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Class 8 in Electrodynamics Based on course by Yuri Lyubarsky and Edited By Avry Shirakov Department of Physics, Ben-Gurion University, Beer-Sheva 84105, Israel This exercise pool is intended for an undergraduate course in “Electrodynamics 1”. Electromagnetic Waves Maxwell’s eq’s: 1 ρf ∇·E= ∇·B=0 ∇×E=− ∂B ∂t ∇ × B = µJ + µ ∂E ∂t If ρf , J = 0 then we have electromagnetic wave eq: ∇2 E = 1 ∂2E c2 ∂t2 It also can be written as E(r, t) = Re E˜0 ei(k·r−ωt) while E˜0 is a complex amplitude, while the magnetic field is: B(r, t) = for k = n (k × E) c 2π . Once again, the field boundary conditions: λ 1 E1⊥ − 2 E2⊥ = σf B1⊥ − B2⊥ = 0 k k E1 − E2 = 0 1 k 1 k B1 − E = kf × n̂ µ1 µ2 2 Snell’s law: n1 sin(θ1 ) = n2 sin(θ2 ) Intensity: I= 1 c0 E 2 =< S > 2 Waves: TE (transverse electric)→ Ez = 0; TM (transverse magnetic)→ Bz = 0; TEM (transverse electro-magnetic)→ Ez , Bz = 0 (cannot occur in hollow wave-guides). 2 Resonant Cavities Placing end plane surfaces of a cylindrical wave-guide, radius R with an infinite conductivity. The cavity is filled with lossless dielectric with µ, . Find the fundamental mode of the waveguide. Solution The reflections at the end surfaces, the ẑ dependence of the fields is same as of standing waves, of the form: A sin(kz) + B cos(kz) the plane boundary surfaces are at z = 0 and z = L, so the boundary conditions can be satisfied at each surface when k=s π for s = 0, 1, 2, ... L For TM fields, vanishing of Et at z = 0 and z = L sπz Ez = f (x, y) cos for s = 0, 1, 2, ...) L and the transverse fields are sπz sπ Et = − 2 sin (∇f )t Lγ L Ht = sπz iω cos z × (∇f )t γ2 L For TE field, vanishing of Hz at z = 0 and z = L requires that sπz Hz = f (x, y) sin for s = 1, 2, 3, ...) L and the transverse fields are sπz iµω Et = − 2 sin z × (∇f )t γ L Ht = sπz sπ cos (∇f )t Lγ 2 L (the boundary conditions are now satisfied), the constant γ 2 is: ∇2t + γ 2 f = 0 while γ 2 = µω 2 − k 2 γ 2 = µω 2 − sπ 2 L and as a result the eigenfrequency is: sπ 2 1 2 ωλs = γλ2 + µ L A practical resonant cavity is a circular cylinder, with an inner radius R and length L. For a TM mode the transverse wave eq. for f = Ez , with the boundary condition Ez = 0 for ρ = R has the solution: f (ρ, φ) = E0 Jm (γmn ρ)e±imφ where γmn = xmn is the n’th root of Jm (x) = 0 x0n = 2.405 5.520 8.654 x1n = 3.832 7.016 10.173 x2n = 5.136 8.417 11.620 xmn R 3 while m = 0, 1, 2, .. and n = 1, 2, 3, ... The TM mode is of the form r x2mn s2 π 2 1 + 2 ωmns = √ 2 µ R L the resonant frequencies form a discrete set of values, for m = 0, n = 1, s = 0 we get the lowest TM mode, ω010 = 2.405 . √ µR The explicit fields are 2.405ρ −iωt Ez = E0 J0 e R r Hphi = −i E0 J1 µ 2.405ρ R e−iωt the result is independent of L (length of the cylinder), there is no tuning for the eigenfrequency. ∂f x0 For a TE mode the basic solution still implies, but the boundary conditions Hz |R = 0 makes γmn = mn where ∂ρ R 0 x0mn is the n’th root of Jm (x) = 0 x00n = 3.832 7.016 10.173 x01n = 1.841 5.331 8.536 x02n = 3.054 6.706 9.970 again, the TE modes are of the form s 0 2 xmn s2 π 2 1 + 2 ωmns = √ 2 µ R L for m = 1, n = 1, s = 1 we get the lowest TE mode, ω111 1.841 = √ µ r 1+ 2.912R2 L2 The explicit field expression are πz 1.841ρ Hz = H0 J1 e−iωt cos(φ) sin R L . For L > 2.03R the resonance frequency ω111 is smaller than that for TM mode (ω010 ). Then the T E111 mode is the fundamental oscillation of the cavity. Problem 0355 An infinitely long thin cylindrical shell (inner radius a, thickness d << a) made of a conductor with the conductivity σ is placed in a magnetic field Re(B0 eiωt ) parallel to the cylinder axis. Find the magnetic field inside the cylinder. Solution The magnetic field outside is: B= B0 iωt e + e−iωt 2 inside the conducting layer the field is µ0 σ ∂B = ∇2 B ∂t where we used Jf = σE and ∇ × (∇ × B) = ∇ (∇ · B) − ∇2 B = −∇2 B. We guess that the solution for B is of the form : B(r) = B + (r)eiωt + B i (r)e−iωt (ẑ) 4 plugging in the solution ∂2 + B = iσωµB + ∂r2 ∂2 − B = iσωµB − ∂r2 while the B + = C1 eik1 r + C2 e−ik1 r and B − = D1 eik2 r + D2 e−ik2 r while r k1 = µσω (1 + i) and k2 = 2 r µσω (1 − i) 2 from the convergence of the magnetic field B0 = C1 eik1 (d+a) + C2 e−ik1 (d+a) 2 B0 = D1 eik2 (d+a) + D2 e−ik2 (d+a) 2 as we see, inside the field is the solution of Poison’s eq, with constant boundary conditions. + iωt − −iωt Bin = Bin e + Bin e Bin = C1 eik1 a + C2 e−ik1 a On the other hand, in the inner bound: ∇ × B = µσE and H E · dl = − dΦmag dt For B + ∇×B=− ∂ Az = ik1 C1 eik1 a − C2 e−ik1 a ∂r I Edl = Eφ 2πa dΦmag d + = πa2 Bin = iωπa2 C1 eik1 a + C2 e−ik1 a dt dt summing up, ik1 C1 eik1 a − C2 e−ik1 a = −µσiωπa2 C1 eik1 a + C2 e−ik1 a here we find the radio between C1 and C2 , same mechanism is applied for finding B − .