Astronomy 9620a / Physics 9302a 2nd Problem List and Assignment

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1 Astronomy 960a / Physics 930a nd Problem List and Assignment Your solution to problems 3, 7, and 11 has to be handed in on Tue., Nov. 19, Paramagnetic materials. Let us consider an atom that has one unpaired electron, which moves in an orbit about the nucleus (classical model). a) Show that the magnetic dipole moment m of the electron can be expressed as m = q L, (1.1) M where ( q), M, and L are the electron s charge, mass, and angular momentum respectively. b) This magnetic dipole is then subjected to an external magnetic induction B = Be z. Consider the torque felt by the magnetic dipole, and show that, under these conditions, L will precess about the z-axis (i.e., B ) with constant angle (let us call it θ ), and at the Larmor frequency ω L defined by ω L = qb M. (1.) c) Paramagnetic materials generally consist of atoms, molecules, or ions that have a net nonzero angular momentum. Under the influence of an external magnetic induction, each components of a gas will have a potential energy U = m B == mbcos( θ). (1.3) The probability P( θ)dθ that the magnetic moment lies between the angles θ and θ + dθ is proportional to the Boltzmann factor and the solid angle π sin( θ)dθ βmb cos θ P( θ)dθ e sin( θ)dθ, (1.4) where β = 1 kt. Show that the average value cos( θ) is given by [Hint. You will need the following relation cos( θ) = coth( βmb) 1 βmb. (1.5) 1

2 ] α ln d) Assume that βmb 1, and show that = eα x x dx eα x dx eα x dx. (1.6) cos( θ) = 1 βmb. (1.7) 3 = m cos( θ) is the only non-vanishing component of the mean Furthermore, since m z dipole moment, show that if we define a paramagnetic susceptibility χ m = M B, then χ m = nm 3kT, (1.8) with n the number density of molecules. (Please note that the magnetic susceptibility is usually defined by M = χ m H.) e) Using L!, and your best guess for the other quantities involved in equation (1.8), provide an order-of-magnitude calculation for χ m. Solution. a) The magnetic dipole moment is given by m = 1 x J d 3 x. (1.9) For an unpaired electron in orbit, we have J = qvδ ( x x 0 ), and equation (1.9) becomes m = q v xδ ( x x 0 ) d 3 x = q x 0 v = q ( M x 0 Mv) = q M L. (1.10) b) The torque on the dipole is given by N = m B. (1.11) But we also know that N dl dt, combining this relation with equation (1.11) we write

3 dl dt = q L B, (1.1) M any variation in L is then perpendicular to both B and L. If B = Be z, then dl x dt dl y dt dl z dt = q M L yb = q M L xb = 0, (1.13) or from the last equation L z = cste. We now define a new variable η = L x + il y such that we can transform equations (1.13) to The solution to equation (1.14) is straightforward with or dη dt = i qb η. (1.14) M η = Ae iω Lt, (1.15) L x = Acos ω L t L y = Asin( ω L t), (1.16) where ω L = qb M (1.17) is the Larmor frequency. Since both A and L z are constants, and that equations (1.16) are that of a circle of radius A, we see that the angular momentum vector precesses about the z- or B-axis at the Larmor frequency. The angle θ made by L relative to B is given by βmb cos θ c) Given that P( θ)dθ e βmb cos θ sin( θ)dθ = e A θ = tan 1. (1.18) L z d cos θ, then 3

4 cos( θ) = = 1 1 βmb cos θ e cos( θ)d cos θ 1 1 βmb βmb cos θ e ln eβmb cos θ 1 d cos θ 1 { d cos ( θ ) }. (1.19) Since 1 e βmb cos( θ) d cos( θ) 1 = 1 βmb eβmb cos( θ) βmb sinh βmb =, 1 1 (1.0) then cos( θ) = βmb sinh βmb = coth( βmb) 1 βmb. cosh( βmb) sinh ( βmb ) βmb ( βmb) (1.1) d) When βmb 1 (or kt mb ) equation (1.1) simplifies to cos( θ)! ( βmb) βmb + 1 ( 1 6 βmb βmb )3! 1 βmb 1+ 1 βmb βmb 1 (1.)! 1 3 βmb. The total mean magnetization M = m z because of their cos ϕ or sin( ϕ) dependency) is given by e z (the other components average to zero M = n m z = nm B 3kT e z, (1.3) and the paramagnetic susceptibility is 4

5 χ m = nm 3kT. (1.4) e) With L! = J s, q = C, and M = kg, we have m q! M = C s -1 m. (1.5) Furthermore, taking T = 300 K, n = 10 5 m -3 (an approximate value for air), and k = J K -1, the paramagnetic susceptibility is χ m (1.6) This value is higher than what can be calculated ( χ m 10 4 ) with a more careful analysis.. Diamagnetic materials. As we saw in the Problem 1, electrons moving around a nucleus in an atom will also precess with angular frequency when subjected to an external magnetic induction B = Be z. a) Show that the current I resulting from this precession is ω L = qb M, (.1) I = Zq B 4π M, (.) where Z is the number of electrons in the atom, and that the corresponding dipole moment is m = Zq B 4M ρ, (.3) where ρ = x + y is the mean square radius of the electrons orbit when projected on the plane perpendicular to B. b) Assume spherical charge distributions for the electrons, and show that the diamagnetic susceptibility χ is given by 5

6 χ = n m B = nzq r where n is still the number density, and r = x + y + z. 6M, (.4) c) It is found that the magnitude of the diamagnetic susceptibility is in general much smaller than that of the paramagnetic susceptibility. Provide at least two more fundamental differences between the two quantities. Solution. a) The precession will produce an electric current I = charge revolutions per unit time, and if there are Z precessing electrons, then I = ( Zq) ω L π = ( Zq) 1 π = Zq B 4π M. qb M (.5) The corresponding dipole moment is the current time the area. So, if ρ = x + y is the mean square radius of the electrons orbit when projected on the plane perpendicular to B, then m = Iπ ρ = Zq B 4M ρ. (.6) b) Assuming spherical orbits for the electrons, we set r = x + y + z as the mean square distance of the electrons from the nucleus so that ρ = x + y = 3 r. (.7) Substituting this relation into equation (.6) yields and the diamagnetic susceptibility is given by m = Zq 6M r B, (.8) 6

7 with n is still the number density. χ = n m B = nzq r 6M, (.9) c) Three more fundamental differences between paramagnetic and diamagnetic susceptibilites are: 1) The diamagnetic susceptibility is always negative, contrary to the paramagnetic susceptibility. ) Diamagnetism is a property of all matter, whereas paramagnetism requires that elementary components have a net nonzero angular momentum. 3) The diamagnetic susceptibility does not depend on temperature. The 1 T dependence of the paramagnetic susceptibility expresses Curie s Law. 3. Suppose that E( x,t) = 1 4πε 0 B( x,t) = 0, q r H vt r e r (3.1) where H ( x) is the Heaviside or step distribution (see equation (1.14) of the lecture notes), and r = x. Show that these fields satisfy all of Maxwell s equations, and determine ρ and J. Describe the physical situation that gives rise to the fields. Solution. This is the field of a point charge q located at the origin, out to an expanding spherical shell of radius vt ; outside this shell the field is zero. It follows that the shell carries a total charge +q. We first calculate the charge density ρ with E = q 1 H ( vt r) 4πε 0 r e r + 1 r e r H ( vt r) = q H ( vt r) 1 4πε 0 r + 1 r e r H ( vt r) = q 4π H ( vt r)δ ( r) 1 4πε 0 r δ ( vt r) = q 4π H ( t)δ ( r) 1 4πε 0 r δ ( vt r), (3.) 7

8 where equation (1.16) of the lecture notes was used. From Coulomb s law, the charge density is ρ = qδ ( r) H ( t) + q 4πr δ ( vt r ). (3.3) This equation mathematically verifies our physical interpation of the fields. The current density J is easily evaluated from J = ρv = qv 4πr δ ( vt r)e, (3.4) r since v = 0 at the origin for the (stationary) negative charge. Evidently, B = 0, and E + B = 0, since E has only one component along e r that is independent of θ and ϕ. The only remaining law to verify is For this, we have B = µ 0 J + µ 0 ε 0 E = 0. (3.5) E = qv 4πε 0 r δ ( vt r)e, (3.6) r from equation (3.1), and ε 0 E = J = ρv as a comparison with equation (3.4) readily shows. 4. A uniform magnetic induction field B( t) = B( t)e z, fills a circular region of radius R located in the xy-plane. Let s assume that the magnetic induction is increasing in time. Use Lenz law to qualitatively characterize the induced electric field, and then evaluate it. Solution. Lenz law states that the electric field induced (and the current potentially created) will oppose the change in magnetic induction. Using the right hand rule, the magnetic induction generated by the induced current must be oriented downward (i.e., along e z ). The induced electric field must then run clockwise in the circumferential direction. Mathematically, 8

9 π! E dl = R E ϕ dϕ = ( π R)E ϕ = df 0 dt = d dt S B e z da db t = π R dt. (4.1) Therefore, E = R dt db t e ϕ. (4.) This equation verifies our interpation using Lenz law. 5. A line of charge λ is uniformly glued onto the rim of a wheel of radius b, which is then suspended horizontally, as shown in Figure 1. Although it is originally at rest, the wheel is free to rotate (the spokes are made of some non-conducting material wood, maybe). In the central region, out to a radius a ( < b), there is initially a uniform magnetic induction B 0, pointing up. Now someone turns off the field off. a) Use Lenz law, and whatever conservation law(s) that might be needed, to qualitatively describe what happens to the wheel after the magnetic induction field is turned off. Does the wheel stay at rest or is it set in some sort of mechanical motion? b) Now, quantify the answer you gave in a), again using the needed laws (either from conservation arguments, and/or electromagnetic considerations). Does the final state of the wheel depend on the details of how B 0 is turned off? c) If you determined that the wheel is set in some sort of motion by the disappearance of the magnetic induction, then specify exactly what was the source of the motion, and the agent responsible for this motion. Be specific. Solution. a) The changing magnetic induction field will induce an electric field (Faraday s law of induction), curling around the axis of the wheel. This electric field exerts a force on the charges glued to the rim, and the wheel starts to turn. According to Lenz law, it will rotate in such a direction that its (magnetic induction) field tends to restore the initial upward flux. The rotational motion, then, is counterclockwise, as viewed from above. b) Quantitatively, Faraday s law states that 9

10 ! E dl = df dt = d dt B n da S = πa db dt. (5.1) The torque on a segment of length dl along the rim of the wheel is given by dn = r ( E λdl), (5.) where E is the induced electric field. Integrating over the rim, we get the total torque on the wheel N dl dt =!! = e z bλ E dl x ( E λdl) = e z λbπa db dt. (5.3) The total angular momentum of the rim, after the initial magnetic induction has been turned off, is L = N dt = e z λbπa db = e z λbπa B 0, 0 B 0 (5.4) Figure 1 The wheel of Problem 5. A line charge λ is uniformly glued to the wheel at a radius b, and a magnetic induction field B 0 initially exists out to a radius a, until it is turned off. The wheel is free to rotate. 10

11 and the wheel is turning clockwise when seen from above, as stated in a). Since we must also have L = I z ω e z, with I z the principal moment of inertia of the wheel about its symmetry axis, then ω = λπa b I z B 0. (5.5) That is, the speed of the wheel is independent of the way in which B 0 is turned off. c) There must be conservation of angular momentum for this system. Since the wheel has acquired a net quantity of angular momentum, the electromagnetic fields must have lost an equal amount. So, the source of the motion was the initial angular momentum present in the electromagnetic fields, and the agent for the motion of the source is the electric field induced by the cancellation of B An alternating current I = I 0 cos(ωt) flows down a long straight wire, and urns along a coaxial connecting tube of radius a. a) Using (among others) Ampère s Law as an approximation, determine in what direction does the induced electric field point (radial, circumferential, or longitudinal)? b) Assuming that the electric field goes to zero as r, find E r,t radial distance from the wire., where r is the c) Using the result obtained in b) calculate the displacement current density J d. d) Integrate the equation for J d to get the total displacement current I d. e) Calculate the ratio of I d I. Set a = 1 mm ; can you advance a hypothesis as to why Faraday did not discover displacement currents? Solution. a) Using Ampère Law (in its integral form in the magnetostatics approximation) it is straightforward to determine that the magnetic induction field is circumferential (i.e., B = Be ϕ ). Turning to Faraday s Law of induction (again its integral form), we find that the induced electric field is longitudinal. b) We will be again using Ampère s and Faraday s Law. For the former, we define a surface of radius s that is perpendicular to the wire/tube pair. Whether we choose that s < a or s > a we find that the total current crossing the surface is I or zero, respectively. Then the magnetic induction field can be calculated to be µ 0 I B = πs e, s < a ϕ 0 s > a. (6.1) 11

12 For Faraday s Law, we choose a rectangular surface that has two sides of length l parallel to the wire; one side is inside the tube at a distance r from the wire, while the other side is very far away outside the tube ( r ). Applying Faraday s Law we have! E dl = d dt B nda El = d a µ 0 I dt πs l ds, r (6.) which yields E = µ 0 di π dt ln a r = µ 0 I 0ω π e z sin ( ωt )ln a r e z, r a. (6.3) c) The displacement current density is given by E J d = ε 0 = µ 0ε 0 π ω I 0 cos ( ωt )ln a r = µ 0ε 0 π ω I ln a r e z. e z (6.4) d) The total displacement current I d = J nda = ω I c = ω I c 0 a ln a = ω Ia 4c. = µ 0ε 0 ω I π r ln( a) r ln( r) r 0 a r ln r ln a r πr dr dr r a 4 0 (6.5) e) The ratio of the two currents is I d I = ω a 4c. (6.6) 1

13 With a = 1 mm, we see that even if we set the angular frequency as high as ω = c = rad/s, then I d I = Obviously, a frequency of the order of 100 MHz was not accessible to Faraday, and he therefore did not have a chance of experimentally discovering displacement currents. 7. Consider the Jefimenko formulae for the electromagnetic fields E( x,t) = 1 4πε 0 B( x,t) = µ 0 e R ρ ( x, t ) R + e R ρ x, t cr t J ( x, t ) 4π e R R + J x, t t 1 J x, t c R t e R cr d 3 x d 3 x (7.1) when applied to the case of point charge where = qδ ( x, t ) = qv t ρ x, t J x r t δ x r ( t ). (7.) The position and velocity of the charge at time t are specified by r ( t ) and v ( t ), respectively. In evaluating expressions involving the arded time, one must put t = t R c, where R = x r t x inside delta functions within integrands). ( R = x a) As a preliminary to deriving the Heaviside-Feynman expressions for the electric and magnetic induction fields of a point charge, show that δ x r t d 3 x = 1 κ, (7.3) where κ = 1 e R v c. Take note that κ is evaluated at the arded time. b) Starting with equations (7.1), use equations (7.) and the result of part a) to show that c) Show that E( x,t) = q 4πε 0 4π B( x,t) = q µ 0 e R κ R v e R κ R + 1 c + 1 c e R κ R v e R κ R 1 c. v κ R (7.4) [ f ] = 1 f κ t where f is any well-behaved function of r ( t ) and t., (7.5) 13

14 d) Finally, use equations (7.4) and (7.5) to derive the Heaviside-Feynman expressions for the electric and magnetic induction fields of a point charge. That is, show that Solution. E( x,t) = q 4πε 0 B( x,t) = q µ 0 e R R v e R 4π κ R [ + R] c + e R R 1 c[ R] + 1 c v e R κ [ e R ]. (7.6) a) Referring to Figure, let us consider an infinitesimal rectangle of area ab (we work in two dimensions for simplicity) that is not moving relative to the observer. If the same element is set in motion at a velocity v relative to the observer, then it will not appear as a rectangle anymore (as shown in the figure), since electromagnetic signals emanating from different points and arriving at the same time at the position of the observer must leave the source at different times (while it is continuously moving). More importantly, (again referring to Figure ) a signal originating from point 1 must leave before another leaving point, if they are to arrive at the same time at the observer, and as the source is moving we will find that b b (while a = a, since it is along a direction perpendicular to v ). More precisely, during the time interval Δ t that elapses between the emissions of the two signals, the signal from the point 1 travels a distance cδ t in the direction of the observer cδ t = b v v e. (7.7) R Figure A infinitesimally small rectangle as seen from an observer located a long distance away along the e R unit vector, when the rectangle is not moving (left) or moving at a velocity v (right) relative to the observer. 14

15 During the same time interval, the surface element will travel the distance v Δ t = ( b b). (7.8) Combining equations (7.7) and (7.8), we get b = b 1 e R v c. (7.9) A generalization to three-dimension is straightforward, and it follows that if d 3 x is the infinitesimal volume element at rest, then d 3 x κ, with κ = 1 e R v c, is the corresponding apparent volume element seen by the observer when the volume element is moving at a velocity v. Notably, δ x r t d 3 x = 1 κ. (7.10) There is at least one other way of deriving this result (from Rybicki & Lightman, pp ). Consider the following integral (for the total charge) = ρ ( x, t ) Q x,t d 3 x. (7.11) We are certainly free to rewrite this equation as = d 3 Q x,t x d t ρ ( x, t )δ t t + x x c. (7.1) If we substitute the first of equations (7.) into equation (7.1), and integrate over space, we have = q δ Q x,t ( t t + x r ( t ) c) d t = q δ ( t t + R ( t ) c) d t, (7.13) where R t t = But = x r ( t ) and R ( t ) = R ( t ). We now make the change of variable ( t ) c in the time integral of equation (7.13), which implies that t t + R d t = d t + 1 c dr t d t d t. (7.14) 15

16 dr t d t = R t = dr t d t = R t dr ( t ) = R t d t dr ( t ) d t ( t ), v (7.15) or, alternatively dr t d t Inserting equations (7.16) and (7.14) in equation (7.13), we get = q κ δ ( t ) Q x,t which yields the desired result, with κ as defined above. = e R v ( t ). (7.16) d t = q κ, (7.17) b) Before integrating in equations (7.1), it is important to realize that since t = t R c, where R = x x is not a function of the arded time t (it only becomes a function of t after the integration, see equations (7.1) and (7.13)), then in the integrands d dt! [ ] = d (!) d t. (7.18) From this equation, and the definitions for the current and the charge densities of equations (7.), it is straightforward to obtain equations (7.4) from the Jefimenko formulae for the electromagnetic fields. c) It is now essential to realize that after the integrations R becomes a function of the arded time. More precisely, R ( t ) = x r ( t ), and therefore equation (7.18) does not apply anymore. So if t = t + R ( t ) c, then we have that dt = ( 1+ c 1 dr ( t ) d t )d t. It follows from equation (7.16) that and, therefore, that dt = 1 e R v c = κd t, (7.19) 16

17 d dt! [ ] = 1 κ d d t (!). (7.0) d) We start with the last term on the right-hand side of the first of equations (7.4) 1 c v κ R = 1 c 1 κ R R t = 1 1 e R c κ t = 1 c = 1 c = 1 c = 1 c [ e R ] + [ e R ] [ e R ] + c [ e R ] 1 c R 1 κ t R e R κ R e R t ( R) ( κ R e v) R e R ( κ R κ 1) e R R + 1 c e R κ R. (7.1) The last term of this equation will cancel with the second term on the right-hand side of the first of equations (7.4). We now transform the second term on the right-hand side of equation (7.1) with 1 c e R R = 1 c = 1 c e R R R e R R = e R κ 1 R κ = e R R [ R] + R e R κ R [ + R] c [ ] [ ] + R c e R R e R R e R R. (7.) Substituting equations (7.1) and (7.) into the first of equations (7.4) yields the Feynman formula for the electric field due to a moving point charge E( x,t) = q 4πε 0 e R R [ + R] c For the magnetic induction field, we start with e R R + 1 c [ e R ]. (7.3) 17

18 1 c v e R κ R = 1 1 c [ R] = = = 1 c[ R] 1 c[ R] 1 c[ R] v e R κ v e R κ v e R κ v e R κ + v e R κ 1 v e R c κ R v e R κ 1 κ R κ + v e R κ R 1 R [ R] v e R κ R. (7.4) Inserting this result in the second of equations (7.4) yields the Heaviside formula for the magnetic induction field due to a moving point charge B( x,t) = q µ 0 v e R 4π κ R + 1 c[ R] v e R κ. (7.5) 8. Conservation of angular momentum. Show that the differential and integral forms of the law of conservation of angular momentum are and d dt ( L mech + L field ) + M! = 0, (8.1) d 3 x + n M! da = 0, (8.) ( L mech + L ) field V where the field angular momentum density is S L field = x g = 1 c x ( E H ), (8.3) and the flux of angular momentum is described by the tensor Solution.! M =! T x. (8.4) The mechanical force density F on a charge density ρ (of current density J ) subjected to electromagnetic fields is 18

19 The corresponding density of the mechanical torque N is F = ρe + J B. (8.5) N L mech = x F = x ( ρe + J B). (8.6) Referring to equations (4.78) to (4.85) (pp ) of the lecture notes, it can be shown that equation (8.6) can be transformed to L mech = x T! g, (8.7)! where T and g are the Maxwell stress tensor, and the electromagnetic momentum density vector, respectively. If we define the field angular momentum density with equation (8.3), then ( L mech + L field ) = T! x. (8.8) We now evaluate the term on the right-hand side of equation (8.8) ( T! x) = ε i ijk m T jm x k ε ijk T jm m x k ε ijk T jm δ km = m ε ijk T jm x k = m ε ijk T jm x k (8.9) ε ijk T jk = m ε ijk T jm x k = m ( ε ijk T jm x k ), since ε ijk T jk = 0, from the symmetry of the Maxwell stress tensor (i.e., T ij = T ji ; see equation (4.8) of the lecture notes). Inserting equation (8.9) in equation (8.8) yields ( L mech + L field ) + M! = 0, (8.10) with! M =! T x (according to equation (8.9),! M has two indices i and j, and is therefore a second rank tensor). Finally, integrating equation (8.10) over space will give us d dt d 3 x + M! d 3 x = 0, (8.11) ( L mech + L ) field V V 19

20 or, using the divergence theorem, d dt d 3 x + n M! da = 0. (8.1) ( L mech + L ) field V 9. Suppose you had an electric charge q e and a magnetic monopole q m (as far as we know, they don t exist; see Jackson sections 6.11 and 6.1). The field of the electric charge is S E( r) = 1 q e 4πε 0 r e, (9.1) r with the electron located at the origin, and the field of the magnetic monopole is B( r) = µ 0 q m 4π r e r, (9.) with r = r de z (the magnetic monopole is located a distance d away from the electron and on the z-axis ). Find the total angular momentum stored in the fields. This is a static problem. Solution. With the problem as stated E( r) = q e 4πε 0 µ B( r) = q 0 m r r 3 r 4π r = q 3 m µ 0 4π ( r de z ) r + d 3, rd cos θ (9.3) with z = r cos( θ). The electromagnetic momentum density is g = ε 0 ( E B) = q e q m d µ 0 16π r 3 ( r e z ) r + d 3. rd cos θ (9.4) The angular momentum density is 0

21 L = r g = q e q m d µ 0 16π r 3 r ( r e z ) r + d 3 rd cos θ = q e q m d µ r 0 cos( θ)e r e z 16π r 3 r + d 3. rd cos θ (9.5) But since e r has components along e x and e y, and that are proportional, respectively, to sin( θ)cos( ϕ) and sin( θ)sin( ϕ), then L x r sin( θ) dr dθ dϕ dϕ L x = cos ϕ = 0 L y r sin( θ) dr dθ dϕ L y = sin ϕ = 0. π 0 π 0 dϕ (9.6) We are left with L z = L z r sin( θ) drdθdϕ cos ( θ) 1 = q e q m d µ π π r sin θ 0 dϕ dθ dr 16π r + d rd cos( θ) We make the change u cos( θ), then (9.7) The radial integral gives L z = q e q m d µ 1 0 du( 1 u r ) dr 8π. (9.8) 1 0 r + d rdu 3 r dr = 0 r + d rdu d 1 u 3 = ( ru d) r + d rdu 0 1 d( 1 u), (9.9) and 1

22 µ L z = q e q 0 m 8π µ = q e q 0 m 8π u 1 u du ( 1 + u)du (9.10) = q e q m µ 0 4π. The total angular momentum is therefore L = q e q m µ 0 4π e z. (9.11) [Note: This result is independent of the separation between the charges. From quantum mechanics, we know that the angular momentum comes in half-integer multiples of!. So, this result suggests that if magnetic monopoles exist, electric and magnetic charges must be quantized. That is, q e q m µ 0 4π = n!, for n = 1,, 3, ; an idea proposed by Dirac in If even only one monopole exists somewhere in the Universe, this would explain why the electric charge comes in distinct units (from Griffiths, p. 36).] 10. A classical atomic electron of charge q circles about a nucleus of charge Q on a stable orbit of radius r. The centripetal acceleration provided by the Coulomb attraction between opposite charges is counterbalanced by the centrifugal force mv r due to the motion of the electron; m and v are the mass and the speed of the electron, respectively. a) Show that the kinetic energy of the electron is given by T 1 mv = 1 8πε 0 qq r. (10.1) b) A very small magnetic induction field db is slowly turned on, perpendicular to the plane of the orbit. Assume that this change causes the electron to move at a new speed v 1 = v + dv on a new orbit of radius r 1 = r + dr with dv v and dr r, show that the kinetic energy becomes T 1 1 mv 1! 1 qq 1 dr 8πε 0 r r + 1 (10.) qvrdb. c) Use Faraday s Law to show that the increase in kinetic energy dt T 1 T is just right to sustain the circular motion at the same radius. That is, show that dr = 0.

23 [Hints: i) Keep your calculations to first order, while assuming that db is of first order (e.g., dvdb! 0 ). ii) When considering a change dr on the orbital radius r, you can assume that it is infinitesimal such that, for example, 1 + dr r for the speed.] Solution. 1! 1 dr r. The same would be true a) Since the system is initially in equilibrium, we must have from Lorentz force mv r = 1 4πε 0 qq r, (10.3) where m and v are, respectively, the mass and the speed of the electron, and therefore the kinetic energy T of the electron is T = 1 mv = 1 8πε 0 qq r. (10.4) b) After the magnetic induction field is turned on, the electron is on a new orbit r 1 = r + dr, with the assumption that dr r, and moves at a new velocity v 1. Therefore, the Lorentz force tells us that mv 1 = 1 qq r 1 4πε 0 r + qv 1dB, (10.5) 1 and to first order T 1 = 1 mv 1 = 1 qq + 1 8πε 0 r 1 qv r db 1 1 = 1 8πε 0 qq ( r + dr) + 1 q( v + dv) ( r + dr)db! 1 qq 1 dr 8πε 0 r r + 1 qvrdb. (10.6) c) The change in kinetic energy is dt = T 1 T! qvr db 1 qq dr. (10.7) 8πε 0 r 3

24 However, according to the Faraday law of induction we have for the induced electric field! E dl = πre = d dt B n da db = πr dt, (10.8) or E = r So, the force imparted on the electron by the induced electric field is or m dv dt The increase in kinetic energy is therefore = qe = qr mdv = qr db dt. (10.9) db dt, (10.10) db. (10.11) dt = d 1 mv = mvdv = qvr db, (10.1) and comparing equations (10.7) and (10.1) we find that dr = 0. (10.13) Therefore, the size of the orbit remains the same after the apparition of the (infinitesimal) magnetic induction field db. 11. Electromagnetic Fields Inside a Conductor. To make a current flow in a conductor a force must be applied. It is usually found that for most substance the current density J and the force per unit charge f are related through J = σ f, (11.1) 4

25 with σ the conductivity of the substance. But when the force driving the current is electromagnetic in nature, equation (11.1) takes the form of the so-called Ohm s Law J = σ ( E + v B). (11.) It is normally the case that the mean velocity of the charges within the conductor is small enough that Ohm s Law is approximated to J = σe. (11.3) a) We know that in electrostatic the electric field and the charge density within a conductor are zero, but this is not necessarily the case for time varying fields. This, in fact, should be obvious from Ohm s Law. Assume that we are given a charge density ρ and a current density J, and use Gauss Law, Ohm s Law, and the continuity equation to show that for a homogeneous medium of permittivity ε the time evolution of the charge density is ρ( t) = ρ( 0)e σ ε t. (11.4) Equation (11.4) implies that any charge density will eventually dissipate in a time inversely proportional to the conductivity. b) Since we know that any charge density will eventually dissipate, set ρ = 0 in Maxwell s equations and use Faraday s Law and the Ampère-Maxwell Law to show that the medium allows the propagation of attenuated waves such as where k and κ are collinear and E( x,t) = E 0 e κ x e i( k x ωt), (11.5) k = ω µε κ = ω µε 1+ σ ωε 1+ σ ωε (11.6) Equations (11.5) and (11.6) show that electromagnetic fields can only exist inside a conductor for a distance of the order of the so-called skin depth δ, with δ 1 κ. (11.7) 5

26 Also, show that the electric and magnetic fields are transverse to the direction of propagation and perpendicular to each other. c) Assume that a plane wave propagating with a wave vector k is incident on the conductor s surface, whose outward normal unit vector is n. This plane wave will excite current and charge motions in the conductor such that electromagnetic fields (as expressed by equation (11.5)) will be induced. Consider the boundary conditions stemming from Maxwell s equations, and show that for a good conductor (i.e., where σ ωε ) the electric and magnetic fields E 1 and H 1 outside the conductor are related at its surface by with Z s the surface impedance of the conductor defined with Solution. E 1 surface = Z s H 1 surface, (11.8) Z s = ( 1 i) δσ. (11.9) a) Using Gauss Law, equation (11.3) for Ohm s Law, and the continuity equation we find or E = 1 σ J = 1 ρ σ = ρ ε, (11.10) ρ( t) = ρ( 0)e σ ε t. (11.11) b) Faraday s Law and the Ampère-Maxwell Law (using Ohm s Law) can be written as E = B B = µσe + µε E. (11.1) Taking the curl of both equations we get ( E) E = µσ E µε E ( B) B = µσ B µε B. (11.13) 6

27 But since B = E = 0 (from equation (11.11) for the electric field) then E = µε E B = µε B E + µσ B + µσ. (11.14) It is natural to test whether these equations allow harmonic fields of the type E( x,t) = E 0 e i (!k x ωt ) (11.15) as a solution. Inserting equation (11.15) into the first of equations (11.14) we find which will yield an acceptable solution if If we define then!k E = µ ( ω ε + iωσ )E, (11.16)!k = µ ( ω ε + iωσ ). (11.17)!k = k + iκ, (11.18)!k =!k!k = k κ + ik κ = ω µε + iωσµ. (11.19) But since k κ = ωµσ for any frequency, and that µ and σ are assumed constant and isotropic, then k and κ must be collinear (that is, k κ = kκ ) and k k = ω µε κ κ = ω µε 1+ σ ωε 1 + σ ωε (11.0) It follows that the electric field can be expressed as 7

28 E( x,t) = E 0 e κ x e i( k x ωt), (11.1) with k and κ collinear. But because Gauss Law specifies that! k E = 0, and since furthermore B =!k B = 0, then it must also be that the fields are transverse to the direction of propagation. Finally, the Faraday and Maxwell-Ampère Law dictate that!k E = ωb!k B = µ ( ωε + iσ )E, (11.) which imply that E and B are also perpendicular to one another. c) We now assume that a plane wave propagating with a wave vector k is incident on the conductor s surface, whose outward normal unit vector is n. This plane wave excites currents and charge motions in the conductor such that electromagnetic fields like expressed by equation (11.1) are induced. If the conductor is not perfect (although it is good), then there will not be any surface current on its surface. Instead a current density J = σe will be induced, with E the electric field inside the conductor. We first need to work out the relationship between the wave vectors k and k that respectively define the directions of propagation for the induced and incident waves, and n the unit normal at the boundary of the conductor. If we denote the electric and magnetic fields outside the conductor by E 1 and H 1 (made of the incident fields and whatever other fields that might be scattered by the conductor), then the boundary condition brought up by Faraday s Law states that n ( E 1 E) = 0. (11.3) The tangential components of the fields are continuous. If we call θ and θ the respective angles made by k and k with n, then at a point on the surface of the conductor some distance r b away (from the arbitrary origin in the direction of a unit tangential vector t ) the phases of the fields inside and outside the surface of the conductor will respectively be kr b sin( θ) and k r b sin ( θ ), which must be equal everywhere on the boundary at all times. This will only be achieved if k sin( θ) = For a good conductor such that σ ωε we have k sin ( θ ). (11.4) k! κ! ωµσ! 1 δ, (11.5) and at a given frequency ω we find that 8

29 k lim σ k = ω c 0. (11.6) ωµσ That is, the wave induced in the conductor must travel in a direction almost normal to the boundary (of course, this procedure is equivalent to applying Snell s Law for the refraction of waves). In other words, we can to a good degree of precision make the following substitution k k + k sin ( θ )t, (11.7) and the amplitude of the induced electric and magnetic fields E 0 and B 0 induced in the conductor are practically parallel to the boundary. So, if E! is the amplitude of the tangential component of the electric field at the exterior of the conductor s surface, then the electric field within the conductor must be (from equation (11.3))! E " e κ x e i k x+ E x,t k sin ( θ )t x ωt. (11.8) Now, since for a good conductor the fields are confined to a very small skin depth δ, it is tempting to assume that a surface current density K is induced on the conductor. We find this surface current density by integrating the component of the current density J = σe (which is basically tangential to the surface of the conductor) along a direction perpendicular to the boundary. That is, K = σ E! e κ x e i k x+ = 0 σ κ ik E! ei = δσ 1 i E! ei k sin ( θ )t x ωt dx n k sin ( θ )t x ωt k sin ( θ )t x ωt, (11.9) where the notation dx n was used to make it apparent that the integration is along a path normal to the boundary. Alternatively, with E 1 the electric field exterior to the conductor, we can invert equation (11.9) to write that at the surface of the conductor where we introduced the surface impedance of the conductor E 1 surface = Z sk, (11.30) Z s = ( 1 i) δσ = ( 1 i) ωµ σ. (11.31) Finally, the boundary condition stemming from the Ampère-Maxwell Law states that 9

30 n H 1 = n H = K, (11.3) or E 1 surface = Z s H 1 surface. (11.33) 1. Find all the elements of the Maxwell stress tensor! T for a monochromatic plane wave traveling in the z direction and linearly polarized in the x direction. Does your result make sense, considering that! T is the momentum flux density, and that! T da is the rate at which momentum crosses an area da? How is the momentum flux related to the energy density? Solution. The components of the Maxwell stress tensor are given by With the electric field given by then it must be that T ij = ε 0 E i E j + c B i B j 1 δ ij E k E k + c B k B k. (1.1) E( x,t) = e x E 0 e i( kz ωt), (1.) E B( x,t) = e 0 ( y c ei kz ωt ). (1.3) With these fields, all the off-diagonal components (i.e., for i j ) in equation (1.1) will be zero. The diagonal components are T xx = ε 0 E x 1 E + c c E = 0 c T yy = ε 0 c E y 1 E + c c E = 0 T zz = ε 0 E + c c E = ε 0 E = u, (1.4) where u is the energy density (see equation (4.71) of the lecture notes). So, we have 30

31 T zz = ε 0 E 0 cos ( kz ωt), (1.5) with all the other components zero. The momentum of these fields is in the z direction, and it is being transported in the same direction. So, it makes sense that T zz should be the only not zero component of the Maxwell stress tensor. We also know that! T da is the rate at which the momentum crosses (i.e., leaving) an area da. Here we have no momentum transported in the x and y directions. The momentum per unit time per unit area flowing across the surface oriented in the z direction is T zz = u = g c (see equations (4.81) and (4.8) of the lecture notes), so Δp = g caδt, and hence Δp Δt = g ca = momentum per unit time crossing area A. Evidently, the momentum flux is the energy density. 13. A localized electric charge distribution produces an electrostatic field, E = Φ. Into this field is placed a small, localized time-independent current density J( x), which generates a magnetic field H. a) Show that the momentum of these electromagnetic fields can be transformed to P field = 1 c ΦJ d 3 x (13.1) provided that the product ΦH falls off rapidly enough at large distances. How rapidly is rapidly enough? b) Assuming that the current distribution is localized to a region small compared to the scale of variation of the electric field, expand the electrostatic potential in a Taylor series and show that P field = 1 c E ( 0 ) m, (13.) where E( 0) is the electric field at the location of the current density and m is the magnetic moment caused by the current density. c) Suppose the current distribution is placed instead in a uniform electric field E 0 (filling all space). Show that, no matter how complicated is the localized current density J, the result in part a) is augmented by a surface integral contribution from infinity equal to minus one-third the result of b), yielding P field = 3c E 0 m. (13.3) 31

32 Solution. a) From Equation (4.86) of the Lecture Notes we have for the electromagnetic linear momentum P field = 1 c E H d 3 x = 1 c Φ H d 3 x (13.4) Φ H = 1 c ΦH d 3 x. Upon using Equation (1.31) of the Lecture Notes the first integral can be transformed into a surface integral such that P field = 1 c n ( ΦH) da + 1 c Φ H d 3 x = 1 c Φ H d 3 x (13.5) = 1 c ΦJ d 3 x, since the surface integral will vanish at a large distance r if ΦH falls faster than r when r and there is no displacement current because the electric field is timeindependent. b) Expanding the scalar potential in a Taylor Series about x = 0 (where the current density is located) we have! Φ( 0) + x Φ( x) x=0! Φ( 0) x E( 0) Φ x (13.6) when the scale of variation for E x Equation (13.6) into Equation (13.5) we find P field = Φ( 0) c is large compared to the size of J( x). Inserting J d 3 x E i 0 c x i J d 3 x. (13.7) However, we know from the discussion following Equation (3.33) of the Lecture Notes that J d 3 x = 0 (13.8) 3

33 and E i ( 0) x i J d 3 x = 1 E( 0) ( x J)d 3 x = E( 0) m, (13.9) where the magnetic dipole moment is given by Equation (3.40) of the Lecture Notes. It follows from equations (13.7) to (13.9) that P field = 1 c E ( 0 ) m. (13.10) c) If the electric field is uniform, then the scalar potential must be linear in x and Equation (13.6) is exact (when E( 0) is replaced by E 0 ). We must, however, start with Equation (13.4) instead of the last of Equation (13.5) since there is no guarantee that ΦH will fall faster than r at large distances (in fact, at large distances the field due to the magnetic dipole moment m, which goes as r 3, should dominate and therefore ΦH r ). 1st way (as required) We then proceed with Equation (13.4) P field = 1 c ( ΦH) d 3 x + 1 c ΦJ d 3 x H d 3 x d 3 x = 1 Φ 0 c E 0i x i H + 1 c ΦJ d 3 x (13.11) = E 0i c n ( x i H) da + 1 c ΦJ d 3 x. But at a very large distance r, we have and therefore H( x) = 1 4πr 3n( n m) m (13.1) 3 n ( x i H) = x i ( n H) = x i 4πr 3 { m } n 3n n m (13.13) = x i ( n m). 4πr 3 33

34 We can then write E 0i n ( x i H) da x = E i 0i 4πr 3 ( n m) r dω (13.14) x = m E i 0i 4πr n dω. By choosing the surface of integration to be a sphere centered at the origin (and of radius r ) we can write n = e r, and since dω = sin( θ)dθdϕ the only term from E 0 x that will yield a non-zero integral is E 0z z = E 0z r cos( θ). Moreover, we can also write E 0z z = E 0r cos θ which when inserted in Equation (13.14) yields E 0θ sin( θ) r cos ( θ ), (13.15) E 0i n ( x i H) da π E 0r cos ( θ) E 0θ sin θ = 1 m e r = 1 3 E 0r ( m e r ) 0 cos( θ) sin( θ)dθ (13.16) = 1 3 E 0 m. From Equations (13.5), (13.9), (13.11), and (13.16) we have nd way We proceed from and transform Equation (13.4) to P field = 3c E 0 m. (13.17) P field = 1 c ( ΦH) d 3 x + 1 c ΦJ d 3 x H d 3 x d 3 x = 1 Φ 0 c E 0i x i H = E 0i c ( x i H) d 3 x + 1 c ΦJ d 3 x + 1 c ΦJ d 3 x (13.18) where Equation (13.8) was used. Instead of transforming the first integral into a surface integral evaluated, we will keep it as a volume integral and transform it as follows 34

35 d 3 x x i H = ε kmn m x i H n k d 3 x d 3 x = ε kmn δ mi H n + x i m H n = ε kin H n d 3 x + ε kmn x i m H n d 3 x, (13.19) and E 0i ( x i H) d 3 x = E 0 H d 3 x + E 0i x i H d 3 x. (13.0) But we also know from the discussion leading to, and from, Equation (3.53) that which implies that (also using Equation (13.9)) Hd 3 x = m, (13.1) 3 E 0i ( x i H) d 3 x = E 0 3 m + E 0i x i J d 3 x = 3 E 0 m E 0 m (13.) = 1 3 E 0 m. It follows from the result obtained in b) that 3 rd way P field = 3c E 0 m. (13.3) Equation (13.3) could have been derived directly from the first of Equation (13.4) and Equation (13.1) since the electric field is uniform over all space. 35