Tentamen i ELEKTROMAGNETISK FÄLTTEORI

Transcription

1 Karlstads Universitet Fysik Tentamen i ELEKTROMAGNETISK FÄLTTEORI [ VT 17, FYGB3] Datum: Tid: Lärare: Jürgen Fuchs Tel: / Total poäng: 5 Godkänd / 3: 5 Väl godkänd: : : 4 Tentan består av delar som inlämnas separat: Del 1: 1 p. Del : 4 p. Hjälpmedel: Del 1 & : Ordbok/ordlista engelska svenska Del (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 & Mathematics Handbook Endast en uppgift per sida. Svaren måste vara väl motiverade. FYGB3 Tentamen

2 Problem Electrostatics: Electric field 6 p. a It is natural to work with cylindrical coordinates, with the symmetry axis as 4 p. the z-axis and the center of the disk as the origin. Then in the general formula V( r) = 1 4πǫ we have to use S ρ s r r ds ds = r dr dϕ and r r = z e z r e r. This gives V(,,z) = ρ s 4πǫ π dϕ d r z +(r ) dr = ρ s ǫ [ z +(r ) ] d = ρ s ( z ǫ +d z ). The electric field is then obtained by taking the gradient, which by symmetry only has a z-component: dv E(,,z) = e z dz = ρ ( s z ) e ǫ z +d ±1 z, where the upper sign applies for z> and the lower for z<. Note that for the calculation it matters whether one is above or below the charged disk. b For very large values of z we can approximate 1 p. z ( ) ± z +d 1+ d 1/ d 1 z z (i.e. neglect all higher terms in the Taylor expansion). This gives E(,,z) ± d ρ s ǫ z e z. c In terms of the total charge Q=πd ρ s on the disk, the electric field for large 1 p. z reads E(,,z) Q 4πǫ z (± e z). Recognizing ± e z to be the same as e r on the positive and negative z-axis, respectively, this is the same as the electric field of a point charge Q at the origin. Thus, as could have been expected, very large away from the disk we only can see the total charge. FYGB3 Tentamen

3 Problem 3 Electrostatics: Image charges 5 p. For each of the two point charges there is a corresponding image charge: q 1 = Q and q = Q, located at r 1 =(a,, a) and at r =( a,, a), respectively. a We have to sum up the forces that the second real charge and the two image 3 p. charges exert on the first real charge. Using that the relevant distances are r 1 r 1 = r 1 r = a and r 1 r = a, application of Couolmb s law yields ( F 1 = Q 1 4πǫ 4a e x 1 4a e z + 1 e x + e ) z 8a = Q 16ǫ a ( 1 1 ) ( e x + e z ). b Now we must sum up the electric field intensities of all four charges the two p. real and the two image charges. The relevant distances are Thus r 3 r 1 = r 3 r 1 = a and r 3 r = r 3 r = 5a, E( r 3 ) = Q ( e 4πǫ a z ( e z ) 1 5 = Q ( 1 1 ) πǫ a 5 e z. 5 e x + e z + 1 e x e ) z Note that the field is in z-direction, in agreement with the fact that it can be regarded as the superposition of the fields of two vertical dipoles. The surface charge density is then ρ s ( r) = D( r) = ǫ E( r) for every point r on the conducting plane. FYGB3 Tentamen

4 Problem 4 Conductors 5 p. One can describe the situation as a circuit involving three resistors, with the following resistances: R 1 = resistance of the cable between one end and the isolation fault, R = resistance of the cable between the other end and the isolation fault, R 3 = R fault = resistance between cable and ground at the isolation fault. The resistance R measured at the first end is thus obtained as a series coupling of R 1 with a circuit consisting of parallely coupled resistance R and R fault, so that R = R R fault ) 1. ( 1 R + 1 R fault ) 1. Analogously, RL = R + ( 1 R 1 + Using that the total resistance is of the cable is R 1 +R =: R = L σs, this constitutes two equations expressing the resistance of the fault and its location, which is R 1 L/R, in terms of the various parameters. These equations can e.g. be solved by first rewriting them as = and =. R R fault R R 1 R 1 R fault R L R 5 p. Subtracting these from each other gives a linear equation for R 1 if R =R L and a quadratic equation otherwise, with solution 1 R = R if R =R L, R 1 = 1 ( R (R L R) ) R R L (R R)(R L R) else. R L R (By considering the limit that R and R L are very close one can see that the second solution of the quadratic equation is unphysical.) Inserting this result into one of the two original equations determines the default resistance to be R 1 (R R 1 ) = R(R R) if R =R L, R R fault = 1 R 4(R R ) 1 ( (R +R L )R 1 +R R) else. R L R FYGB3 Tentamen

5 Problem 5 Magnetostatics 6 p. a Describe the field B as a superposition B= B 1 + B, where B 1 is the field of a cylindrical conductor with current density J and B is the field of a cylindrical conductor with current density J that is entirely contained in the first conductor and has its axis parallel to the axis of the former. 5 p. Choose Cartesian coordinates such that the axis of the first conductor is the z-axis and the axis of the second conductor is the parallel line (x,y) = (d,). The field B 1 has a simple expression in cylindrical coordinates. The total current in the cylindrical conductor (without cavity) is I =πb J and hence B 1 (r,ϕ,z) = 1 µ J r e ϕ. For the field B there is in principle a simple expression, too, namely B ( r, ϕ,z) = 1 µ J r e ϕ, but in different coordinates, namely cylindrical coordinates for which the z-axis is at (x,y) = (d,). Working with cylindrical coordinates is therefore inconvenient. However, one arrives at equally simple formulas when instead expressing the fields in Cartesian coordinates: B 1 (x,y,z) = 1 µ J ( y e x +x e y ), B ( x,ỹ,z) = 1 µ J ( ỹ e x + x eỹ). The two Cartesian coordinate systems are simply related by x=x J and ỹ=y; hence B( r) = B 1 + B = 1 µ J d e y. Thus in particular the field in the cavity is homogeneous. b There are many applications in which it is desirable to have a homogeneous magnetic flux density in a region without material, so that one can place some test object in that region. The cavity as considered here is the simplest way to realize such a situation. 1 p. Typically one wants to have a large magnetic flux density, and hence wants d to be as large as possible, Since d 1 b this requires a cylinder with a large radius, and the larger that radius the more expensive the instrument becomes. Thus one tries to make d large, but not larger than necessary for the purpose one has in mind. FYGB3 Tentamen

6 Problem 6 Induction 6 p. a Denoting the current in the straight wire by i 1 and the flux resulting from it 1 p. in the loop by Φ 1 one has L 1 = Φ 1(t) i 1 (t) = b d+a d B 1 (t) d+a i 1 (t) dr = b d µ πr dr = µ b π ln(1+ a d ). b Applying Kirchhoff s voltage law to the loop one gets, with a suitable 1 p. convention on the direction of the current i in the loop, di 1 L 1 dt = L di dt +R i (t). c p. According to the given constraints, the explicit form of the current i 1 (t) is i 1 (t) = I (θ(t t 1 ) θ(t t )) with θ the Heavyside step function. Thus the differential equation reads L 1 I (δ(t t 1 ) δ(t t )) = L di dt +R i (t). The solution that satisfies the initial condition to be zero for times smaller than t 1 is given by i (t) = L 1I L (e R (t t 1 )/L θ(t t 1 ) e R (t t )/L θ(t t )). This is seen by the following calculation: di (t) d L = L 1 I dt dt (e R (t t 1 )/L θ(t t 1 ) e R (t t )/L θ(t t )) = L 1 I (e R (t t 1 )/L δ(t t 1 ) e R (t t )/L δ(t t )) R L L 1 I (e R (t t 1 )/L θ(t t 1 ) e R (t t )/L θ(t t )) = L 1 I (δ(t t 1 ) δ(t t )) Ri (t). d The work done is p. W = R (i (t)) dt = R ( L1 I L ) ( t 1 e R (t t 1 )/L dt + = L 1I L (1 e R (t t 1 )/L ). t e R (t t )/L dt ) e R (t t 1 t )/L dt t FYGB3 Tentamen

7 Problem 7 Elektromagnetic power 7 p. a The electric field E and potential in the region between the conductors 3 p. are the same as the one of a constant line charge density, i.e. in cylindrical coordinates (r, e ϕ,z) E( r) = E(r) = c r e r and V(r) = c lnr for some constant c. The value of tis constant follows from giving V(a) V(b) = V, E(r) = V (ln(b/a))r e r. The magnetic field B is in cylindrical coordinates B( r) = B(r) = µ I πr e ϕ = µ V πrr e ϕ, with I the current in the inner conductor. b Thus the Poynting vector in this region is p. P( r) = E( r) H( r) = 1 E( r) V B( r) = µ πr ln(b/a) 1 r e z. c By integrating this expression over a cross section the region between the 1 p. conductors, one gets P = b a P ds = V πr ln(b/a) π dφ b a 1 r rdr e z e z = V R. d Interpretation: This coincides with the power that is dissipated by a current 1 p. in a resistor of resistance R whose ends have a potnetial difference of V. FYGB3 Tentamen

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