{K}\times \hat{n}) [/tex]
So I decided to use this for my purpose. Writing B in terms of the vector potential A, we have
[tex] \vec{\nabla}\times \vec{A_2}-\vec{\nabla}\times \vec{A_1}=\mu_o(\vec{K}\times \hat{n}) [/tex]
Now writing in terms of the Cartesian components explicitly and collecting x,y,z components, we have
[tex]\hat{x}\left[\frac{\partial A_{2z} }{\partial y}- \frac{\partial A_{2y} }{\partial z}-
\frac{\partial A_{1z} }{\partial y}+\frac{\partial A_{1y} }{\partial z}\right]
-\hat{y}\left[\frac{\partial A_{2z} }{\partial x}-\frac{\partial A_{2x} }{\partial z}-
\frac{\partial A_{1z} }{\partial x}+\frac{\partial A_{1x} }{\partial z}\right]
+\hat{z}\left[\frac{\partial A_{2y} }{\partial x}-\frac{\partial A_{2x} }{\partial y}
-\frac{\partial A_{1y} }{\partial x}+\frac{\partial A_{1x} }{\partial y}\right]
=\mu_o(\vec{K}\times \hat{n}) [/tex]lets call this equation 1where K is the surface current. Now I take
[tex] \vec{K} = K\hat{x} [/tex]
and
[tex] \hat{n}=\hat{z} [/tex]
so that
[tex]\mu_o(\vec{K}\times \hat{n}) = \mu_o K(-\hat{y}) [/tex]
so I take the dot products of the equation 1 with [itex]\hat{x},\hat{y},\hat{z}[/itex].
There will be three equations in all.
[tex]\frac{\partial A_{2z} }{\partial y}+\frac{\partial A_{1y} }{\partial z}=
\frac{\partial A_{2y} }{\partial z}+\frac{\partial A_{1z} }{\partial y} [/tex][tex]\frac{\partial A_{2y} }{\partial x}+\frac{\partial A_{1x} }{\partial y}=
\frac{\partial A_{2x} }{\partial y}+\frac{\partial A_{1y} }{\partial x} [/tex]
[tex]\frac{\partial A_{2z} }{\partial x}-\frac{\partial A_{2x} }{\partial z}
-\frac{\partial A_{1z} }{\partial x}+\frac{\partial A_{1x} }{\partial z}=
\mu_o K [/tex]Now after this , I am totally lost. We actually know two more things about the vector
potential.
[tex]\vec{\nabla}\cdot \vec{A} = 0 [/tex]and at the boundary, A is continuous.
[tex]\vec{A_2}=\vec{A_1} [/tex]
But I don't know how to use this information to prove the result I am seeking. Any guidance
will be appreciated.
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