Magnetic Boundary Conditions

Magnetic Boundary Conditions

AU : Dec.-04, 11, 14, 17, May-17

• The conditions of the magnetic field existing at the boundary of the two media when the magnetic field passes from one medium to other are called boundary conditions for magnetic fields or simply magnetic boundary conditions. When we consider magnetic boundary conditions, the conditions of  are studied at the boundary. The boundary between the two different magnetic materials is considered. To study conditions of  at the boundary, both the vectors are resolved into two components ;

a) Tangential to boundary and

b) Normal (perpendicular) to boundary.

• Consider a boundary between two isotropic, homogeneous linear materials with different permeabilities µ₁ and µ₂ as shown in the Fig. 8.8.1. To determine the boundary conditions, let us use the closed path and the Gaussian surface.

 

1. Boundary Conditions for Normal Component

• To find the normal component of   , select a closed Gaussian surface in the form of a right circular cylinder as shown in the Fig. 8.8.1. Let the height of the cylinder be Ah and be placed in such a way that Δh/2 is m medium 1 and remaining Δh/2 is in medium 2. Also the axis of the cylinder is in the normal direction to the surface.

• According to the Gauss's law for the magnetic field,

• The surface integral must be evaluated over three surfaces,

i) Top, ii) Bottom and iii) Lateral.

• Let the area of the top and bottom is same, equal to ΔS.

• As we are very much interested in the boundary conditions, reduce Δh to zero. As Δh →   0, the cylinder tends to boundary and only top and bottom surfaces contribute in the surface integral. Thus surface integrals are calculated for top and bottom surfaces only. These surfaces are very small. Let the magnitude of normal component of  be B_N1 and B_N2 in medium 1 and medium 2 respectively. As both the surfaces are very small, we can assume B _N1 and B _N2 constant over their surfaces. Hence we can write,

For top surfaces

• Note that the negative sign is used for one of the surface integrals because normal component in medium 2 is entering the surface while in medium 1 the component is leaving the surface. Hence B _N1 and B _N2 are in opposite direction.

• From equation (8.8.6), we can write,

B _N1 ΔS = B _N2 ΔS

i.e. B _N1 = B _N2  ... (8.8.7)

• Thus the normal component of  is continuous at the boundary.

• As the magnetic flux density and the magnetic field intensity are related by,

• Thus equation (8.8.7) can be written as,

• Hence the normal component of   is not continuous at the boundary. The field strengths in two media are inversely proportional to their relative permeabilities.

 

2. Boundary Conditions for Tangential Component

• According to Ampere's circuital law,

• Consider a rectangular closed path abcda as shown in the Fig. 8.8.1. It is traced in clockwise direction as a-b-c-d-a. This closed path is placed in a plane normal to the boundary surface. Hence  can be divided into 6 parts.

• From the Fig. 8.8.1 it is clear that, the closed path is placed in such a way that its two sides a-b and e-d are parallel to the tangential direction to the surface while the other two sides are normal to the surface at the boundary. This closed path is placed in such a way that half of its portion is in medium 1 and the remaining is in medium 2. The rectangular path is an elementary rectangular path with elementary height Ah and elementary width Δw. Thus over small width Δw,   can be assume constant say H_tan1 in medium 1 and H_tan2 in medium 2. Similarly over a small height can be assumed

constant say H_N1 in medium 1 and H_N2 in medium 2. Now assume that  is the surface current normal to the path. Also from the Fig. 8.8.1 it is clear that the normal and tangential components in medium 1 and medium 2 are in opposite direction. Thus equation (8.8.10) can be written as,

• To get conditions at boundary, Δh → 0. Thus,

K• dw = H_tan1 (Δw) - H_tan2 (Δw)

H_tan1 - H_tan2

In vector form, we can express above relation by a cross product as

• where  is the unit vector in the direction normal at the boundary from medium 1 to medium 2.

• For , the tangential components can be related with permeabilities of two media using equation (8.8.12),

• Consider a special case that the boundary is free of current. In other words, media are not conductors; so K = 0. Then equation (8.8.12) becomes

H_tan1 - H_tan2 = 0

Or H_tan1 - H_tan2 ….. (8.8.15)

 • For tangential components of  we can write,

• From equations (8.8.15) and (8.8.16) it is clear that tangential component of  are continuous, while tangential component of  are discontinuous at the boundary, with the condition that the boundary is current free.

• Let the fields make angles a₁ and a₂ with the normal to the interface as shown in the Fig. 8.8.2.

• Interms of angle a₁ and a₂, we can write relationship between normal components and tangential components of .

In medium 1,

tan ɑ1 = B_tan1 / B_N1 ….. (8.8.17)

• Similarly in medium 2,

tan ɑ2 = B_tan2 / B_N2 ….. (8.8.17)

As we know, B_N1 = B_N2

• Consider an interface between air (medium 1) and soft iron (medium 2). For air, µ_r1 = 1. For soft iron, let µ_r2 = 7000. Then

B_tan1 / B_tan2 = tan ɑ1 / tan ɑ2 = 1 / 7000

If ɑ2 = 85°, then ɑ1 = 0.093° and B_tan1 = 0.

• Thus practically when fields cross a medium of high µr to low µr, then the magnetic fields   are always perpendicular to the boundary.

 

Ex. 8.8.1 Given that  in region y - x - 2 ≤ 0 where µ1 = 5µ0, calculate   in the region y - x – 2 ≥  0 where µ1 = 2µ0

Sol. : As y – x – 2 ≤ 0 is a plane , the region 1 is given as y - x ≤ 2 , i.e. y ≤ x + 2

 Let the surface be defined as,

f(x,y) = y – x - 2

Then the unit vector normal to the plane is given by,

 

Ex. 8.8.2 There exists a boundary between two magnetic material at z = 0, having permittivities µ1 = 4 µ0 for region 1 where z > 0 and µ2 = 7 µ0 for region 2 where z < 0. For a field  in region 1, find tangential component of magnetic field intensity in region 2. Surface current density at boundary z = 0 is 60  A/m.

Sol. : From the given data, it is clear that z = 0 plane separates two magnetic materials. That means z-axis is normal to the boundary. Then normal component of  is given by

Now above z = 0 there exists 1st magnetic material with µ1 = 4 µ0  while below z = 0, there exists 2nd magnetic material 1 to material 2.

• According to boundary condition, the normal component is continuous. Hence we can write,

Now according to boundary condition, for tangential component we can write,

Thus from equations (4) and (8), the magnetic field in 2nd material is given by,

 

Ex. 8.8.3 In region 1, as shown in the Fig. 8.8.4

Determine  in other medium and also calculate the angles made by the fields with the normal.

Sol. : Assume that the boundary is current free.

In medium 1,

As per the boundary shown in the Fig. 8.8.4, x and y are tangential, while z component is normal. So according to the boundary conditions for current free boundary, the tangential component for  remain same. The normal component can be calculated as

From equation (1), the normal component i.e. component in z-direction is

Then the magnetic flux density in medium 2 is given by

As z-direction is perpendicular to boundary, in medium 1, we can write,

The fields make angle 74.49° in medium 1 while 13.52° in medium 2.

 

Ex. 8.8.4 A current sheet  A/m is located at z = 0. The region 1 which is at z < 0 has µ_r1 =4 and region 2 which is at z > 0 has µ _r2 = 3. Given 

Sol. : From the given data, z-axis is normal to the boundary. The normal component of  i.e. H_N2 = 8 A/m. Similarly the tangential component of  = 4.5 A/m.

As there is a current sheet at z = 0, tangential component of  is not continuous at the boundary. For the boundary containing current sheet we must write,

where  is unit vector normal to the boundary drawn from medium 1 to medium 2. In this case

For the boundary containing a current sheet, normal component of  is discontinuous. So we can write,

 

Examples for Practice

Ex. 8.8.5 Let the permittivity be 5 uH/m in region A where x < 0 and 20 uH/m in the region B where x > 0. If there is surface current density

Ex. 8.8.6 In the Fig. 8.8.5 shows is incident on xy plane (z = 0). The medium 1 at z > 0 has µ₁ = 4 µ₀ z < 0 has µ₂ = 7µ₀

Ex. 8.8.7 Region 1 is the semi-infinite space in which 2x-5y>0, while region 2 is defined by 2x -5y < 0. Let µ_r1= 3, µ_r2 = 4 and

Ex. 8.8.8 Two regions are separated by a surface 3x - 2y + 5z = 0 . Region 1 has permeability µ₁ = 2µ₀ and region 2 has µ₂ = 5µ₀. The point ϕ (2,2,2) lies in the region 2 for a field

Review Questions

1. Write a short note on boundary conditions in magnetic field.

2. Derive the boundary conditions to explain the behaviour of magnetic field at the interface of two magnetic media.

AU : Dec.-11, Marks 16; Dec.-14, 17, May-17, Marks 8