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# 10.2: Boundary Conditions

## 10.2.1 The Tangential Components of the Electric Field.

Apply Stokes’ theorem to the Maxwell equation

$\operatorname{curl}(\vec{\text{E}})=-\frac{\partial \vec{\text{B}}}{\partial \text{t}} \nonumber$

and the small loop whose sides are L long and $$\delta$$ long as shown in Figure (10.1.2):

$\oint \vec{\text{E}} \cdot \vec{\text{d} \text{L}}=-\frac{\partial}{\partial \text{t}} \int \int_{\text {Area }} \vec{\text{B}} \cdot \text{d} \vec{\text{A}}. \nonumber$

One then takes the limit as the sides $$\delta$$ shrink to zero. The line integral of the electric field gives

$\oint \vec{\text{E}} \cdot \text{d} \vec{\text{L}}=\left(\text{E}_{\text{t} 1}-\text{E}_{\text{t} 2}\right) \text{L}, \nonumber$

where Et1 is the field component parallel with L in material number 1 (vacuum in this case) and Et2 is the electric field component parallel with L in material number 2. The flux of the magnetic field through the loop goes to zero as $$\delta$$ goes to zero, therefore

$\left(\text{E}_{\text{t} 1}-\text{E}_{\text{t} 2}\right)=0 \nonumber$ Figure $$\PageIndex{3}$$: The Maxwell equation $$\operatorname{curl}(\vec{\text{H}})=-\partial \vec{\text{D}} / \partial \text{t}$$ requires the tangential components of $$\vec H$$ to be continuous across any interface. See the text. Figure $$\PageIndex{4}$$: The Maxwell equation $$\operatorname{div}(\vec{\text{B}})=0$$ requires the normal component of $$\vec B$$ to be continuous across any interface. See the text.

or

$\text{E}_{\text{t} 1}=\text{E}_{\text{t} 2} . \label{10.23}$

At the boundary between two materials the transverse components of $$\vec E$$ must be continuous.

## 10.2.2 The Tangential Components of the Magnetic Field.

Apply Stokes’ theorem to a small loop as shown in fig(10.2.3):

$\operatorname{curl}(\vec{\text{H}})=\frac{\partial \vec{\text{D}}}{\partial \text{t}} , \nonumber$

where it has been assumed that there are no free currents in either material, and no surface free current density on the interface between material number(1) and material number(2). Therefore

$\oint_{C} \vec{\text{H}} \cdot \text{d} \vec{\text{L}}=\frac{\partial}{\partial \text{t}} \int \int_{A r e a} \vec{\text{D}} \cdot \vec{\text{d} \text{S}} . \nonumber$

Upon taking the limit as $$\delta$$ shrinks to zero the surface integral over $$\vec D$$ gives nothing and

$\left(\text{H}_{\text{t} 1}-\text{H}_{\text{t} 2}\right) \text{L}=0 , \nonumber$

that is

$\text{H}_{\text{t} 1}=\text{H}_{\text{t} 2} . \label{10.24}$

The transverse components of the magnetic field $$\vec H$$ must be continuous across the boundary between two materials.

## 10.2.3 The Normal Component of the Field B.

The normal component of the magnetic field $$\vec B$$ must be continuous across any interface as a consequence of the Maxwell equation $$d i v(\vec{\text{B}})=0$$; see Figure (10.2.4). In Figure (10.2.4) Gauss’ theorem is applied to a small pill-box that spans an arbitrary surface. The height of the pill-box, $$\delta$$, is taken to be so small that any contributions to the surface integral from the sides of the box can be neglected. The continuity of the normal component of $$\vec B$$ is then forced by the requirement that the surface integral of $$\vec B$$ over the pill-box be zero:

$\text{B}_{\text{n} 1}=\text{B}_{\text{n} 2} . \label{10.25}$