Unit IV Boundary Conditions

Electromagnetic Boundary Conditions

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 C onditions ( 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 𝜀 2 . The fields E 1 and E 2 in media 1 and media 2 can be decomposed as E 1 = E t1 + E n1 E 2 = E t2 + E n2 Consider the closed path abcda shown in the below figure . By conservative property  

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 and Consider a rectangular path and gaussian surface to determine the boundary conditions According to Ampere’s Law,  

5. + + + = Here rectangular path height = E tan1 ( ) + E N1 ( ) + E N2 ( ) – E tan2 ( ) – E N2 ( ) – E N2 ( ) = 0 At the boundary , = 0 ( /2- /2) E tan1 ( ) – E tan2 ( ) = 0 E tan1 ( ) = E tan2 ( ) E tan1 = E tan2 The tangential components of Electric field intensity are continuous across the boundary. In vector form, ( tan1 - tan2 ) X N12 = 0 Since D = E, the above equation can be written as = , =  

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 According to Gauss’s law = Q  

7. + + = Q = Top Bottom Lateral At the boundary, = 0,So only top and bottom surfaces contribute in the surface integral The magnitude of normal component of is D N1 and D N2 For top surface = = D N1 = D N1 For bottom surface = = D N2 = D N2 For Lateral surface = = . = 0 D N1 = Q = B N1  

8. D N1 = Q = In vector form, ( N1 - N2 ) . N12 = For perfect dielectric, = 0 N1 - N2 = 0 N1 = N2 The normal components of the electric flux density are continuous across the boundary if there is no free surface charge density. Since D = 2 E n2  

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 + + + = I= I encl = J ; K=Surface current density normal to the Path ( Here rectangular path height = J = H tan1 ( ) + H N1 ( ) + H N2 ( ) – H tan2 ( ) – H N2 ( ) – H N2 ( ) At the boundary , = 0 ( /2- /2) J = H tan1 ( ) – H tan2 ( ) In vector form, tan1 - tan2 = X N12   J = H tan1 – H tan2

11. For the tangential component can be related with Permeabilities of two media B = H, B tan1 = H tan1 & B tan2 = 2 H tan2 = H tan1 and = H tan2 - = J Special Case : The boundary is of free of current then media is not a conductor, So K = 0 H tan1 = H tan2 For tangential component of = , = / - = 0 = =  

12. (ii) Boundary Conditions for Normal Component Closed Gaussian surface in the form of circular cylinder is consider to find the normal component of According to Gauss’s law for magnetic field = 0 The surface integral must be evaluated over 3 surfaces (Top, bottom and Lateral) + + = 0 Top Bottom Lateral At the boundary, = 0,so only top and bottom surfaces contribute in the surface integral The magnitude of normal component of is B N1 and B N2 For top surface = = B N1 = B N1  

13. For bottom surface = = B N2 = B N2 For Lateral surface = = . = 0 B N1 = 0 B N1 B N1 Thus the normal component of is continuous at the boundary we know that = For medium 1and 2 1 1 N 1 = 2 N 2 = Hence the normal component of is not continuous at the boundary. The field strength in two medias are inversely proportional to their relative permeabilities  

14. = Hence the normal component of 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  

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