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TE Waves

E_z & = & 0
B_zn|_S & = & 0 The electric field is strictly transverse, but the magnetic field in the $ z$ -direction can be nonzero. Doing exactly the same algebra on the same two equations as we used in the TM case, we get instead: \Vec{H} _&perp#perp;= ±k&mu#mu;&omega#omega;(\Hat{z} ×\Vec{E} _&perp#perp;) along with \Vec{B} _&perp#perp;= ±ik(&mu#mu;&epsi#epsilon;&omega#omega;^2 - k^2)_&perp#perp; &psi#psi; where $ \psi(x,y) e^{ikz} = B_z$ and (&nabla#nabla;_&perp#perp;^2 + (&mu#mu;&epsi#epsilon;&omega#omega;^2 - k^2))&psi#psi;= 0 and the boundary conditions for a TE wave: &psi#psi;n|_S = 0



Robert G. Brown 2017-07-11