Tentamen i ELEKTROMAGNETISK FÄLTTEORI

Transcription

1 Karlstads Universitet Fysik Tentamen i ELEKTROMAGNETISK FÄLTTEORI [ VT 216, FYGB3] Datum: Tid: Lärare: Jürgen Fuchs c/o Mattias Flygare Tel: Total poäng: 5 Godkänd / 3: 25 Väl godkänd: : : 42 Tentan består av 2 delar som inlämnas separat: Del 1: 1 p. Del 2: 4 p. Hjälpmedel: Del 1 & 2: Ordbok/ordlista engelska svenska Del 2 (efter del 1 har inlämnats dessutom: Ett handskrivet A4 ark med valfritt innehåll (skrivet på ena sidan, ej maskinskriven eller maskinkopierad inlämnas tillsammans med tentan Physics Handbook Endast en uppgift per sida. Svaren måste vara väl motiverade. FYGB3 Tentamen

2 Karlstads Universitet Fysik Tentamen i ELEKTROMAGNETISK FÄLTTEORI FÖR CIVILINGENJÖRER [ VT 216, FYGB3] Datum: Tid: Lärare: Jürgen Fuchs c/o Mattias Flygare Tel: Total poäng: 5 Godkänd / 3: 25 Väl godkänd: : : 42 Tentan består av 2 delar som inlämnas separat: Del 1: 1 p. Del 2: 4 p. Hjälpmedel: Del 1 & 2: Ordbok/ordlista engelska svenska Del 2 (efter del 1 har inlämnats dessutom: Ett handskrivet A4 ark med valfritt innehåll (skrivet på ena sidan, ej maskinskriven eller maskinkopierad inlämnas tillsammans med tentan Physics Handbook Endast en uppgift per sida. Svaren måste vara väl motiverade. FYGB3 Tentamen

3 Problem 2 Electrostatics: Electric field 5 p. Choose Cartesian coordinates such that the semicircle lies in the x-y-plane with its center at the origin and is symmetric about the y-axis. Then by symmtery the electric field at the center has only a y-component. The contribution to E by a small line element located at r is in radial direction. It is important to realize that e r is not a constant; explicitly, the contribution of a line element at angle ϕ to the y-component is 5 p. de y = d sinϕ ρ l 4πǫ d 2 dϕ. Hence E = e y ϕ=π ϕ= de y = ρ l 2πǫ d e y. FYGB3 Tentamen

4 Problem 3 Electrostatics: Capacitance 6 p. a The sources of the elictric displacement D are the macroscopic charges. Using the spherical symmetry, Gauss law gives D(r = Q 4πr e 2 r for r 1 r r 2 (assuming that r 1 <r 2. The electric field intensity E satisfies E= D/ǫ with ǫ=ǫ ǫ r, hence E(r = (1+drQ 4πǫ cr 2 e r for r 1 r r 2. b The potential difference between the two spheres is r2 V = E dl Q ( 1 = 1 +d ln r 2. 4πǫ c r 1 r 2 r 1 r 1 Thus the capacitance is C = Q V = 4πǫ cr 1 r 2 r 2 r 1 +dr 1 r 2 ln(r 2 /r 1. c The polarization is given by P = D ǫ E = (ǫ ǫ ( E = ǫ c E 1+dr 1 = Q ( 1 1+dr e 4πr 2 r. c Hence the polarization charge density is (using the divergence in spherical coordinates ρ P = P = 1 r 2 r (r2 P r = 1 ( drq = dq r 2 r 4πc 4πcr. 2 1 p. d The surface polarization charge density is ρ Ps = P e n. Here e n = e r, hence 1 p. ρ Ps (r 1 = Q ( 1 1+dr 1 and ρ 4πr1 2 Ps (r 2 = Q ( 1 1+dr 2. c 4πr2 2 c FYGB3 Tentamen

5 Problem 4 Conductors 6 p. a ( This problem is essentially exercise 5 6 in Cheng s book. 3 p. For the solution see also pages 218 and 219 of the book. There are no macroscopic charges, hence the potential V satisfies the Laplace equation. It makes sense to aasume that, in cylindrical coordinates, the potential does not depend on r or z, since definitely each of the end faces of the wire should be an equipotential. Using the Laplacian in cylindrical coordinates it thus follows that d 2 V dϕ 2 =, and this is solved by V being a linear function of ϕ. Moreover, with such a function one can satisfy the boundary conditions. By the uniqueness of the solution to the Laplace equation with Dirichlet boundary conditions, this is the solution, and hence in particular also the assumptions are justified. b Since the potential is linear in ϕ, the electric field is in ϕ-direction with constant magnitude. Explicitly, V(ϕ = 2 π V ϕ+const and E = V = 1 r with V the potential difference between the two ends. V ϕ = 2 πr V e ϕ The current density is J =σe and hence the current in the wire is I = J ds h b = σ dz dr E e ϕ ϕ=const a = 2σ π V h b a 1 2σh dr = r π V ln b a, with a and b=a+h the inner and outer radious of the quarter circle. Thus the resistance is R = V π = I 2σh ln(b/a. c If just the piece of wire is present, then the situation is unphysical, since any potential difference in a conductor is quickly reduced to zero by a rearrangement of charges. However, the situation can be physically realized as part of a larger system, e.g. a circuit with a voltage source. 1 p. FYGB3 Tentamen

6 Problem 5 Magnetostatics 7 p. a Using cylindrical coordinates (r,ϕ.z, with the z-direction coinciding with the axis of the transmission line, the magnetic B-field is (including also the outside region r a+d+b 4 p. µ Ir 2πa e 2 ϕ for r a, µ I 2πr e ϕ for a r a+d, B(r,ϕ,z = µ I 2πr r 2 (a+d+b 2 (a+d+b 2 (a+d 2 e ϕ for a+d r a+d+b, for r a+d+b. b The magnetic vector potential A is related to B by 3 p. B = A = 1 e r r e ϕ e z r r ϕ z. A r ra ϕ A z When B is of the special form B= B ϕ (r e ϕ, then the only partial derivatives contributing to the curl are A r z and A z r. We can therefore assume that the ϕ-component is zero and that A r and A z do not depend on the other coordinates, and moreover can take (say A r = (in this assumption one uses implicitly that A can be prescribed arbitrarily. Then A is the form A(r,ϕ,z=A z (r e z and we get ( µ I r 2 4πa +c 2 1 e z for r a, ( µ I 2π lnr +c 2 e z for a r a+d, A( r = e z B ϕ (rdr = ( µ I 2πb(2a+2d+b [1 2 r2 (a+d+b 2 lnr]+c 3 e z c 4 e z for a+d r a+d+b, for r a+d+b, where the additive constants c i are determined by imposing continuity of A. FYGB3 Tentamen

7 Problem 6 Induction 5 p. The magnetic flux through the loop is Φ = B(x,y,z,t ds, S where S is the surface bounded by the square, i.e. d ds = dxdy e z and dxdy = dx Thus d Φ = B cos(ωt e z e z dx = B d cos(ωt 2d ( πy π sin 2d The resistance of the loop is S d d dy. ( πy cos dy 2d d = 2 π B d 2 cos(ωt. R = L wire σs = 4d σa with S the cross section and σ the conductivity of the wire. Thus the current induced in the wire is I = V R = 1 R dφ dt = 1 2π B dωσa sin(ωt. 5 p. FYGB3 Tentamen

8 Problem 7 Electric circuits / Maxwell s equations 6 p. a The voltage drops at the resistor and at the capacitor add up to zero, hence we have Q dq = V = RI = R C dt. This differential equation for Q together withthe initial condition Q(=Q. is solved by Q(t=Q e t/rc. Hence Thus E(t = 1 ǫ E = Q ǫs D(t = σ(t ǫ and = Q(t ǫs = Q ǫs e t/rc. τ = RC = ǫrs d. The direction of E is orthogonal to the plates. b Choose cylindrical coordinates for which the z-axis is the symmetry axis of the capacitor, so that the electric field is in z-direction. By symmetry, the magnetic field only has a ϕ-component, and this component only depends on r. By comparison with the Maxwell equation B = µ ǫ E t and 1 with the curl in cylindrical coordinates (the relevant part being e z r it follows that B(r,t = µ ǫre(t e ϕ 2τ (for r<b, and B== E outside the capacitor. (rb ϕ r c The Poynting vector is P(r,t = E(t H(r,t = ǫre2 (t e r. 2τ Thus the total energy flowing through the surface of the cylindrical region is W = 2π b dp( b,tdt = π b 2 dq 2 ǫτ S 2 e 2t/τ dt = b 2 dq 2 2πǫb 4. FYGB3 Tentamen

9 Problem 8 Elektromagnetic waves 5 p. a 1 p. The exponential term in the expression for E( r must be equal to exp( i k r = exp( i(k x x+k y y+k z z, so that one can read off that the wave vector k, whose direction gives the direction of propagation of the wave, is k = a( ex e y. Note that this is perpendicular to E (and to B. b The complex magnetic flux density B( r is given by 1 B( r = 1 H( r = k E a E = exp(ia (x y ( e x + e y 2 e z. µ ωµ ωµ c The Poynting vector is P( r,t = E( r,t H( r,t ( ( = R E( r exp(iωt R H( r exp(iωt = E2 a [ ( 2( ex R exp(ia (x yexp(iωt] + e y + e z ( e x + e y 2 e z ωµ = 3E2 a ωµ cos 2 ( a(x y+ωt( e x e y. (Note that this is in direction of k, as it should be. The time average of the function cos 2 is 1, hence the time average of the 2 Poynting vector is P( r = 3E2 a 2ωµ ( e x e y. FYGB3 Tentamen