Boundary conditions for magnetic fields

[lys04]

Homework Statement

Deriving the boundary conditions for the tangential and normal components of the magnetic field at a current sheet

Relevant Equations

Ampere's law: $$\oint \vec B \cdot d \vec l = \mu_{0} I_{enc}$$

In this diagram, why is the H vector/ B vector (They differ by a constant of $$ \mu_{0} $$) pointing in the same direction on opposite sides of the current sheet?
Also, I'm a bit confused on how did they go from $$ K \Delta w = H_{1,t} \Delta w - H_{2,t} \Delta w $$ to $$ (\vec H_{1} - \vec H_{2} ) \times \hat n_{12} = \vec K $$

The former is only got to do with magnitude whereas the latter involves directions right?
Any help is appreciated!!

 

[kuruman]

The former is only got to do with magnitude whereas the latter involves directions right?

Right.

 

[lys04]

Right.

So do you mind explaining how did they go from the one with magnitudes only to the one involving vectors?
And how how could they make the assumption that H points in the same direction both above and below the surface?

And also how does the magnitude version apply to an infinite sheet of current?

For example, if I have an infinite sheet with surface current flowing in the x-direction, the magnetic field above and below is going to point in the y-direction, although in opposite directions, they have the same magnitude.
When I take the difference then it'll give me 0.

 

[kuruman]

So do you mind explaining how did they go from the one with magnitudes only to the one involving vectors?

Study the drawing where all the quantities and vectors are defined. $H_{1t}$ is the tangential component of $\mathbf {H}_1$. What do you get when you take the cross product $\mathbf {H}_1\times \mathbf {\hat n}_{12}$?

And how how could they make the assumption that H points in the same direction both above and below the surface?

The point in the same general direction. That's because the normal component of $\mathbf B$ is continuous across the boundary.

When I take the difference then it'll give me 0.

So what? The difference between the magnitudes of two vectors is not the difference between the two vectors.

Edited for LaTeX typo.

 

[lys04]

What do you get when you take the cross product H1×n^12?

Get another vector thats perpendicular to both

 

[lys04]

normal component of B is continuous across the boundary.

Yeah I get this, but couldn't the H vector below the surface be pointing upwards but going to the left instead?

 

[lys04]

So what? The difference between the magnitudes of two vectors is not the difference between the two vectors.

But doesn't the equation $$ K \Delta w = H_{1,t} \Delta w - H_{2,t}\Delta w $$ also hold?

 

[kuruman]

Get another vector thats perpendicular to both

And what is the name of that vector? Study the figure.

Yeah I get this, but couldn't the H vector below the surface be pointing upwards but going to the left instead?

No, because the tangential component discontinuity is equal to the surface current density. That's what the equation is all about.

But doesn't the equation $$ K \Delta w = H_{1,t} \Delta w - H_{2,t}\Delta w $$ also hold?

What makes you think it does not? It is another way to say what I just wrote above about the tangential component discontinuity of $\mathbf H$ at the interface.

 

[lys04]

And what is the name of that vector? Study the figure.

$$ (\vec H_{1} - \vec H_{2}) \times \hat n_{1,2} = \vec K $$
$$ (\vec H_{1,t} + \vec H_{1,n} - \vec H_{2,t} - \vec H_{2,n}) \times \hat n_{1,2} = \vec K $$
$$ \vec H_{1,t} \times \hat n_{1,2} + \vec H_{1,n} \times \hat n_{1,2} - \vec H_{2,t} \times \hat n_{1,2} - \vec H_{2,n} \times \hat n_{1,2} = \vec K $$

Since
$$\vec H_{1,n} $$ and $$\hat n_{1,2}$$ are in the same direction, their cross product is 0 so I'm left with

$$ \vec H_{1,t} \times \hat n_{1,2} - \vec H_{2,t} \times \hat n_{1,2} = \vec K $$

so

$$ (\vec H_{1,t} - \vec H_{2,t}) \times \hat n_{1,2} = \vec K $$

So $$ \vec H_{1} \times \hat n_{1,2} $$ gives a vector in the direction of K but not necessarily K?

 

[kuruman]

So $$ \vec H_{1} \times \hat n_{1,2} $$ gives a vector in the direction of K but not necessarily K?

Yes, but you have to consider the difference between tangential components$$(\mathbf H_{1} - \mathbf H_{2}) \times \mathbf{\hat n}_{1,2} = \mathbf K. $$ It's the discontinuity (difference) of the tangential component at the interface that is equal to the surface current density.

This is analogous to the boundary condition for the electric displacement vector at a dielectric interface $$(\mathbf D_{1} - \mathbf D_{2}) \cdot \mathbf{\hat n}_{1,2} = \sigma$$ which says that the discontinuity of the normal of component of $\mathbf D$ at the interface is equal to the surface free charge density.

I think you got it.

 

[lys04]

Yes, but you have to consider the difference between tangential components(H1−H2)×n^1,2=K. It's the discontinuity (difference) of the tangential component at the interface that is equal to the surface current density.

Yeah alright, but I still don’t understand why this implies that the tangential component needs to point in the same direction?

 

[kuruman]

Yeah alright, but I still don’t understand why this implies that the tangential component needs to point in the same direction?

The same direction as what? The boundary condition $$(\mathbf H_{1} - \mathbf H_{2}) \times \mathbf{\hat n}_{1,2} = \mathbf K$$ says that the tangential component of the difference $~(\mathbf H_{1} - \mathbf H_{2})~$ has the same magnitude and direction as the surface current density $\mathbf K.$ That's the only "same" direction implicit in the equation.

 

[lys04]

The same direction as what? The boundary condition

In the initial diagram both the tangential component is pointing to the right

 

[kuruman]

In the initial diagram both the tangential component is pointing to the right

The boundary condition in scalar form $$H_{1t}-H_{2t}=K \tag{1}$$ is derived from the line integral around the closed loop abcda. The integration is clockwise which means that positive direction of the normal to the loop is into the screen. The positive direction of $K$ is also into the screen. In the limit $\Delta h \rightarrow 0$, there are no contributions from the sides and the closed loop integral is $$\oint\mathbf H\cdot d\mathbf l=\int_a^b\mathbf H_1\cdot d\mathbf l_1+\int_c^d\mathbf H_2\cdot d\mathbf l_2. $$ Now let's define the positive $x$-axis to the right, so that $d\mathbf l_1=dx~\mathbf{\hat x}.$ Then $$\mathbf H_1\cdot d\mathbf l_1=(H_1\cdot\mathbf{\hat x})~dx=H_{1t}~dx\implies \int_a^b\mathbf H_1\cdot d\mathbf l_1=H_{1t}\Delta w.$$ In this expression $H_{1t}$ is the x-component of $\mathbf H_1$ that can be positive or negative.

Noting that $d\mathbf l_2=-dx~\mathbf{\hat x}$,$$\mathbf H_2\cdot d\mathbf l_2=-(H_2\cdot\mathbf{\hat x})~dx=-H_{2t}~dx\implies \int_c^d\mathbf H_2\cdot d\mathbf l_2=-H_{2t}\Delta w$$ where, as before, $H_{2t}$ is the x-component of $\mathbf H_2$ that can be positive or negative. Putting all this together gives equation (1).

In equation (1) $H_{1t}$ and $H_{2t}$ can be in any direction that pleases you and there are four choices. However, if you want to make a drawing, you have to choose one of the four. Once you make that choice and you want to draw in the direction of $\mathbf K$, it better be consistent. If $H_{1t}-H_{2t}>0$, the current should be into the screen and if $H_{1t}-H_{2t}<0$ the current should be out of the screen.

The drawing that you posted shows, as best as I can tell, that $H_{1t}=H_{2t}$ which means that $K=0$, yet it shows a surface current into the screen. I didn't draw it, so don't ask me why it looks that way.

 

[lys04]

In equation (1) H1t and H2t can be in any direction that pleases you and there are four choices. However, if you want to make a drawing, you have to choose one of the four. Once you make that choice and you want to draw in the direction of K, it better be consistent. If H1t−H2t>0, the current should be into the screen and if H1t−H2t<0 the current should be out of the screen.

Oh alright that makes sense.

The drawing that you posted shows, as best as I can tell, that H1t=H2t which means that K=0, yet it shows a surface current into the screen. I didn't draw it, so don't ask me why it looks that way.

Yeah I realised that too.

Oh by the way in vector form $$ (\vec H_{1} - \vec H_{2}) \times \hat n_{1,2} = \vec K $$, does the unit normal vector need to point in a specific direction? Like pointing from surface 1 to 2 or from surface 2 to 1?

By the way thanks so much for the long reply! Really appreciate it.

 

[kuruman]

Oh by the way in vector form $$ (\vec H_{1} - \vec H_{2}) \times \hat n_{1,2} = \vec K $$, does the unit normal vector need to point in a specific direction? Like pointing from surface 1 to 2 or from surface 2 to 1?

There is only one surface, the interface between two different media. The normal needs to be consistent with the boundary condition and the sense of integration. In this case the sense of integration is clockwise. This makes $\mathbf K$ positive when it's into the screen. Now if you write $$(\mathbf H_{1} - \mathbf H_{2}) \times \mathbf{\hat n}_{1,2} = \mathbf K$$ this says that the vector on the left hand side must be equal to the vector on the right hand side. This happens if you choose the normal from medium 1 into medium 2 so that $~\mathbf H_{1}\times \mathbf{\hat n}~$ is in the same direction as $\mathbf K$ which puts, of course, $-\mathbf H_{2}\times \mathbf{\hat n}~$ in a direction opposite to $\mathbf K$. If you choose the direction of the normal to be from medium 2 into medium 1, then you have to reverse the order of the difference in the left hand side of the boundary condition.

By the way thanks so much for the long reply! Really appreciate it.

You are welcome.

 

[lys04]

There is only one surface, the interface between two different media.

O alright!

In this case the sense of integration is clockwise. This makes K positive when it's into the screen. Now if you write (H1−H2)×n^1,2=K this says that the vector on the left hand side must be equal to the vector on the right hand side. This happens if you choose the normal from medium 1 into medium 2 so that H1×n^ is in the same direction as K which puts, of course, −H2×n^ in a direction opposite to K. If you choose the direction of the normal to be from medium 2 into medium 1, then you have to reverse the order of the difference in the left hand side of the boundary condition.

So I guess it depends on the direction of $$ \vec H_{1} - \vec H_{2}$$ ? If we go by the direction of $$ \vec K $$ given in the diagram, using the right hand rule gives me that if $$ \vec H_{1} - \vec H_{2}$$ points to the right then the normal vector has to point from medium 2 to medium 1 whereas if $$ \vec H_{1} - \vec H_{2}$$ points to the left then the normal vector needs to point from medium 1 to medium 2?

 

[kuruman]

You got it backwards. The direction of the cross product of the difference with the normal must be into the screen which is the known direction of $\mathbf K$.

(Edited for clarity.)