..

1. R.M.K. COLLEGE OF ENGINEERING AND TECHNOLOGY
RSM Nagar, Poudyal - 601 206
DEPARTMENT OF ECE
EC 8451
ELECTRO MAGNETIC FIELDS
UNIT IV
TIME - VARYING FIELDS AND MAXWELL’S EQUATIONS
Dr. KANNAN K
AP/ECE

2. Electromagnetic Boundary Conditions
(i) Electric Boundary condition
(ii) Magnetic Boundary condition

3. BOUNDARY CONDITIONS FOR ELECTROMAGNETIC FIELDS
In homogeneous media, electromagnetic quantities vary smoothly and continuously. At a boundary
between dissimilar media, however, it is possible for electromagnetic quantities to be discontinuous.
Continuities and discontinuities in fields can be described mathematically by boundary conditions
and used to constrain solutions for fields away from these boundaries.
Boundary conditions are derived by applying the integral form of Maxwell’s equations to a small
region at an interface of two media
BOUNDARY CONDITIONS FOR ELECTRIC FIELDS
Consider the E field existing in a region that consists of two different dielectrics characterized by 𝜀
1 and 𝜀2 . The fields E1 and E2 in media 1 and media 2 can be decomposed as
E1 = Et1 + En1
E2 = Et2 + En2
Consider the closed path abcda shown in the below figure . By conservative property
𝑬 ⋅ 𝒅𝑳 = 𝟎

4. When magnetic field enter from one medium to another medium, there may be discontinuity in the
magnetic field, which can be explained by magnetic boundary condition
To study the conditions of H and B at the boundary, both the vectors are resolved into two components
(i) Tangential to the boundary (Parallel to boundary)
(ii) Normal to the boundary (Perpendicular to boundary)
These two components are resolved or derived using Ampere’s law and Gauss’s law
Consider two isotropic and homogeneous linear materials at the boundary with different permeabilities
𝜇1 and 𝜇2
Consider a rectangular path and gaussian surface to determine the boundary conditions
According to Ampere’s Law,
𝑬. 𝒅𝑳 = 0

5. 𝑎
𝑎
𝐸. 𝑑𝐿 = 𝑎
𝑏
𝐸. 𝑑𝐿 + 𝑏
𝑐
𝐸. 𝑑𝐿 + 𝑐
𝑑
𝐸. 𝑑𝐿 + 𝑑
𝑎
𝐸. 𝑑𝐿 = 0 ∆𝑤
Here rectangular path height = ∆ℎ 𝑎𝑛𝑑 𝑊𝑖𝑑𝑡ℎ = ∆𝑤
Etan1(∆𝑤) + EN1(
∆ℎ
2
) + EN2(
∆ℎ
2
) – Etan2(∆𝑤) – EN2(
∆ℎ
2
) – EN2(
∆ℎ
2
) = 0
At the boundary , ∆ℎ = 0 (∆ℎ/2- ∆ℎ/2)
Etan1(∆𝑤) – Etan2(∆𝑤) = 0
Etan1(∆𝑤) = Etan2(∆𝑤)
Etan1 = Etan2
The tangential components of Electric field intensity are continuous across the boundary.
In vector form,
(𝑬 tan1 - 𝑬 tan2 ) X 𝐚 N12 = 0
Since D = 𝜖E, the above equation can be written as
D tan1
𝜖𝟏
=
D tan𝟐
𝜖2
,
Dtan1
Dtan2
=
𝜖1
𝜖2

6. Consider a cylindrical Gaussian Surface (Pill box) shown in the Figure , with height ∆h and with top
and bottom surface areas as ∆s
Boundary Conditions for Normal Component
Closed Gaussian surface in the form of circular cylinder is consider to find the normal component of 𝐵
According to Gauss’s law
𝑆
𝐷. 𝑑𝑠 = Q

7. 𝑆
𝐷. 𝑑𝑠 + 𝑆
𝐷. 𝑑𝑠 + 𝑆
𝐷. 𝑑𝑠 = Q = 𝝆𝒔∆𝒔
Top Bottom Lateral
At the boundary, ∆ℎ = 0,So only top and bottom surfaces contribute in the surface
integral
The magnitude of normal component of 𝐷 is DN1 and DN2
For top surface 𝑇𝑂𝑃
𝐷. 𝑑𝑠 = 𝑇𝑂𝑃
DN1 𝑑𝑠
= DN1 𝑇𝑂𝑃
𝑑𝑠
= DN1 ∆𝑠
For bottom surface 𝐵𝑜𝑡𝑡𝑜𝑚
𝐷. 𝑑𝑠 = 𝐵𝑜𝑡𝑡𝑜𝑚
DN2 𝑑𝑠
= DN2 𝐵𝑜𝑡𝑡𝑜𝑚
𝑑𝑠
= DN2 −∆𝑠 = − DN2 ∆𝑠
For Lateral surface 𝐿𝑎𝑡𝑒𝑟𝑎𝑙
𝐷. 𝑑𝑠 = 𝐷 𝐿𝑎𝑡𝑒𝑟𝑎𝑙
𝑑𝑠
= 𝐷 . ∆ℎ = 0
∴ DN1 ∆𝒔 − DN2 ∆𝒔 = Q = 𝝆𝒔∆𝒔

8. DN1 − DN2 = Q = 𝝆𝒔
In vector form,
(𝑫 N1 - 𝑫 N2 ) . 𝐚 N12 = 𝝆𝒔
For perfect dielectric, 𝝆𝒔 = 0
𝑫 N1 - 𝑫 N2 = 0
𝑫 N1 = 𝑫 N2
The normal components of the electric flux density are continuous across the boundary if
there is no free surface charge density.
Since D = 𝜀 𝐸
𝜀 1En1 =
𝜀2 En2
𝐄𝐧𝟏
En2
=
𝜺2
𝜺𝟏

9. BOUNDARY CONDITIONS FOR MAGNETIC FIELDS
Consider a magnetic boundary formed by two isotropic homogenous linear materials with permeability 𝜇1
and 𝜇 2

10. (i) Boundary conditions for Tangential Component
According to Ampere’s Circuital law 𝐻 ⋅ 𝑑𝐿 = 𝐼
𝑎
𝑎
𝐻. 𝑑𝐿 = 𝑎
𝑏
𝐻. 𝑑𝐿 + 𝑏
𝑐
𝐻. 𝑑𝐿 + 𝑐
𝑑
𝐻. 𝑑𝐿 + 𝑑
𝑎
𝐻. 𝑑𝐿 = I= Iencl ∆𝑤
= J ∆𝑤 ; K=Surface current density normal to the Path (∆𝑤) ∆ℎ
Here rectangular path height = ∆ℎ 𝑎𝑛𝑑 𝑊𝑖𝑑𝑡ℎ = ∆𝑤
J ∆𝑤 = Htan1(∆𝑤) + HN1(
∆ℎ
2
) + HN2(
∆ℎ
2
) – Htan2(∆𝑤) – HN2(
∆ℎ
2
) – HN2(
∆ℎ
2
)
At the boundary , ∆ℎ = 0 (∆ℎ/2- ∆ℎ/2)
J ∆𝑤 = Htan1(∆𝑤) – Htan2(∆𝑤)
In vector form,
𝐇 tan1 - 𝐇 tan2 = 𝑱 X 𝐚 N12
J = Htan1– Htan2

11. For 𝐵, the tangential component can be related with Permeabilities of two media
B = 𝜇 H, B tan1 = 𝜇1Htan1 & B tan2 = 𝜇2 H tan2
∴
B tan1
𝜇1
= Htan1 and
B tan2
𝜇2
= Htan2
B tan1
𝝁𝟏
-
B tan𝟐
𝝁2
= J
Special Case :
The boundary is of free of current then media is not a conductor, So K = 0
Htan1 = Htan2
For tangential component of 𝐵 = 𝜇 𝐻 , 𝐻 = 𝐵/ 𝜇
B tan1
𝜇1
-
B tan2
𝜇2
= 0
B tan1
B tan2
=
𝜇1
𝜇2
=
𝜇𝑟1
𝜇𝑟2

12. (ii) Boundary Conditions for Normal Component
Closed Gaussian surface in the form of circular cylinder is consider to find the normal component of 𝐵
According to Gauss’s law for magnetic field
𝑆
𝐵. 𝑑𝑠 = 0
The surface integral must be evaluated over 3 surfaces (Top, bottom and Lateral)
𝑆
𝐵. 𝑑𝑠 + 𝑆
𝐵. 𝑑𝑠 + 𝑆
𝐵. 𝑑𝑠 = 0
Top Bottom Lateral
At the boundary, ∆ℎ = 0,so only top and bottom surfaces contribute in the surface integral
The magnitude of normal component of 𝐵 is BN1 and BN2
For top surface 𝑇𝑂𝑃
𝐵. 𝑑𝑠 = 𝑇𝑂𝑃
BN1 𝑑𝑠
= BN1 𝑇𝑂𝑃
𝑑𝑠
= BN1 ∆𝑠

13. For bottom surface 𝐵𝑜𝑡𝑡𝑜𝑚
𝐵. 𝑑𝑠 = 𝐵𝑜𝑡𝑡𝑜𝑚
BN2 𝑑𝑠
= BN2 𝐵𝑜𝑡𝑡𝑜𝑚
𝑑𝑠
= BN2 −∆𝑠 = − BN2 ∆𝑠
For Lateral surface 𝐿𝑎𝑡𝑒𝑟𝑎𝑙
𝐵. 𝑑𝑠 = 𝐵 𝐿𝑎𝑡𝑒𝑟𝑎𝑙
𝑑𝑠
= 𝐵 . ∆ℎ = 0
∴ BN1 ∆𝑠 − BN2 ∆𝑠 = 0
BN1 ∆𝑠 = BN2 ∆𝑠
BN1 = BN2
Thus the normal component of 𝑩 is continuous at the boundary
we know that 𝐵 = 𝜇 𝐻
For medium 1and 2
𝜇1 𝐻1N1 = 𝜇2 𝐻2N2
𝐻1N1
𝐻2N2
=
𝜇2
𝜇1
=
𝜇r2
𝜇 r1
Hence the normal component of 𝑯 is not continuous at the boundary.
The field strength in two medias are inversely proportional to their relative permeabilities

14. 𝐻1N1
𝐻2N2
=
𝜇2
𝜇1
=
𝜇r2
𝜇 r1
Hence the normal component of 𝐻 is not continuous at the boundary.
The field strength in two medias are inversely proportional to their relative permeabilities
The Electromagnetic boundary conditions are concluded as

15. THANK YOU