Problem Set #5: 5.2, 5.4, 5.8, 5.12, 5.15, 5.19, 5.24, 5.27, 5.35 (Due Tuesday, April 8th)

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1 Chapter 5 Magnetostatics Problem Set #5: 5.2, 5.4, 5.8, 5.12, 5.15, 5.19, 5.24, 5.27, 5.35 (Due Tuesday, April 8th 5.1 Lorentz Force So far we were concerned with forces on test charges Q due to static configurations of charges F Coulomb = QE. (5.1 The electrostatic force is known as Coulomb force and is conveniently described by the electric field E. It turns out that the charges moving with velocity v can also experience another force known as Lorentz force F Lorentz = Q(v B. (5.2 As we will see shortly, the force is due to moving charges which generatethe so-called magnetic field B. But before we discuss mechanics which generates magnetic field, let us study the combined effect of the two forces on the test charges F = Q (E + v B (5.3 The main difference is that the Lorentz force acts in the direction perpendicular to the plane spanned by the velocity vector v and the magnetic field vector B, when the Coulomb force always acts in the direction of the electric field vector E. Thus, it should be clear that the Lorentz force does not do work dw = F Lorentz dl = Q (v B vdt =. Consider an example of a point charge moving with initial velocity v = vŷ 55

2 CHAPTER 5. MAGNETOSTATICS 56 (or upward in statics magnetic field B = Bẑ (or out of the page. Then according to (5.3 thelorenzforce ˆx ŷ ẑ F = Q det v = QvBˆx (5.4 B would create a centripetal acceleration. Since the acceleration would always point orthogonal to the direction of motion with magnitude a = F m = QvB m (5.5 the charge will move on a circular orbit (in the clockwise direction with centripetal acceleration also given by From (5.5 and(5.6 wefindthattheradiusis a = v2 R. (5.6 R = mv. (5.7 It is easy to generalize this result if the initial velocity vector has a nonvanishing z-component v = vŷ + v ẑ (5.8 which is in the direction of B. This would not change the circular motion in the x y plan, but in addition the particle would keep moving with a constant velocity v in the ẑ direction. The overall effect would be a helical motion. Another example is a particle with mass m and charge Q in a uniform electric field E = Eẑ (5.9 and a uniform magnetic field initially at rest in the origin B = Bˆx (5.1 ẋ( = ẏ( = ż( = (5.11 x( = y( = z( =. (5.12

3 CHAPTER 5. MAGNETOSTATICS 57 Then the Coulomb and Lorentz forces or F = Q Eẑ +det ˆx ŷ ẑ ẋ ẏ ż B = Q (Eẑ + Bżŷ Bẏẑ = Q (E Bẏ ẑ + żŷ (5.13 ẍ = (5.14 ÿ = m ż (5.15 z = QE m m ẏ. (5.16 The solution of (5.14 withinitialconditions(5.11 and(5.12 is Moreover, from (5.15 and(5.16 with solutions x(t =. ( z = m ÿ = ( m 2 ż. (5.18 ( ( z(t =C 1 cos m t + C 2 sin m t + C 3 (5.19 and from (5.16 and(5.19 ( ( m ẏ(t = z + QE = ( ( ( C 1 cos m m m t + C 2 sin m t + E (5.2 B ( ( y(t = C 2 cos m t + C 1 sin m t + E B t + C 4. (5.21 The integration constants C 1,C 2,C 3 and C 4 are set by the initial conditions (5.11 and(5.12, =z( = C 1 + C 3 =y( = C 2 + C 4 =ż( = C 2 m =ẏ( = C 1 m + E B (5.22

4 CHAPTER 5. MAGNETOSTATICS 58 or C 1 = 2 me C 2 = C 3 = 2 me C 4 = (5.23 thus ( y(t = 2 me sin m t z(t = 2 me cos ( m t + E t. (5.24 B + 2 me. (5.25 To simplify the answer we can define R me 2 (5.26 ω m (5.27 then or y(t = R sin (ωt+rωt. (5.28 z(t = R cos (ωt+r. (5.29 (y Rωt 2 +(z R 2 = R 2. (5.3 This is nothing but a formula for a circle of radius R whose center is moving with a constant speed ωr = E B. (5.31 The trajectory of the particle is called a cycloid. We can also consider a collection of moving charges (e.g. in a wire, then it is convenient to define current as the rate of flow of charge per unit time [I] = Charge Time.

5 CHAPTER 5. MAGNETOSTATICS 59 Then the magnetic force which is felt by the wire will be given by the sum (or integral of the forces felt by each charge F = (v Bdq = (v Bλdl = (I B dl = (dl B I. If the current in the wire is constant then F = I (dl B. Then one can have an interesting situation when a gravitational force of a wire is balanced by the magnetic force mg = IlB given that the magnetic field B and the wire of length l point in the directions so that the magnetic force points upwards. If you increase the current further, then the wire will start moving upwards, but it is not the magnetic field that does the work, but the battery which generates the current! 5.2 Biot-Savart Law In magnetostatics the charges are allowed to move but in a way that the created magnetic field is static corresponding to a steady state distribution of charges, ρ =. (5.32 t Consider a total charge Q inside some volume V.ThenthechangeofQ with time can be only due to the flow of charges through the boundary, i.e. Q t = J ˆn da, (5.33 where J is the current density measured in units of charge per unit area pre unit time Charge [J] = Area Time. (5.34 Using the divergence theorem we can rewrite (5.33 as ρ(rd 3 r = J(rd 3 r (5.35 t

6 CHAPTER 5. MAGNETOSTATICS 6 which must be satisfied for an arbitrary volume element. Thus, wegeta differential equation ρ + J = (5.36 t known as the continuity equation. In magnetostatics the equation is further simplified due to the steady state condition (5.32, J =. (5.37 It is an experimental fact that the steady states (5.32 produceanobservable magnetic phenomena which depends on the current density J. Consider awireoflength l, crosssectionalarea a, pointinginˆl and carrying an electric current I, then at a distance r from the wire there is a magnetic field which is -directlyproportional l, -directlyproportionaltoi, -inverselyproportionaltothesquareofthedistancer and -pointsinthedirectionnormaltotheplanespannedbyˆl and ˆr, i.e. In a fixed coordinate system (and in SI units, B = k I l r 2 (ˆl ˆr. (5.38 B(r = µ J(r (r r a l, (5.39 4π 3 where the current density is J = aˆl. I (5.4 To obtain an integral expression we replace a l with a volume integral over d 3 r, B(x = µ J(r (r r d 3 r. (5.41 4π 3 This is the Biot-Savart Law of magnetostatics (which is analogous to the Coulomb s Law of electrostatics discovered first by Oersted, and elaborated by Biot and Savart and later by Ampere. Ampere s experiment showed that the force on a current element dl in a magnetic field is given by df = I (dl B. (5.42

7 CHAPTER 5. MAGNETOSTATICS 61 This implies that the total force on a current density distribution is F = J(r B(rd 3 r (5.43 and the total torque N = r (J(r B(r d 3 r. (5.44 For example, if the magnetic field B is generated by a closed current loop #2 then the force which a closed current loop #1 experiences can be calculated by substituting (5.41 into(5.42. Note that the volume integral is replaced by line integral since J = Iδ(r dl. (5.45 and the resulting force is F = µ 4π I dl1 (dl 2 (l 1 l 2 1I 2 l 1 l 2 3 = µ 4π I (dl1 dl 2 (l 1 l 2 +dl 2 (dl 1 (l 1 l 2 1I 2 l 1 l 2 3 = µ 4π I 1I 2 (dl1 dl 2 (l 1 l 2 l 1 l µ 4π I 1I 2 the second term vanishes for closed loops, i.e. F = µ 4π I 1I 2 ( dl 2 dl 1 1 (5.46 l 1 l 2 (dl1 dl 2 (l 1 l 2 l 1 l 2 3. (5.47 Thus, the parallel currents attract and antiparallel current repel. 5.3 Ampere s Law Using the product rule for curl ( J(r = J(r J(r = J(r 1 = J(r (r r (5.48 3

8 CHAPTER 5. MAGNETOSTATICS 62 we can rewrite (5.41 as B(r = µ 4π = µ 4π J(r (r r 3 d3 r ( J(r d3 r. (5.49 Then it is convenient to define a vector potential A = µ J(r 4π d3 r (5.5 such that B = A (5.51 and, thus, the Gauss Law for magnetic fields (in the absence of magnetic charges must be satisfied B = ( A =. (5.52 As an aside, note that there is a freedom in defining A which is called the gauge freedom or gauge symmetry. Indeed, if we redefine A new (r =A old (r+ ψ(r, (5.53 where ψ(x is an arbitrary function, then the magnetic field would not change, B new = A new = (A old + ψ = A old + ψ = A old = B old. (5.54 So, we have a freedom of choosing A which would make the calculations simple without effecting physically observable quantities such as B. This is similar to the freedom we had in choosing the electric potential since redefinitions V new (r =V old (r+c (5.55 would not change the physically observable electric field E new = V new = (V new + C = V old = E old. (5.56 The new (gauge transformation or new (gauge symmetry is much bigger since it involves and arbitrary function ψ(r and not just a constant C. For example one can choose a Coulomb gauge such that A =. (5.57

9 CHAPTER 5. MAGNETOSTATICS 63 (This is still not uniquely defined since we can add to the vector potential a gradient of an arbitrary scalar function which satisfies Laplace equation. Of course to fully appreciate the gauge symmetry of electrodynamics we need to look at the quantum electrodynamics where the vector potential appears as a more fundamental quantity. For example in the double slit experiment a shift in the interference pattern is observed if there is a non-vanishing magnetic flux in-between the two trajectories B ˆn da = ( A ˆn da = A dl. (5.58 The shift is known as the Aharonov-Bohm effect which can be explained using quantum mechanics where the same electron can simultaneously travel along two different trajectories. This shows that vector potential is more fundamental then magnetic field when it comes to quantum physics. If we take a curl of (5.49 andmakeuseof and A = ( A 2 A ( = 1 (5.6 = 4πδ(r r (5.61 then we get B(r = µ ( 4π J(r d3 r = µ ( 4π J(r d3 r µ ( 1 J(r 2 d 3 r 4π = µ 4π J(r 1 d3 r + µ J(r = µ 4π J(r 1 d3 r + µ J(r. (5.62 In the case of magnetostatics (5.37we get the Ampere s law in differential form, B = µ J(x, (5.63 or in Coulomb gauge A = µ J (5.64 ( A 2 A = µ J ( A = µ J (5.66

10 CHAPTER 5. MAGNETOSTATICS 64 and by integrating over some surface we obtain the Ampere s law in integral form, B n da = µ J n da (5.67 B dl = µ I. (5.68 Note that (5.63 impliesthatthemagneticfieldhasanon-vanishingcurling only if there is a current; and in a region with no currents the magnetic field is curl free and, thus, completely determined by the boundary conditions. 5.4 Magnetic moments Similarly to the multipole of expansion of the (scalar electric potential, V, one can construct the multipole expansion of the vector (magnetic potential, A. Considerthevectorpotentialatr due to a localized distribution of currents (near origin parametrized by r, where If we expand then A(r = µ 1 4π r A(r = µ 4π J(r dr (5.69 B A (5.7 B = µ J ( = 1 r + r r +... (5.72 r 3 J(r dr + µ 1 4π r 3 (r r J(r dr +... (5.73 For an arbitrary pair of function f(r and g(r and localized J(r we have gj fd 3 r = f (gjfd 3 r + fgj n da (5.74 gj fd 3 r = fg Jfd 3 r fj gd 3 r (5.75 or gj fd 3 r + fg Jfd 3 r + fj gd 3 r =. (5.76

11 CHAPTER 5. MAGNETOSTATICS 65 By substituting in (5.76, f(r =1and g(r =r i we get + r i Jfd 3 r + J i d 3 r =, (5.77 J i d 3 r = r i Jfd 3 r (5.78 and f(x =r i and g(x =r j we get r i J jd 3 r + r i r j Jfd 3 r + r i J jd 3 r + r j J id 3 r = (5.79 r j J id 3 r = r i r j Jfd 3 r (5.8. However for the steady states of magnetostatics the current is divergence-less (i.e. J =and(5.78, (5.8 aresimplified J i d 3 r = (5.81 (r ij j + r jj i d 3 r = (5.82 From (5.81 themagneticmomentofalocalizedcurrentdensity mustvanish and (5.73 becomes A(r = µ 1 (r r J(r dr +... (5.83 4π r 3 Moreover, (5.82 canbeusedtorewrite(5.83 as A(r = µ 1 dr r 4π r 3 j r jj iˆr i i,j=1,2,3 = 1 µ 1 ( d 3 r r 2 4π r 3 j r i J j r jj i ˆx i i,j=1,2,3 = 1 µ 1 d 3 r ˆx (y (x J 2 4π r 3 y y J x z (z J x x J z 1 µ 1 d 3 r ŷ (z (y J 2 4π r 3 z z J y x (x J y y J x 1 µ 1 d 3 r ẑ (x (z J 2 4π r 3 x x J z y (y J z z J y = 1 µ 1 d 3 r r ((y J 2 4π r 3 z z J y ˆx +(z J x x J z ŷ +(x J y y J x ẑ = 1 µ 1 2 4π r r ( r J(r d 3 r. (5.84 3

Problem Set #5: 5.7,5.9,5.13 (Due Monday, April 8th)

Chapter 5 Magnetostatics Problem Set #5: 5.7,5.9,5.13 (Due Monday, April 8th) 5.1 Biot-Savart Law So far we were concerned with static configuration of charges known as electrostatics. We now switch to