Boundary Conditions for Magnetic Fields and Surface Currents

This question asks about the specific boundary condition that describes how the parallel components of the magnetic field $\vec{H}$ change across an interface, especially when there's a free surface current density, $K_f$. We need to consider the relationship between $\vec{H}$ and $\vec{B}$ in linear media.

Understanding Magnetic Field Boundary Conditions

When electromagnetic fields transition from one medium to another across an interface, they often change abruptly. These changes are governed by boundary conditions derived from Maxwell's equations. For magnetic fields, we usually consider two main conditions:

• The normal component of the magnetic flux density ($\vec{B}$) is continuous across the boundary (assuming no magnetic charges): $B_{1n} - B_{2n} = 0$.
• The tangential component of the magnetic field strength ($\vec{H}$) is discontinuous in the presence of a surface current density: $H_{1t} - H_{2t} = K_{f} \times \hat{n}$.

In these equations:

• Subscripts 1 and 2 refer to the fields in the two different media.
• The subscript 'n' denotes the component normal (perpendicular) to the interface.
• The subscript 't' or the parallel symbol ($\parallel$) denotes the component parallel to the interface.
• $K_f$ is the surface density of free currents.
• $\hat{n}$ is the unit normal vector pointing from medium 2 to medium 1.

Relating B and H Fields

The question statement specifically mentions the discontinuity of $\vec{H}$ but the options involve $\vec{B}$. In linear, isotropic media, the relationship between $\vec{B}$ and $\vec{H}$ is given by:

$$ \vec{B} = \mu \vec{H} $$

where $\mu$ is the magnetic permeability of the medium. For the two media, we have:

$$ \vec{B}_1 = \mu_1 \vec{H}_1 \quad \Rightarrow \quad \vec{H}_1 = \frac{1}{\mu_1} \vec{B}_1 $$

$$ \vec{B}_2 = \mu_2 \vec{H}_2 \quad \Rightarrow \quad \vec{H}_2 = \frac{1}{\mu_2} \vec{B}_2 $$

Deriving the Boundary Condition

We are interested in the parallel components. So, we have:

$$ \vec{H}_{1}^{\parallel} = \frac{1}{\mu_1} \vec{B}_{1}^{\parallel} $$

$$ \vec{H}_{2}^{\parallel} = \frac{1}{\mu_2} \vec{B}_{2}^{\parallel} $$

Now, substitute these into the boundary condition for the parallel component of $\vec{H}$ in the presence of surface currents:

$$ \vec{H}_{1}^{\parallel} - \vec{H}_{2}^{\parallel} = \vec{K}_f \times \hat{n} $$

Substituting the expressions for $\vec{H}_1^{\parallel}$ and $\vec{H}_2^{\parallel}$ in terms of $\vec{B}$:

$$ \frac{1}{\mu_1} \vec{B}_{1}^{\parallel} - \frac{1}{\mu_2} \vec{B}_{2}^{\parallel} = \vec{K}_f \times \hat{n} $$

Analyzing the Options

Let's compare our derived condition with the given options:

• Option 1: $\frac{1}{\mu_1} {B}_1^{\parallel} - \frac{1}{\mu_2} {B}_2^{\parallel} = {K}_f \times \hat{n}$ - This matches our derived condition exactly.
• Option 2: ${E}_1^{\parallel} - {E}_2^{\parallel} = 0$ - This is the boundary condition for the parallel component of the electric field ($\vec{E}$), not the magnetic field.
• Option 3: $\frac{1}{\mu_1} B_1^{\parallel} - \frac{1}{\mu_2} B_2^{\parallel} = 0$ - This equation implies continuity of $\frac{1}{\mu} \vec{B}^{\parallel}$, which is incorrect in the presence of surface currents.
• Option 4: $\frac{1}{\mu_1} {B}_1^{\parallel} + \frac{1}{\mu_2} {B}_2^{\parallel} = {K}_f \times \hat{n}$ - The sign is incorrect; it should be a difference, not a sum.

Therefore, the correct boundary condition relating the parallel components of $\vec{B}$ across the interface, considering the effect of free surface current density $K_f$ in linear media, is the one derived.