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Electromagnetic Boundary Conditions Explained
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 MAXWELLβS EQUATIONS
Dr. KANNAN K
AP/ECE
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 Maxwellβs 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 π
1 and π2 . 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
π¬ β π π³ = π
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 Ampereβs law and Gaussβs law
Consider two isotropic and homogeneous linear materials at the boundary with different permeabilities
π1 and π2
Consider a rectangular path and gaussian surface to determine the boundary conditions
According to Ampereβs Law,
π¬. π π³ = 0
5. π
π
πΈ. ππΏ = π
π
πΈ. ππΏ + π
π
πΈ. ππΏ + π
π
πΈ. ππΏ + π
π
πΈ. ππΏ = 0 βπ€
Here rectangular path height = ββ πππ ππππ‘β = βπ€
Etan1(βπ€) + EN1(
ββ
2
) + EN2(
ββ
2
) β Etan2(βπ€) β EN2(
ββ
2
) β EN2(
ββ
2
) = 0
At the boundary , ββ = 0 (ββ/2- ββ/2)
Etan1(βπ€) β Etan2(βπ€) = 0
Etan1(βπ€) = Etan2(βπ€)
Etan1 = Etan2
The tangential components of Electric field intensity are continuous across the boundary.
In vector form,
(π¬ tan1 - π¬ tan2 ) X π N12 = 0
Since D = πE, the above equation can be written as
D tan1
ππ
=
D tanπ
π2
,
Dtan1
Dtan2
=
π1
π2
6. Consider a cylindrical Gaussian Surface (Pill box) shown in the Figure , with height βh and with top
and bottom surface areas as βs
Boundary Conditions for Normal Component
Closed Gaussian surface in the form of circular cylinder is consider to find the normal component of π΅
According to Gaussβs law
π
π·. ππ = Q
7. π
π·. ππ + π
π·. ππ + π
π·. ππ = Q = ππβπ
Top Bottom Lateral
At the boundary, ββ = 0,So only top and bottom surfaces contribute in the surface
integral
The magnitude of normal component of π· is DN1 and DN2
For top surface πππ
π·. ππ = πππ
DN1 ππ
= DN1 πππ
ππ
= DN1 βπ
For bottom surface π΅ππ‘π‘ππ
π·. ππ = π΅ππ‘π‘ππ
DN2 ππ
= DN2 π΅ππ‘π‘ππ
ππ
= DN2 ββπ = β DN2 βπ
For Lateral surface πΏππ‘ππππ
π·. ππ = π· πΏππ‘ππππ
ππ
= π· . ββ = 0
β΄ DN1 βπ β DN2 βπ = Q = ππβπ
8. DN1 β DN2 = Q = ππ
In vector form,
(π« N1 - π« N2 ) . π N12 = ππ
For perfect dielectric, ππ = 0
π« N1 - π« N2 = 0
π« N1 = π« N2
The normal components of the electric flux density are continuous across the boundary if
there is no free surface charge density.
Since D = π πΈ
π 1En1 =
π2 En2
ππ§π
En2
=
πΊ2
πΊπ
9. BOUNDARY CONDITIONS FOR MAGNETIC FIELDS
Consider a magnetic boundary formed by two isotropic homogenous linear materials with permeability π1
and π 2
10. (i) Boundary conditions for Tangential Component
According to Ampereβs Circuital law π» β ππΏ = πΌ
π
π
π». ππΏ = π
π
π». ππΏ + π
π
π». ππΏ + π
π
π». ππΏ + π
π
π». ππΏ = I= Iencl βπ€
= J βπ€ ; K=Surface current density normal to the Path (βπ€) ββ
Here rectangular path height = ββ πππ ππππ‘β = βπ€
J βπ€ = Htan1(βπ€) + HN1(
ββ
2
) + HN2(
ββ
2
) β Htan2(βπ€) β HN2(
ββ
2
) β HN2(
ββ
2
)
At the boundary , ββ = 0 (ββ/2- ββ/2)
J βπ€ = Htan1(βπ€) β Htan2(βπ€)
In vector form,
π tan1 - π tan2 = π± X π N12
J = Htan1β Htan2
11. For π΅, the tangential component can be related with Permeabilities of two media
B = π H, B tan1 = π1Htan1 & B tan2 = π2 H tan2
β΄
B tan1
π1
= Htan1 and
B tan2
π2
= Htan2
B tan1
ππ
-
B tanπ
π2
= 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 π΅ = π π» , π» = π΅/ π
B tan1
π1
-
B tan2
π2
= 0
B tan1
B tan2
=
π1
π2
=
ππ1
ππ2
12. (ii) Boundary Conditions for Normal Component
Closed Gaussian surface in the form of circular cylinder is consider to find the normal component of π΅
According to Gaussβs law for magnetic field
π
π΅. ππ = 0
The surface integral must be evaluated over 3 surfaces (Top, bottom and Lateral)
π
π΅. ππ + π
π΅. ππ + π
π΅. ππ = 0
Top Bottom Lateral
At the boundary, ββ = 0,so only top and bottom surfaces contribute in the surface integral
The magnitude of normal component of π΅ is BN1 and BN2
For top surface πππ
π΅. ππ = πππ
BN1 ππ
= BN1 πππ
ππ
= BN1 βπ
14. π»1N1
π»2N2
=
π2
π1
=
πr2
π r1
Hence the normal component of π» 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