# 5.6 Magnetic Boundary Conditions

In Chapter 4, we derived a set of boundary conditions that describes how, at the boundary between two dissimilar contiguous media, the electric flux and field D and E in the first medium relate to those in the second medium. We now derive a similar set of boundary conditions for the magnetic flux and field B and H. By applying Gauss’s law to a pill box that straddles the boundary, we determined that the difference between the normal components of the electric flux densities in two media equals the surface charge density ρ s. That is,

By analogy, application of Gauss’s law for magnetism, as expressed by Eq. (5.44), leads to the conclusion that

Because B 1=μ 1 H 1 and B 2=μ 2 H 2 for linear, isotropic media, the boundary condition for H corresponding to Eq. (5.79) is

Comparison of Eqs. (5.78) and (5.79) reveals a striking difference between the behavior of the magnetic and electric fluxes across a boundary: Whereas the normal component of B is continuous across the boundary, the normal component of D is not (unless ρ s=0). The reverse applies to the tangential components of the electric and magnetic fields E and H: Whereas the tangential component of E is continuous across the boundary, the tangential component of H is not (unless the surface current density J s=0). To obtain the boundary condition for the tangential component of H, we follow the same basic procedure used in Chapter 4 to establish the boundary condition for the tangential component of E. With reference to Fig. 5-24, we apply Ampère’s law (Eq. (5.47)) to a closed rectangular path with sides of lengths Δ l and Δ h, and then let Δ h→0 to obtain

∮C H⋅d l=∫a b H 1⋅ℓ​ˆ 1 d ℓ+∫c d H 2⋅ℓ​ˆ 2 d ℓ=I,(5.81)

where I is the net current crossing the surface of the loop in the direction specified by the right-hand rule (I is in the direction of the thumb when the fingers of the right hand extend in the direction of the loop C). As we let Δ h of the loop approach zero, the surface of the loop approaches a thin line of length Δ l. The total current flowing through this thin line is I=J s Δ l, where J s is the magnitude of the component of the surface current density J s normal to the loop. That is, J s=J s⋅n ˆ, where n ˆ is the normal to the loop. In view of these considerations, Eq. (5.81) becomes

(H 1−H 2)⋅ℓ ˆ 1 Δ l=J s⋅n ˆ Δ l.(5.82)

The vector ℓ ˆ 1 can be expressed as ℓ ˆ 1=n ˆ×n ˆ 2, where n ˆ and n ˆ 2 are the normals to the loop and to the surface of medium 2 (Fig. 5-24), respectively. Using this relation in Eq. (5.82) and then applying the vector identity A⋅(B×C)=B⋅(C×A) leads to

n ˆ⋅[n ˆ 2×(H 1−H 2)]=J s⋅n ˆ.(5.83)

Since Eq. (5.83) is valid for any n ˆ, it follows that

# Figure 5-24 Boundary between medium 1 with μ 1 and medium 2 with μ 2.

Figure 5 24 Full Alternative Text

This equation implies that the tangential components of H parallel to J s are continuous across the interface, whereas those orthogonal to J s are discontinuous in the amount of J s.

Surface currents can exist only on the surfaces of perfect conductors and superconductors. Hence, at the interface between media with finite conductivities, J s=0 and

# Example 5-7: Tilted-Plane Boundary

In Fig. 5-25 the plane defined by y+2 x=2 separates medium 1 of permeability μ 1 from medium 2 of permeability μ 2. If no surface current exists on the boundary and

B 1=x ˆ 2+y ˆ 3 (T),

find B 2 and then evaluate the result for μ 2=2 μ 1.

# Figure 5-25 Magnetic media separated by the plane y+2 x=2 (Example 5-7).

Figure 5 25 Full Alternative Text

Solution: To apply the boundary conditions at the plane between the two media, we first need to obtain an expression for the unit vector normal to the boundary. To apply the boundary condition given by Eq. (5.84), we need an expression for n ˆ 2, which isthe unit vector pointing away from medium 2 (Fig. 5-25).

From the geometry of the small triangle in Fig. 5-25,

θ=tan−1(1 2)=26.57°.

Hence,

n ˆ 2=x ˆ cos θ+y ˆ sin θ=x ˆ 0.89+y ˆ 0.45.

Magnetic field B 1 has a normal component B 1 n along n ˆ 2 and a tangential component B 1 t:

B 1=B 1 n+B 1 t

with

B 1 n=B 1 n n ˆ 2

and

B 1 n=n ˆ 2⋅B 1=(x ˆ cos θ+y ˆ sin θ)⋅(x ˆ 2+y ˆ 3)=2 cos θ+3 sin θ.

The tangential component of B 1 is

B 1 t=B 1−B 1 n=B 1−B 1 n n ˆ 2=(x ˆ 2+y ˆ 3)−(2 cos θ+3 sin θ)(x ˆ cos θ+y ˆ sin θ)=x ˆ(2−2 cos 2 θ−3 sin θ cos θ)+y ˆ(3−2 cos θ sin θ−3 sin 2 θ).

Boundary conditions require that

B 1 n=B 2 n

and

B 1 t μ 1=B 2 t μ 2,

which leads to

B 2=B 2 n+B 2 t=B 1 n+μ 2 μ 1 B 1 t=B 1 n n^2+μ 2 μ 1 B 1 t=(2 cos θ+3 sin θ)(x^cos θ+y^cos θ)+μ 2 μ 1[x^(2−2 cos 2 θ−3 sin θ cos θ)+y^(3−2 cos θ sin θ−3 sin 2 θ)]=x^[2 cos 2 θ+3 sin θ cos θ +μ 2 μ 1(2−2 cos 2 θ−3 sin θ cos θ)]+y^[2 cos θ sin θ+3 sin 2 θ +μ 2 μ 1(3−2 cos θ sin θ−3 sin 2 θ)]

For θ=26.57° and μ 2=2 μ 1,

B 2=x ˆ 1.2+y ˆ 4.6 (T).