PHYS 110B - HW #4 Fall 2005, Solutions by David Pace Equations Referenced As ”EQ

Home , Displacement current, Maxwell stress tensor, Poynting vector

PHYS 110B - HW #4 Fall 2005, Solutions by David Pace Equations referenced as ”EQ. #” are from Grifﬁths Problem statements are paraphrased

[1.] Problem 8.2 from Griffiths Reference problem 7.31 (figure 7.43). (a) Let the charge on the ends of the wire be zero at t = 0. Find the electric and magnetic fields in the gap, E~ (s, t) and B~ (s, t).

(b) Find the energy density and Poynting vector in the gap. Verify equation 8.14.

(c) Solve for the total energy in the gap; it will be time-dependent. Find the total power flowing into the gap by integrating the Poynting vector over the relevant surface. Verify that the input power is equivalent to the rate of increasing energy in the gap. (Griffiths Hint: This verification amounts to proving the validity of equation 8.9 in the case where W = 0.)

Solution (a) The electric field between the plates of a parallel plate capacitor is known to be (see example 2.5 in Griffiths), σ E~ = zˆ (1) o where I define zˆ as the direction in which the current is flowing.

We may assume that the charge is always spread uniformly over the surfaces of the wire. The resultant charge density on each “plate” is then time-dependent because the ﬂowing current causes charge to pile up. At time zero there is no charge on the plates, so we know that the charge density increases linearly as time progresses.

It σ(t) = (2) πa2 where a is the radius of the wire and It is the total charge on the plates at any instant (current is in units of Coulombs/second, so the total charge on the end plate is the current multiplied by the length of time over which the current has been ﬂowing).

The electric ﬁeld between the ends of the wire is,

~ It E = 2 zˆ (3) oπa

The magnetic ﬁeld is found from Ampere’s law, ! I Z ∂E~ B~ · d~l = µ I + µ · d~a Eq. 7.38 (4) o enc o o ∂t

where the enclosed current is zero, Ienc = 0, and the magnetic ﬁeld is generated entirely by the displacement current.

1 The displacement current, Id (the non-zero term in Ampere’s law above), through the gap must be in the zˆ direction (the only proper current ﬂow is through the wire),

Bφ(2πs) = Id (5)

Solving for the displacement current in the gap gives, ! Z ∂E~ I = µ · d~a (6) d o o ∂t

Z ∂ It = µoo 2 zˆ · d~a (7) ∂t oπa Z s Z 2π µoI = 2 (ˆz) · s ds dφ zˆ (8) πa 0 0

2 s µoI s = 2 (2π) (9) πa 2 0 2 µoIs = (10) a2

Now the magnetic ﬁeld is found from (5),

µoIs B~ = φˆ (11) 2πa2 where it should be noted that the field increases with s. This is only for the region in the gap and inside the wire radius of a. The linear increase of the magnetic field in this region agrees with that we would find in a solid wire with a uniform current density. For s > a the magnetic field is a fringing field that goes to zero as s → ∞.

(b) The energy density is given in terms of the ﬁelds in the gap, 2 1 2 B uem = oE + Eq. 8.13 (12) 2 µo 2 2 2 2 2 1 I t 1 µoI s = o 2 2 4 + 2 4 (13) 2 oπ a µo 4π a 2 2 2 1 I t µos = 2 4 + (14) 2 π a o 4 noting that E2 = E~ · E~ .

The Poynting vector in the gap is, 1 S~ = E~ × B~ Eq. 8.10 (15) µo 1 It µoIs ˆ = 2 zˆ × 2 φ (16) µo oπa 2πa I2ts = − 2 4 sˆ (17) 2oπ a

2 The Poynting vector represents energy ﬂow. Taking special note of the direction found above we see that the energy is ﬂowing into the gap.

Veriﬁcation of equation 8.14 follows (umech = 0 because there are no charges to move in the gap and the energy within the gap is therefore due entirely to that of the ﬁelds), ∂ (u + u ) = −∇~ · S~ Eq. 8.14 (18) ∂t mech em

2 2 2 2 ∂ I t µos 1 ∂ −I ts 2 4 + = − s 2 4 (19) ∂t 2π a o 4 s ∂s 2oπ a I2t 1 I2ts 2 4 = − − 2 4 (20) oπ a s oπ a I2t = 2 4 (21) oπ a

Z 2 1 2 B Uem = oE + dτ Eq. 8.5 (22) 2 µo Z a Z 2π Z w 2 2 2 1 I t µos = 2 4 + s ds dφ dz (23) 2 0 0 0 π a o 4 2 Z a 2 3 I t s µos = 2 4 (2πw) + ds (24) 2π a 0 o 4 2 2 2 4 I w t a µoa = 4 + (25) πa 2o 16 2 2 2 I w t µoa = 2 + (26) 2πa o 8 where this result is for the volume of the cylindrical gap (length w and radius a).

The problem tells us to determine the total power ﬂowing into the gap by integrating the Poynting vector over the surface enclosing it. This is the cylindrical surface occurring at s = a. Technically, this also includes the circular surfaces at each of the plates, but for these surfaces the product S~ ·d~a = 0 so they do not contribute to the solution.

Z 2π Z w Power in = S~ · s dφ dz sˆ (27) 0 0 Z 2π Z w I2ts = − 2 4 s dφ dz (28) 0 0 2oπ a I2ta2 = − 2 4 (2wπ) (29) 2oπ a I2tw = − 2 (30) oπa

3 Finally, we essentially want to verify equation 8.9, Z I dW d 1 2 1 2 1 = − oE + B dτ − E~ × B~ · d~a (31) dt dt V 2 µo µo S where this represents the rate at which work is done on a collection of charges in the volume V, that is enclosed by the surface S.

Since there are no charges in the gap, W = 0 and dW/dt = 0. The equation becomes, Z I d 1 2 1 2 1 oE + B dτ = − E~ × B~ · d~a (32) dt V 2 µo µo S d I Uem = − S~ · d~a (33) dt S

we have solved for Uem in (26), and the integral on the right hand side of the above equation is the power ﬂowing into the gap, given by (30). Continuing with the veriﬁcation of equation 8.9,

2 2 2 2 d I w t µoa I tw 2 + = − − 2 (34) dt 2πa o 8 oπa I2tw I2tw 2 = 2 (35) oπa oπa this equation is veriﬁed.

[2.] Professor Carter Problem Two charged shells, both of radius R and with surface charge density σ, are placed so that there is a distance d > 2R between their centers. Calculate the force of one shell on the other by integrating Maxwell’s stress tensor. (Hint: a good surface to choose includes the plane which passes directly between the two and then closes out at inﬁnity).

Solution Generally, the force on charges within a volume V is,

I ↔ d Z F~ = T · d~a − oµo S~ dτ Eq. 8.22 (36) S dt V

In this problem we will determine the force on one of the spheres due to the other sphere using the equation above. Enclose one of the spheres (call it number 2 for no specific reason) in a volume. In this problem there is no time dependence. The term involving the time derivative of the Poynting ↔ vector is zero. We will find the expression for the Maxwell stress tensor, T , which is determined by the total field in the system (we sum the contributions from sphere 1 and sphere 2, recalling that the stress tensor includes self-fields).

There are no magnetic ﬁelds in this problem so the Maxwell stress tensor is,

 2 E2  Ex − 2 ExEy ExEz   ↔   =  2 E2  T o  EyEx Ey − 2 EyEz  (37)     2 E2 EzEx EzEy Ez − 2

4 The expression in (37) was given directly in lecture on 10/21/2005, but it may also be derived from the general form of the stress tensor. 1 2 1 1 2 Tij ≡ o EiEj − δijE + BiBj − δijB Eq. 8.19 (38) 2 µo 2

I will draw the enclosed region around sphere 2 and determine the force that 1 exerts on it. The electric fields outside of the spheres are determined from Gauss’ law. To best simplify the geometry of the surface integral that must be done, I use a surface that includes the plane exactly between the centers of the spheres. This surface closes at infinity, where the electric field is zero, so the integration needs only be completed along the plane directly between the spheres (see figure 1).

Figure 1: Geometry of problem 2. The force is determined by integrating the Maxwell stress tensor over a surface including the plane directly between the spheres.

Setting the origin of this system at the center of sphere 1, the electric ﬁeld due to sphere 1 is given by,

~ q E1 = 2 rˆ (39) 4πor σR2 = 2 rˆ (40) or where q is the total charge on the sphere (therefore, q = σ(4πR2)).

Writing this in terms of Cartesian coordinates, and setting x = d/2 since we are only interested in the electric ﬁeld along the plane between the charges gives,

2 ! ~ σR 1 (d/2)ˆx + yyˆ + zzˆ E1 = 2 2 2 p (41) o (d/2) + y + z (d/2)2 + y2 + z2 ! σR2 1 (d/2)ˆx + yyˆ + zzˆ = 2 2 2 p (42) o (d /4) + y + z (d2/4) + y2 + z2

5 p using r = (d/2)2 + y2 + z2 as the distance from the center of sphere 1 to the plane. The (d/2)ˆx term comes from my setting sphere 1 at the origin and placing center of the other sphere at x = d. This is equivalent to treating the spheres as point charges located at their centers. Since the electric ﬁeld in question is that at the plane directly between them, this ﬁeld could just as easily be due to point charges as it is due to charged spherical shells.

Once again, stress tensor depends on the total field in the system. As such, the field of the second sphere must be included. The symmetry in this problem allows the electric field due to the second sphere to be written immediately after a translation in the x coordinate. The electric field of sphere 2 is the same as that due to sphere 1, except that x2 = x − d.

2 ! ~ σR 1 (d/2 − d)ˆx + yyˆ + zzˆ E2 = 2 2 2 p (43) o (d/2 − d) + y + z (d/2 − d)2 + y2 + z2 ! σR2 1 (−d/2)ˆx + yyˆ + zzˆ = 2 2 2 p (44) o (−d/2) + y + z (−d/2)2 + y2 + z2 ! σR2 1 (−d/2)ˆx + yyˆ + zzˆ = 2 2 2 p (45) o (d /4) + y + z (d2/4) + y2 + z2

Add these to get the total electric ﬁeld along the plane between them.

E~tot = E~1 + E~2 (46)

σR2 1 3/2 = 2 2 2 [((d/2) − (d/2))ˆx + (y + y)ˆy + (z + z)ˆz] (47) o (d /4) + y + z

2σR2 1 3/2 = 2 2 2 [yyˆ + zzˆ] (48) o (d /4) + y + z

This provides all the information needed to completely write out the stress tensor. First, note the following,

Ex = 0 (49)

2σR2 y E = y 2 2 2 3/2 (50) o ((d /4) + y + z ) 2σR2 z E = z 2 2 2 3/2 (51) o ((d /4) + y + z ) 2 4 2 2 2 4σ R y + z E = 2 2 2 2 3 (52) o ((d /4) + y + z )

Referring back to (36), the force on sphere 2 due to 1 may be written,

Z ∞ Z ∞ ↔ F~ = T ·dy dz (−xˆ) (53) −∞ −∞ where the −xˆ represents the direction of the surface normal vector from the surface enclosing sphere 2.

6 In the following steps I have included the fact that Ex = 0 in order to simplify.

 E2  − 2 0 0    −1  ↔   · (−xˆ) =  2 E2  0 T o  0 Ey − 2 EyEz    (54)     0 2 E2 0 EzEy Ez − 2 E2 = − − xˆ + 0y ˆ + 0z ˆ (55) o 2

2 4 2 2 o 4σ R y + z = · 2 2 2 2 3 xˆ (56) 2 o ((d /4) + y + z ) 2σ2R4 y2 + z2 = 2 2 2 3 xˆ (57) o ((d /4) + y + z )

Putting this back into the integral expression for the force provides,

Z ∞ Z ∞ 2 4 2 2 ~ 2σ R y + z F = 2 2 2 3 xˆ dy dz (58) −∞ −∞ o ((d /4) + y + z ) and it is seen that the ﬁnal answer will put the force in either the positive or negative xˆ. If this were not the case, then we would already know our solution to be incorrect. Consider the top part of sphere 1. It exerts a force on the bottom of sphere 2 that is directed partially along the −zˆ direction. The bottom part of sphere 1, however, exerts a force on the top part of 2 that is directed along the +ˆz direction and cancels out the previously mentioned −zˆ force. This symmetry is preserved for the total force along the y and z directions, resulting in only a non-zero component along the x direction. Furthermore, since the shells have the same surface charge density we know the force between them should be repulsive. The force on 2 must be along the +ˆx direction or we know the answer is incorrect.

The integral in (58) is non-trivial. Since there is no x dependence in the expression for the force along the plane we can use change of coordinate system to obtain a better integral with which to work. p Consider the yz plane to be in cylindrical coordinates. If we let s = y2 + z2 and then incorporate the φ direction into the integral we get,

Z ∞ Z ∞ 2 4 2 2 ~ 2σ R y + z F = 2 2 2 3 xˆ dy dz (59) −∞ −∞ o ((d /4) + y + z ) Z ∞ Z 2π 2σ2R4 s2 = 2 2 3 xˆ s ds dφ (60) 0 0 o ((d /4) + s ) where the new integral is the result of starting over with the geometry considerations and not a mathematical change of variable. Notice that there is an extra factor of s that comes from the da term in cylindrical coordinates. The xˆ dependence is left alone because it came from the tensor work and the change to cylindrical coordinates was made after solving this part.

7 In this case the integral over s is solved using a change of variable and then an integral table.

Z ∞ 3 s 2 2 2 3 ds → Let u = s du = 2s ds (61) 0 ((d /4) + s ) Z ∞ us du = 2 3 (62) 0 ((d /4) + u) 2s 1 Z ∞ u = 2 3 du (63) 2 0 ((d /4) + u) 1 u (d2/4) ∞ = (−1) 2 2 + 2 2 (64) 2 (u + (d /4)) 2((d /4) + u) 0 1 (d2/4) = − 0 + 0 − 0 − (65) 2 2(d2/4)2 1 = (66) 4(d2/4) 1 = (67) d2 The integral over φ results in a factor of 2π and the ﬁnal solution is,

2 4 ~ 2σ R 1 F = (2π) 2 xˆ (68) o d 4πσ2R4 = 2 xˆ (69) od

where you can use the expression q = 4πR2σ to prove to yourself that this force is equivalent to that between two like charged point particles separated by a distance d.

[3.] Problem by Professor Carter Consider a cylindrical capacitor of length L with charge +Q on the inner cylinder of radius a and −Q on the outer cylindrical shell of radius b. The capacitor is filled with a lossless dielectric with dielectric constant equal to 1. The capacitor is located in a region with a uniform magnetic field B, which points along the symmetry axis of the cylindrical capacitor. A flaw develops in the dielectric insulator, and a current flow develops between the two plates of the capacitor. Because of the magnetic field, this current flow results in a torque on the capacitor, which begins to rotate. After the capacitor is fully discharged (total charge on both plates is now zero), what is the magnitude and direction of the angular velocity of the capacitor? The moment of inertia of the capacitor (about the axis of symmetry) is I, and you may ignore fringing fields in the calculation.

Reference the figure below. Solution Use the concept of the conservation of angular momentum to solve this problem quickly. Before the current flow there is a certain amount of angular momentum stored in the EM fields. After the dielectric breaks down, the capacitor discharges until there is no longer an electric field between its plates. Since there is no longer an electric field there is no longer any angular momentum stored in the fields. All of this angular momentum must now exist in the physical rotation of the system.

8 +Q

-Q

It is still possible to solve this problem considering the force on the current, but that would require considerably more work than the method shown here. Furthermore, the problem statement doesn’t say much about the current; is it uniform or localized?

The total angular momentum in the ﬁelds is, Z h i L~ em = o ~r × (E~ × B~ ) dτ (70) where the integrand is the angular momentum density given as equation 8.34 in Grifﬁths.

The magnetic ﬁeld is given in the problem. For a cylindrical capacitor the electric ﬁeld is known to be zero outside the plates. Inside the capacitor we have (a result determined by applying Gauss’ law to this system, the full solution of which is written out elsewhere), λ E~ = sˆ (71) 2πs where λ is the charge per unit length of the inner cylinder and is the dielectric constant of the material between the plates. We are given that = 1.

The term E~ × B~ = 0 everywhere except the region between the plates. The limits of the volume integral in (70) are then decided. Continuing with the solution for the total angular momentum stored in the ﬁelds gives,

Z b Z 2π Z L λ L~ em = ~s × sˆ × Bzˆ s ds dφ dz (72) a 0 0 2πs Z b Z 2π Z L λB = s sˆ × (−φˆ) s ds dφ dz (73) a 0 0 2πs Z b Z 2π Z L λBs = − ds dφ dz zˆ (74) a 0 0 2π λB Z b = − (2π)(L) s ds zˆ (75) 2π a

9 λB = − (b2 − a2)L zˆ (76) 2 QB = − (b2 − a2)z ˆ (77) 2 where the total charge on the inner cylinder is Q = λL.

This is the total angular momentum in the system. When the electric ﬁeld between the plates of the capacitor goes to zero this angular momentum will be entirely contained within the physical rotation of the system. Angular momentum is related to angular velocity as L~ = I~ω, where I is the moment of inertia. The solution is,

L~ QB ~ω = = − (b2 − a2)z ˆ (78) I 2I where this provides both the direction and magnitude of the angular velocity. The cylindrical capacitor will be spinning about its axis as electromagnetic angular momentum is converted to kinetic angular momentum.

[4.] Problem 8.9 from Griffiths Consider a very long solenoid. This solenoid has radius a, turns per unit length n, and a current Is flowing through it. A loop of wire with resistance R is coaxial with the solenoid. The radius of this wire loop, b, is much greater than the radius of the solenoid. The current in the solenoid is then slowly decreased, leading to a current flow, Ir, in the loop.

(a) Find Ir in terms of dIs/dt.

(b) The energy dissipated in the resistor must come from the solenoid. Calculate the Poynting vector just outside the solenoid and verify that it is directed toward the loop. (Griffiths Hint: Use the electric field due to the changing flux in the solenoid and the magnetic field due to the current in the wire loop.) Integrate the Poynting vector over the entire surface of the solenoid to verify that the 2 total energy “emitted” by the solenoid is equal to that dissipated in the resistive wire, Ir R.

Solution (a) This part is a review of the topics covered in Ch. 7 of Grifﬁths. The current through a resistive wire is Ir = E/R, where E is the emf (voltage) across the wire. The emf is calculated according to, dΦ E = − (79) dt d = − µ nI πa2 (80) dt o s

dIs = −µ nπa2 (81) o dt where this takes advantage of the properties of solenoid fields (i.e. the field outside is zero and the field inside is uniform).

The current in the loop is positive, though the direction depends on how we orient the solenoid, let B~ → zˆ. 2 µonπa dIs Ir = (82) R dt

10 The absolute value of the time derivative term is taken because we know the current in the solenoid is decreasing.

(b) Begin by solving for the fields Griffiths tells us to use. The electric field at the surface of the solenoid is found using Faraday’s law, ∇~ × E~ = −dB/dt~ . Since we care about the field at the surface of the solenoid we set s = a. I d Z E~ · d~l = − B~ · d~a (83) dt dΦ E (2πa) = − (84) φ dt

dIs 1 E~ = −µ nπa2 φˆ (85) o dt 2πa

µona dIs = φˆ (86) 2 dt

Notice that the final direction of the electric field is in the positive φˆ. There is a negative sign in Faraday’s law, but we also know that the time derivative of the solenoid current is negative since it is being slowly decreased. We could also use Lenz’s law to determine this direction: the magnetic field of the solenoid is decreasing (and we said that B~ = Bzˆ), so the current induced in the wire will be such as to try and replace this decreasing field. A wire generates a magnetic field in the +ˆz direction with a current in the +φˆ direction.

The magnetic field at the surface of the solenoid is greatly simplified since b a. This means we may treat the entire surface as though it lies along the z-axis of the loop. The magnetic field along the z-axis of a current loop is given in example 5.6 of Griffiths,

2 µoIr b B~ (z) = zˆ Eq. 5.38 (87) 2 (b2 + z2)3/2 where the direction is set by the orientation of the solenoid.

The Poynting vector may now be calculated, 1 1 µ na dI µ I b2 S~ = E~ × B~ = o s φˆ × o r zˆ 2 2 3/2 (88) µo µo 2 dt 2 (b + z )

2 µonab Ir dIs = sˆ (89) 4(b2 + z2)3/2 dt

The Poynting vector is directed toward the resistive loop, as expected. The next step is to integrate this vector over the entire surface of the solenoid (still using s = a).

Z Z 2π Z ∞ µ nab2I dI P ower = S~ · d~a = o r s sˆ · a dφ dz sˆ 2 2 3/2 (90) 0 −∞ 4(b + z ) dt µ na2b2I dI Z ∞ dz = o r s (2π) 2 2 3/2 (91) 4 dt −∞ (b + z )

From integral tables, Z dx x = (92) (f + cx2)3/2 f(f + gx2)1/2

11 µ πna2b2I dI z ∞ P ower = o r s 2 2 2 1/2 (93) 2 dt b (b + z ) −∞ 2 2 µoπna b Ir dIs 2 = (94) 2 dt b2

2 dIs = µoπna Ir (95) dt

Using (82) to rewrite (95), 2 P ower = Ir R (96) and the power directed from the solenoid toward the resistive loop is equal to the energy dissipated in this loop.

[5.] Problem 8.11 from Grifﬁths Treat the electron as a uniformly charged spherical shell (total charge e) of radius R, spinning with angular velocity ω.

(a) Determine the total energy contained in the EM ﬁelds.

(b) Find the total angular momentum contained in the EM ﬁelds.

(c) Let the mass of the electron be described completely in terms of the energy stored in its ﬁelds, 2 Uem = mec . Furthermore, let the spin angular momentum of the electron be due entirely to its ﬁelds, Lem = ~/2. Solve for the angular velocity and radius of the electron in this case. Calculate the value Rω and comment on whether this value makes sense classically.

Solution (a) The total energy in the fields is given by (22). This is the sum of the energy in the electric field plus that of the magnetic field, so I will determine Ue and Um independently. Begin by determining the fields inside and outside the shell.

From electrostatics and Ch. 2 of Griffiths we know that the electric field inside a uniformly charged spherical shell is zero. Also, the field outside is that of a point charge. Therefore,

E~in = 0 (97)

~ e Eout = 2 rˆ (98) 4πor

The magnetic ﬁeld inside a uniformly charged spinning spherical shell is given in example 5.11 of Grifﬁths, 2 B~ = µ σRω zˆ Eq. 5.68 (99) in 3 o This can be rewritten in terms of the variables given in the problem after we solve for the surface charge density, Qtot e σ = = (100) A 4πR2

12 The magnetic field inside the shell is, µoωe B~ = zˆ (101) in 6πR The magnetic field outside of the shell is that of a dipole. This is known from a variety of sources: Your work in PHYS 110A, reading problem 5.36 from Griffiths, or using the given vector potential from example 5.11 in Griffiths to solve for the field directly by way of B~ = ∇~ × A~. Regardless of your method, the magnetic field outside the shell is,

µom B~ = 2 cos θ rˆ + sin θ θˆ (102) out 4πr3

2 µoωeR = 2 cos θ rˆ + sin θ θˆ (103) 12πr3 I have included the value of the dipole moment in (103) because you probably solved for it in 110A and referenced that solution to provide its value here. If you do not remember this value, then you might have solved for it in the following way.

Break the spinning shell into a series of inﬁnitesimal rings. In this case the differential element of the magnetic dipole moment is given by,

d~m = dI ~a (104)

The differential element on the right hand side of (104) must be for the current because the area vector of any individual ring is,

~a = πl2zˆ (105)

= π(r sin θ)2zˆ (106)

= πr2 sin2 θ zˆ (107) where l is the radius of the ring. The direction zˆ is determined by the orientation of the sphere and its rotation and can therefore be set to whatever value we want.

To ﬁnd dI we need to write out the current through an individual ring. This is determined using the surface current density.

dI = K dL where dL is the θˆ component of the spherical length dl

= σv(r dθ)

= σ|~ω × ~r|r dθ

= σωr2 sin θ dθ

Returning to (104), d~m = πσωR4 sin3 θ dθ (108) where I have taken into account the fact that this is a shell so r = R.

13 The magnetic dipole moment is found through an integration of d~m, Z π ~m = πσωR4 sin3 θ dθ zˆ (109) 0 1 π = πσωR4 − cos θ sin2 θ + 2 zˆ (110) 3 0 4πσωR4 = zˆ (111) 3

Replace the charge density in (111) with that from (100) to get the ﬁnal expression for B~ out.

Now begins the calculation of the total energy in the fields. The electric field is zero inside the shell so it contributes nothing to the total energy. The electric field outside the shell contributes, Z 2 oE U = dτ (112) e,out 2

Z ∞ Z π Z 2π 2 o e 2 = 2 r sin θ dr dθ dφ (113) 2 R 0 0 4πor e2 Z ∞ dr = 2 (4π) 2 (114) 32π o R r e2 1∞ = − (115) 8πo r R e2 = (116) 8πoR On to the magnetic ﬁeld energy inside the shell. Since the ﬁeld inside the shell is uniform the integral may be skipped. The energy inside the shell is simply the magnetic energy density multiplied by the interior volume. 4 U = u · πR3 (117) m,in m,in 3

1 µoωe2 4 = · πR3 (118) 2µo 6πR 3

2 2 µoω e R = (119) 54π Calculating the energy stored in the magnetic ﬁeld outside of the shell will illustrate why I wrote the dipole ﬁeld in terms of spherical coordinates. Once again, recall that B2 = B~ · B~ . Z ∞ Z π Z 2π 2 2 2 4 1 µoω e R 2 2 2 Um,out = 2 6 (4 cos θ + sin θ)r sin θ dr dθ dφ (120) R 0 0 2µo 144π r 2 2 4 Z ∞ Z π µoω e R 1 2 2 = 2 (2π) 4 (4 cos θ + sin θ) sin θ dr dθ (121) 288π R 0 r 2 2 4 ∞ Z π µoω e R 1 2 2 = − 3 (4 cos θ + sin θ) sin θ dθ (122) 144π 3r R 0 2 2 Z π µoω e R = (4 cos2 θ + sin2 θ) sin θ dθ (123) 432π 0

14 This is another integral that you may solve in any manner. Here I make the substitution 4 cos2 θ = 4 − 4 sin2 θ. The integral then reduces to, Z π Z π (4 cos2 θ + sin2 θ) sin θ dθ = (4 − 3 sin2 θ) sin θ dθ (124) 0 0 Z π = 4(2) − 3 sin3 θ dθ (125) 0 1 π = 8 − 3 − cos θ sin2 θ + 2 (126) 3 0

= 8 − 4 = 4 (127)

Returning to the energy expression we have,

2 2 µoω e R U = · 4 (128) m,out 432π

2 2 µoω e R = (129) 108π

The total energy in the ﬁelds is the sum of (116), (119), and (129).

2 2 2 2 2 e µoω e R µoω e R Utot = + + (130) 8πoR 54π 108π

2 2 2 e µoω e R = + (131) 8πoR 36π

(b) The angular momentum stored in the fields is found using (70). The zero electric field inside the shell means that we only need be concerned with the region outside of the shell. All of the fields have already been found, so we begin by calculating the angular momentum density in the fields and then we’ll integrate that over the volume outside of the shell. h i ~lem = o ~r × (E~out × B~ out) (132)

2 e µoωeR ˆ = o ~r × 2 rˆ × 3 2 cos θ rˆ + sin θ θ (133) 4πor 12πr 2 e µoωeR ˆ = o ~r × 2 · 3 sin θ φ (134) 4πor 12πr

2 2 µoωe R = sin θ (−r θˆ) (135) 48π2r5

2 2 µoωe R = − sin θ θˆ (136) 48π2r4

15 On to the total,

Z ∞ Z π Z 2π 2 2 ~ µoωe R ˆ 2 Lem = − 2 4 sin θ θ r sin θ dr dθ dφ (137) R 0 0 48π r 2 2 Z ∞ Z π 2 µoωe R sin θ ˆ = − 2 (2π) 2 dr dθ θ (138) 48π R 0 r 2 2 ∞ Z π µoωe R 1 = − − sin2 θ dθ θˆ (139) 24π r R 0 2 Z π µoωe R = − sin2 θ dθ θˆ (140) 24π 0

The integral in (140) is not as easy as it looks. The θˆ vector changes with the value of θ. Integrating over θ from 0 to π means the θˆ can be replaced with its z component. This is a mathematical fact and you will be able to ﬁnd it elsewhere (proven rigorously). The z component of θˆ is (− sin θ).

2 Z π 2 Z π µoωe R µoωe R − sin2 θ dθ θˆ = − sin2 θ dθ (− sin θ zˆ) (141) 24π 0 24π 0 2 Z π µoωe R = sin3 θ dθ zˆ (142) 24π 0

The solution to the integral is shown in (125) and (126).

2 µoωe R L~ = zˆ (143) em 18π

~ L = (144) em 2

2 µoωe R ~ = (145) 18π 2

18π~ ωR = 2 (146) 2µoe 9π(1.05 × 10−34) = (147) (4π × 10−7)(1.60 × 10−19)2

= 9.23 × 1010 (148)

This product represents the physical speed (in m/s) of a point on the equator of the shell. It is considerably faster than the speed of light and therefore makes no sense physically, demonstrating the need for quantum mechanics in the explanation of various properties of the electron.

To solve for the values independently use (148) and the mass relation given in the problem state- 2 ment, Uem = mec . This provides two equations with which you can solve for the two unknowns.

16 The numerical values are approximately:

R = 2.96 × 10−11 m (149)

ω = 3.12 × 1021 s−1 (150)

[6.] Problem 9.6 from Grifﬁths (a) Write a revised boundary condition (replacing equation 9.27 from Grifﬁths) for the case of a tension T applied across two strings connected with a knot of mass m.

(b) Consider the situation where the knot connecting the strings has mass m and the second string is massless. Find the amplitudes and phases of the reﬂected and transmitted waves.

Solution (a) Begin with,

∂f ∂f = Eq. 9.27 (151) ∂z 0− ∂z 0+ where the + and − subscripts refer to the right and left sides of the knot respectively.

To determine the new boundary condition, refer to the origin of Eq. 9.27 (page 365 of Grifﬁths), ∂f ∂f ∆F =∼ T − (152) ∂z + ∂z −

The above equation is considered at the point z = 0, the location of the knot. Grifﬁths arrives at equation 9.27 by taking the left side of (152) to be zero because the mass of the knot is set to zero. In part (a) of this problem we are asked to consider a knot of some mass, m. This is equivalent to setting ∆F = ma, which for a one dimensional case becomes,

2 ∂ f ∂f ∂f m 2 = T − (153) ∂t ∂z + ∂z −

all of which is evaluated at z = 0. Recall that the function describing the string, f, represents its position. The derivative of f is the velocity, and the second derivative is an acceleration.

(b) This part is a boundary value problem. We will use two equations to solve for the amplitudes and phases of the reﬂected and transmitted waves in terms of the incident values (which may always be assumed to be given). The ﬁrst boundary condition is given in (153). The second boundary condition comes from the fact that the rope itself is continuous and therefore requires that the functions to the left and right of the knot be equal at z = 0.

f(0−, t) = f(0+, t) Eq. 9.26 (154)

The general solution is already known from the properties of waves.

∼ ∼ i(k1z−ωt) i(−k1z−ωt) f− = AI e + AR e (155)

∼ i(k2z−ωt) f+ = AT e (156)

17 ∼ ∼ ∼ where AI , AR, and AT refer to the complex amplitudes of the incident (coming from the left), re- ﬂected, and transmitted waves respectively.

Condition (154) says that we can use either wave function for the time derivative term in (153). Using f+ we have,

∂2f ∼ m = −mω2 A e−iωt (157) ∂t2 T recalling that this is evaluated at z = 0. The shortcut method is to use ∂/∂t = −iω when dealing with waves of this sort.

Computing the right side of (153),

∼ ∼ ∼ ∼ 2 −iωt −iωt −iωt −iωt −mω AT e = iT (k2 AT e − k1 AI e + k1 AR e ) (158)

∼ ∼ ∼ ∼ 2 −mω AT = iT (k2 AT −k1 AI +k1 AR) (159)

This allows the transmitted amplitude to be written in terms of the other amplitudes as,

∼ ∼ ∼ 2 (iT k2 + mω ) AT = iT k1(AI − AR) (160)

Writing out (154) allows us to simplify it,

∼ ∼ ∼ AI + AR = AT (161)

Multiply (161) by iT k1 and add this to (160),

∼ ∼ ∼ iT k1 · (AI + AR = AT ) (162)

∼ ∼ ∼ 2 (iT k2 + mω ) AT = iT k1(AI − AR) (163)

Result :

∼ ∼ 2 2iT k1 AI = iT (k1 + k2) + mω AT (164)

∼ 2iT k1 ∼ AT = 2 AI (165) [iT (k1 + k2) + mω ] ∼ ∼ Putting this expression for AT into (161) and solving for AR gives,

2 ∼ iT (k1 − k2) − mω ∼ AR = 2 AI (166) iT (k1 + k2) + mω Now it is time to make use of the fact that the second string is massless. For waves on strings the velocity is given by s T v = Eq. 9.3 (167) µ

18 where µ is the mass density of the string.

In this problem µ2 = 0 so v2 = ∞. The wave vectors of the two waves are related to the velocities by, k2 v1 = Eq. 9.24 (168) k1 v2 which leads to k2/k1 = 0 in this case.

Return to the expressions for the transmitted and reﬂected amplitudes in terms of the incident amplitude and factor k1 out of the denominators. This allows those expressions to simplify to, ∼ 2 ∼ AT = AI (169) 1 − imω2 k1T 2 ∼ 1 + imω ∼ k1T AR = AI (170) 1 − imω2 k1T Separate the real amplitude from the phase of the waves as follows, 2 iδT iδI AT e = AI e (171) 1 − imω2 k1T 2 1 + imω iδR k1T iδI ARe = AI e (172) 1 − imω2 k1T Thus concludes the setup part of this problem. From this point forward it is all algebra. One method is the following, A eiδT 2 T = iδ imω2 (173) AI e I 1 − k1T imω2 A eiδT 1 + T = k1T iδ imω2 (174) ARe R 1 − k1T If you square both sides of the above equations you will be able to separate out the ratio of the amplitudes from a relation between the phases. Coupling this with the following identity, ei2φ − 1 tan φ = (175) i (ei2φ + 1) will allow you to solve for the desired values in terms of the incident parameters.

After some algebra, the real amplitudes and phases are, 2 AT = AI (176) q m2ω4 1 + 2 2 k1T

AR = AI (177)

2 −1 mω δT = δI + tan (178) k1T

2mω2 δ = δ + tan−1 k1T R I m2ω4 (179) 1 − 2 2 k1T

19 [7.] Problem by Professor Carter A clever cub scout who has studied E&M decides he wants to cheat in the Pinewood derby race by giving his car an extra push using a laser. He happens to have one of the most powerful CW (continuous wave, meaning it can lase continuously) lasers available, with a laser power of 10 kW. He plans to shine this laser on the back of the car, which is coated with a perfect reflector. To figure out is his scheme will help him, calculate the length of time necessary for the laser to accelerate his 0.1 kg car to a speed of 1 m/s. (Note that if the reflector isn’t perfect, then the car will be destroyed long before reaching this speed.)

Solution This is a problem in kinetics. The laser will exert a force on the car that is described according to the radiation pressure of the laser. Since the laser is on continuously, the force on the car will be constant. As such, the car’s acceleration due to the laser will also be constant. The time dependence of an object’s velocity due to a constant acceleration (a) is,

v(t) = vo + at (180)

where vo = 0 in this problem (not explicity stated in the problem, but since this is a race the car must start from rest). The ﬁnal velocity is v = 1.

Pressure, which is force (F ) per unit area (A), is given by, F 2I P = = Eq. 9.64 (181) A c where I is the intensity of the incident light and we need to rewrite this to solve for the force on the car due to the laser. The factor of 2, which is not actually written in Grifﬁths’ form of the equation,

F = PA (182)

2IA = (183) c Intensity is deﬁned as the average power per unit area (see page 381 in Grifﬁths). The laser power (PL) is given in the problem, which brings us to, PL 2A F = (184) A c

2PL = (185) c

2 × 104 = (186) 3 × 108

ma = 0.66 × 10−4 (187)

0.66 × 10−4 a = (188) 0.1

= 0.66 × 10−3 (189)

20 Now we can solve for the time it takes to reach the velocity of 1 m/s.

v = at (190)

1 = 0.66 × 10−3 t (191)

t = 1500 (192)

where this value is in units of seconds. This is equivalent to 25 minutes, meaning that it is unlikely the scout will derive any beneﬁt from this laser propulsion system.

[8.] Problem by Professor Carter

Calculate ∇~ · S~ and ∂Uem/∂t for a linearly polarized electromagnetic plane wave propagating in the zˆ direction and polarized in the xˆ direction. Explain your results physically.

Solution The ”paradigm” for describing such an EM wave is (Eq. 9.48),

E~ (z, t) = Eo cos(kz − ωt + δ)x ˆ (193)

1 B~ (z, t) = E cos(kz − ωt + δ)y ˆ (194) c o

where the direction of propagation for this wave is given according to kˆ = Eˆ × Bˆ =z ˆ

The term ∇~ · S~ is, 1 ∇~ · S~ = ∇~ · E~ × B~ (195) µo E2 = ∇~ · o cos2(kz − ωt + δ)z ˆ (196) µoc E2 ∂ = o cos2(kz − ωt + δ) (197) µoc ∂z −2kE2 = o cos(kz − ωt + δ) sin(kz − ωt + δ) (198) µoc

The term ∂Uem is,

∂Uem ∂ = E2 (199) ∂t ∂t o ∂ = E2 cos2(kz − ωt + δ) (200) ∂t o o

2 = oEo (2ω cos(kz − ωt + δ) sin(kz − ωt + δ)) (201)

2 = 2oωEo cos(kz − ωt + δ) sin(kz − ωt + δ) (202)

21 Determine the physical meaning of these results by comparing the results in (198) and (202),

2 −2kEo ∇~ · S~ µ c cos(kz − ωt + δ) sin(kz − ωt + δ) = o (203) ∂Uem 2 ωE2 cos(kz − ωt + δ) sin(kz − ωt + δ) ∂t o o −k = (204) µooωc Now we must recall the following expressions relating k, ω, and the speed of light. ω = c (205) k 1 = c2 (206) µoo Continuing on,

∇~ · S~ −c2 = (207) ∂Uem c2 ∂t

= −1 (208)

These terms are opposite in magnitude. As the energy in an EM wave decreases in time, the amount of energy ﬂux away from the present location increases. This is another demonstration of energy conservation: whatever is lost within a system must have been radiated away. The topic of radiation arrives soon.

[9.] Problem 9.12 from Grifﬁths Find the elements of the Maxwell stress tensor for a monochromatic plane wave (same one from the previous problem, given by (193) and (194)). Is this the answer you expected? How is the Maxwell stress tensor related to the energy density?

Solution

The Maxwell stress tensor is given by (38). In this problem there are only Ex and By components of Tij. Most of the components can be immediately seen to be zero. The remaining terms are,

 2 2  E2 − E − B 0 0 o x 2 2µo   ↔   2  2 2  T = cos (kz − ωt + δ)  0 − oE + 1 B2 − B 0  (209)  2 µo y 2      2 2 0 0 − oE − B 2 2µo where I have factored out the common cosine term.

2 2 2 2 In this problem Ex = E and By = B , which further simpliﬁes this expression to,

2 2  oE − B 0 0  2 2µo   ↔  2 2  T = cos2(kz − ωt + δ)  0 − oE + B 0  (210)  2 2µo      2 2 0 0 − oE − B 2 2µo

22 Now recall from the previous work on electromagnetic energy that (see Eq. 9.54 for a reminder),

2 2 oE B = (211) 2 2µo

This makes two more of the terms in (210) zero,

 0 0 0    ↔   2  0 0 0  T = cos (kz − ωt + δ)   (212)    2 2  0 0 − oE − B 2 2µo

The Maxwell stress tensor is the negative of the energy density. For EM waves (light waves) the energy density is completely manifested in the momentum density.