This final is a three hour open book, open notes exam. Do all four problems.

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Hilary Heath
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1 Physics 55 Fll 27 Finl Exm Solutions This finl is three hour open book, open notes exm. Do ll four problems. [25 pts] 1. A point electric dipole with dipole moment p is locted in vcuum pointing wy from nd distnce d wy from the flt surfce of semi-infinite dielectric with permittivity ɛ. [15] ) Find the electric potentil Φ everywhere. This problem cn be solved by the method of imges for dielectric. Recll tht for point chrge q locted distnce d from the flt surfce of semi-infinite dielectric ε ε P R 2 R 1 d q q ( q ) the electric potentil is given by where Φ(z > ) = 1 4πɛ nd where the position vectors re Φ(z < ) = 1 q 4πɛ R 1 ( ) q + q R 1 R 2 ( ) ɛ q = q, q 2ɛ = q R 1 = (x, y, z d), R2 = (x, y, z + d) Here we hve ssume tht the physicl chrge q is locted t (,, d) nd the observer is t point P given by (x, y, z). By substituting in q nd q, the potentil is more explicitly written s Φ(z > ) = q ( 1 ɛ ɛ ) 1 4πɛ R 1 R 2 Φ(z < ) = q ( ) (1) 2ɛ 1 4πɛ R 1 z

2 Since point electric dipole my be obtined by tking two chrges q nd +q seprted by distnce l in the limit l, the dipole problem my be solved by liner superposition ε ε P R 2 R 1 q q d q q ( q q ) z Since the electric potentil for dipole in free spce is given by Φ = 1 p r 4πɛ r 3 the electric dipole generliztion of the point chrge solution (1) is then ( ) Φ(z > ) = p R1 + ɛ ɛ R2 4πɛ Φ(z < ) = p 4πɛ ( R 3 1 2ɛ R1 R 3 1 R 3 2 ) (2) where p = pẑ is pointing wy from the dielectric. Note tht, ccording to the figure, the imge dipole points in the opposite direction s the physicl one, so long s we define the direction to be from q to +q. This is wht ccounts for the sign difference between the two terms in the first lines of (1) nd (2). In relity, however, since the imge chrge q hs the opposite sign s q (ssuming ɛ > ɛ ), the imge dipole ctully points in the sme direction s the physicl one. This physicl result is consistent with the plus sign in the first line of (2), which shows tht both dipoles point in the sme direction. [1] b) Wht is the electric potentil if the dipole is insted oriented prllel to the surfce of the dielectric? Note tht the orienttion of the imge dipole is different for the prllel configurtion ε ε P R 2 q q ( q ) R 1 q q ( q ) d z

3 As result, the potentil is given insted by ( Φ(z > ) = p R1 4πɛ Φ(z < ) = p 4πɛ ( R 3 1 2ɛ R1 R 3 1 ɛ ɛ R2 where p = pˆx is pointing prllel to the surfce of the dielectric. Note tht the imge solution cn be generlized for dipole t n rbitrry ngle reltive to the surfce of the dielectric. [25 pts] 2. A wire coil is wound round the surfce of solid sphere of rdius nd reltive permebility µ r. The coil is designed in such wy tht it crries surfce current density K = ˆφ(I/) sin θ. ) R 3 2 ) Find the mgnetic induction B everywhere. Despite the presence of surfce current, this problem my be solved using mgnetic sclr potentil pproch. The trick is to relize tht the two seprte regions r < nd r > re both current-free regions of spce. This llows us to introduce inside (r < ) nd outside (r > ) potentils H in = Φ in M, H out = Φ out M where Φ in M nd Φout M solve Lplce s eqution, 2 Φ M =. Using sphericl coordintes, nd tking zimuthl symmetry into ccount, we my write Φ in M = l Φ out M = l A l r l P l (cos θ) B l r l+1 P l(cos θ) (3) The effect of the surfce current K = ˆφ(I/) sin θ shows up in the mtching conditions t r = ˆr ( B out B in ) r= =, ˆr ( H out H in ) r= = K In explicit components, these conditions re B in r = Br out, Hθ in Hθ out = I sin θ (t r = )

4 Given (3), the pproprite components of the mgnetic induction nd mgnetic field re Br in = µ la l r l 1 P l (cos θ) l nd B out r H in θ = µ (l + 1) B l r l+2 P l(cos θ) = l l A l r l 1 P l (cos θ) sin θ H out θ = l B l r l+2 P l (cos θ) sin θ By orthogonlity, the mtching conditions must independently hold for ech vlue of l. Noting tht I sin θ = I P 1(cos θ) sin θ we see tht the mtching conditions re (l + 1)B l + µ r la l 2l+1 = B l A l 2l+1 = I l+1 δ l,1 These equtions re homogeneous, except for l = 1. As result, only the l = 1 mode contributes, with solution A 1 = I The mgnetic sclr potentil is then µ r /2, B 1 = I 2 µ r /2 1 + µ r /2 Φ in M = I µ r /2 r cos θ = I µ r /2 z, Φ out M = I 2 µ r /2 1 + µ r /2 This gives rise to mgnetic induction B in = µ I 1 r 2 cos θ = I2 µ r /2 1 + µ r / µ r /2ẑ B out = µ I 2 µ r /2 1 + µ r /2 3(z/r)ˆr ẑ r 3 The interior field is uniform, while the exterior field is tht of mgnetic dipole. z r 3

5 [25 pts] 3. A semi-infinite coxil cble consists of n inner conductor of rdius surrounded by n outer conductor of rdius b. A dielectric with permittivity ɛ nd permebility µ fills the volume between the conductors. V [5] ) If constnt (sttic) potentil difference V is pplied between the conductors, wht is the electric field inside the cble? Ignore fringe effects. It is nturl to use cylindricl coordintes for this problem. For the electrosttics problem, n elementry ppliction of Guss lw gives n electric field E = 1 λ 2πɛ ρ ˆρ where λ is the chrge per unit length on the inner conductor. Since the potentil difference between conductors is V, we hve b b V = E d λ dρ l = 2πɛ ρ = λ ( ) b 2πɛ log As result, the electric field is given in terms of V by V E = ρ log(b/) ˆρ [15] b) Show tht, if sinusoidl potentil difference V (t) = V e iωt is pplied t the end of the cble, then Mxwell s equtions dmit trveling wve solution B = ˆφB(ρ)e i(kz ωt), E = ˆρE(ρ)e i(kz ωt) where z is the direction long the xis of the cble. Find B(ρ) nd E(ρ) in terms of V. The sinusoidl potentil difference is of hrmonic form. Thus we my exmine the hrmonic Mxwell s equtions. Firstly, Guss lw for mgnetism, B =, is trivilly stisfied for the bove solution. For the source-free Guss lw, E =, we hve 1 ρ ( ) C ρe(ρ) = E = ρ ρ for some constnt C. This llows us to write E = ˆρ C ρ ei(kz ωt) Frdy s lw, E iω B =, then gives ik C ρ iωb = B = C ρ k ω

6 so tht B = ˆφ C ρ k ω ei(kz ωt) The remining eqution to verify is the Ampère-Mxwell eqution, B + iµɛω E =, which gives i C ρ k 2 ω + ic ρ µɛω = k = µɛ ω As result, Mxwell s equtions re solved provided we impose the stndrd dispersion reltion k = µɛ ω. Note tht, if we tke z = to be the end of the cble, we my solve for the constnt C by imposing b b V e iωt = E z= d C l = ρ e iωt dρ = Ce iωt log This gives C = V / log(b/), so tht ( ) b V E(ρ) = ρ log(b/), µɛ B(ρ) = V ρ log(b/) Note tht trveling wves in the z direction (s well s superpositions of wves) re lso possible. [5] c) Wht is the impednce Z (given by the complex Ohm s lw, V = IZ) of the cble? The impednce is given by Z = V/I. The potentil t the end of the cble (z = ) is lredy given, so ll we need is the current. The current my be obtined from Ampère s lw in integrl form B d l = µi where we integrte long circle of rdius < ρ < b locted t z =. This gives I = 1 µɛ V µ ρ log(b/) ɛ (2πρ) = 2π µ V log(b/) The impednce is then Z = 1 µ 2π ɛ log ( ) b which is rel nd independent of frequency. This is feture of coxil trnsmission lines. Note tht the displcement current term is in the ˆρ direction (since this is the direction of the electric field) nd does not contribute to the bove ppliction of Ampère s lw. Alterntively, we my clculte the current using the

7 mtching condition ˆn H S = K (where S denotes the surfce of the conductor) to obtin the surfce current density K nd they by I = (circumference) K. [25 pts] 4. A plne polrized electromgnetic wve of frequency ω in free spce is normlly incident on the flt surfce of n excellent conductor (µ = µ, ɛ = ɛ nd σ ωɛ ) which fills the region z >. Assume the incident wve is given by E = ˆxE i e i(kz ωt) [1] ) Wht is the current density J inside the conductor (in the limit σ ωɛ )? Express your result in terms of the skin depth δ = 2/µ σω. For normlly incident plne wve, we tke the incoming wve to be in medium with index of refrction n nd the trnsmitted wve to be in medium with index of refrction n. Then incident: E = ˆxEi e i(kz ωt) trnsmitted: E = ˆxE i ( 2n n + n ) e i(k z ωt) For this problem, we hve n = 1. For n, we use the excellent conductor pproximtion n = 1 + i σ i σ = 1 + i σ = (1 + i) c ωɛ ωɛ 2 ωɛ δω where δ = 2/µ σω is the skin depth. The trnsmitted wvenumber k is Hence k = ω c n 1 + i δ E 2 = ˆxE i 1 + n ei(k z ωt) 2 ˆxE i n ei(k z ωt) ˆxE i (1 i) δω c ei(z/δ ωt) e z/δ The current density inside the conductor is then J = σe ˆxE i (1 i) δωσ c ei(z/δ ωt) e z/δ = ˆx E i (1 i) 2 µ δ ei(z/δ ωt) e z/δ (4) [5] b) Now ssume tht the conductor is perfect. Solve for the reflected wve in the limit of perfect conductor. (Note tht E vnishes t the surfce of perfect conductor.) If wve E = ˆxE i e i(kz ωt) is normlly incident on perfect conductor, the reflected wve will hve the form E = ˆxE e i( kz ωt). The totl electric field t z = (the surfce of the conductor) is then ˆx(E i + E )e iωt. Since this is in the prllel direction, it must vnish. Hence E = E i. The reflected wve is then E = ˆxE i e i( kz ωt)

8 This is interpreted s 18 phse shift. [1] c) Compute the idelized surfce current density K on the surfce of the perfect conductor, nd show tht it stisfies the reltion K = J dz where J is the current density found in prt. The surfce current density is given by K = ẑ ( H H H ) z= = 1 µ ẑ ( B + B ) z= where we used the fct tht the trnsmitted wve B vnishes in perfect conductor. The incident nd reflected mgnetic inductions re B = µ ɛ ẑ E, B = µ ɛ ( ẑ) E Hence K = ẑ ( ẑ ( E µ E ) ) z= = ( E µ E ) z= Substituting in E nd E gives K = ˆx µ (2E i )e ωt We now compre this with the current density found in prt. From (4) we hve Jdz = ˆx E i (1 i) 2 µ δ e iωt e (1 i)z/δ dz = ˆx E i (1 i) 2 δ µ δ e iωt 1 i = ˆx 2E i e iωt µ So we see tht K = J dz is indeed stisfied.

Candidates must show on each answer book the type of calculator used.

Candidates must show on each answer book the type of calculator used. UNIVERSITY OF EAST ANGLIA School of Mthemtics My/June UG Exmintion 2007 2008 ELECTRICITY AND MAGNETISM Time llowed: 3 hours Attempt FIVE questions. Cndidtes must show on ech nswer book the type of clcultor