phys 110b - hw #4lief.if.ufrgs.br/~ambusher/griffiths/fall2005_4.pdf · the gap, e~(s,t) and...
TRANSCRIPT
PHYS 110B - HW #4Fall 2005, Solutions by David PaceEquations referenced as ”EQ. #” are from GriffithsProblem 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 inthe 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 intothe gap by integrating the Poynting vector over the relevant surface. Verify that the input power isequivalent to the rate of increasing energy in the gap. (Griffiths Hint: This verification amounts toproving 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 2.5in Griffiths),
~E =σ
εoz (1)
where I define z 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. Theresultant charge density on each “plate” is then time-dependent because the flowing current causescharge to pile up. At time zero there is no charge on the plates, so we know that the charge densityincreases linearly as time progresses.
σ(t) =It
πa2(2)
where a is the radius of the wire and It is the total charge on the plates at any instant (current isin units of Coulombs/second, so the total charge on the end plate is the current multiplied by thelength of time over which the current has been flowing).
The electric field between the ends of the wire is,
~E =It
εoπa2z (3)
The magnetic field is found from Ampere’s law,∮~B · d~l = µoIenc + µoεo
∫ (∂ ~E
∂t
)· d~a Eq. 7.38 (4)
where the enclosed current is zero, Ienc = 0, and the magnetic field is generated entirely by thedisplacement current.
1
The displacement current, Id (the non-zero term in Ampere’s law above), through the gap must bein the z direction (the only proper current flow is through the wire),
Bφ(2πs) = Id (5)
Solving for the displacement current in the gap gives,
Id = µoεo
∫ (∂ ~E
∂t
)· d~a (6)
= µoεo
∫∂
∂t
(It
εoπa2
)z · d~a (7)
=µoI
πa2
∫ s
0
∫ 2π
0(z) · s ds dφ z (8)
=µoI
πa2
(s2
2
)s
0
(2π) (9)
=µoIs2
a2(10)
Now the magnetic field is found from (5),
~B =µoIs
2πa2φ (11)
where it should be noted that the field increases with s. This is only for the region in the gap andinside the wire radius of a. The linear increase of the magnetic field in this region agrees with thatwe would find in a solid wire with a uniform current density. For s > a the magnetic field is afringing field that goes to zero as s →∞.
(b) The energy density is given in terms of the fields in the gap,
uem =12
(εoE
2 +B2
µo
)Eq. 8.13 (12)
=12
(εo
I2t2
ε2oπ2a4
+1µo
µ2oI
2s2
4π2a4
)(13)
=12
(I2
π2a4
)[t2
εo+
µos2
4
](14)
noting that E2 = ~E · ~E.
The Poynting vector in the gap is,
~S =1µo
~E × ~B Eq. 8.10 (15)
=1µo
(It
εoπa2z × µoIs
2πa2φ
)(16)
= − I2ts
2εoπ2a4s (17)
2
The Poynting vector represents energy flow. Taking special note of the direction found above wesee 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 andthe energy within the gap is therefore due entirely to that of the fields),
∂
∂t(umech + uem) = −~∇ · ~S Eq. 8.14 (18)
∂
∂t
(I2
2π2a4
)[t2
εo+
µos2
4
]= −1
s
∂
∂s
(s−I2ts
2εoπ2a4
)(19)
I2t
εoπ2a4= −1
s
(− I2ts
εoπ2a4
)(20)
=I2t
εoπ2a4(21)
(c) Since we have the energy density in the gap we are ready to determine the total energy.
Uem =12
∫ (εoE
2 +B2
µo
)dτ Eq. 8.5 (22)
=12
∫ a
0
∫ 2π
0
∫ w
0
(I2
π2a4
)[t2
εo+
µos2
4
]s ds dφ dz (23)
=I2
2π2a4(2πw)
∫ a
0
[t2s
εo+
µos3
4
]ds (24)
=I2w
πa4
[t2a2
2εo+
µoa4
16
](25)
=I2w
2πa2
[t2
εo+
µoa2
8
](26)
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 Poyntingvector 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.
Power in =∫ 2π
0
∫ w
0
~S · s dφ dz s (27)
=∫ 2π
0
∫ w
0− I2ts
2εoπ2a4s dφ dz (28)
= − I2ta2
2εoπ2a4(2wπ) (29)
= − I2tw
εoπa2(30)
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Finally, we essentially want to verify equation 8.9,
dW
dt= − d
dt
∫V
12
(εoE
2 +1µo
B2
)dτ − 1
µo
∮S
(~E × ~B
)· d~a (31)
where this represents the rate at which work is done on a collection of charges in the volume V , thatis enclosed by the surface S.
Since there are no charges in the gap, W = 0 and dW/dt = 0. The equation becomes,
d
dt
∫V
12
(εoE
2 +1µo
B2
)dτ = − 1
µo
∮S
(~E × ~B
)· d~a (32)
d
dtUem = −
∮S
~S · d~a (33)
we have solved for Uem in (26), and the integral on the right hand side of the above equation is thepower flowing into the gap, given by (30). Continuing with the verification of equation 8.9,
d
dt
I2w
2πa2
[t2
εo+
µoa2
8
]= −
(− I2tw
εoπa2
)(34)
I2tw
εoπa2=
I2tw
εoπa2(35)
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 isa distance d > 2R between their centers. Calculate the force of one shell on the other by integratingMaxwell’s stress tensor. (Hint: a good surface to choose includes the plane which passes directlybetween the two and then closes out at infinity).
Solution
Generally, the force on charges within a volume V is,
~F =∮S
↔T · d~a− εoµo
d
dt
∫V
~S dτ Eq. 8.22 (36)
In this problem we will determine the force on one of the spheres due to the other sphere using theequation above. Enclose one of the spheres (call it number 2 for no specific reason) in a volume. Inthis 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 thestress tensor includes self-fields).
There are no magnetic fields in this problem so the Maxwell stress tensor is,
↔T = εo
E2
x − E2
2 ExEy ExEz
EyEx E2y − E2
2 EyEz
EzEx EzEy E2z − E2
2
(37)
4
The expression in (37) was given directly in lecture on 10/21/2005, but it may also be derived fromthe general form of the stress tensor.
Tij ≡ εo
(EiEj −
12δijE
2
)+
1µo
(BiBj −
12δijB
2
)Eq. 8.19 (38)
I will draw the enclosed region around sphere 2 and determine the force that 1 exerts on it. Theelectric fields outside of the spheres are determined from Gauss’ law. To best simplify the geometryof the surface integral that must be done, I use a surface that includes the plane exactly betweenthe centers of the spheres. This surface closes at infinity, where the electric field is zero, so theintegration 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 stresstensor 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 givenby,
~E1 =q
4πεor2r (39)
=σR2
εor2r (40)
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 inthe electric field along the plane between the charges gives,
~E1 =σR2
εo
(1
(d/2)2 + y2 + z2
)((d/2)x + yy + zz√(d/2)2 + y2 + z2
)(41)
=σR2
εo
(1
(d2/4) + y2 + z2
)((d/2)x + yy + zz√(d2/4) + y2 + z2
)(42)
5
using r =√
(d/2)2 + y2 + z2 as the distance from the center of sphere 1 to the plane. The (d/2)xterm 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 electricfield in question is that at the plane directly between them, this field could just as easily be due topoint 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 secondsphere must be included. The symmetry in this problem allows the electric field due to the secondsphere to be written immediately after a translation in the x coordinate. The electric field of sphere2 is the same as that due to sphere 1, except that x2 = x− d.
~E2 =σR2
εo
(1
(d/2− d)2 + y2 + z2
)((d/2− d)x + yy + zz√(d/2− d)2 + y2 + z2
)(43)
=σR2
εo
(1
(−d/2)2 + y2 + z2
)((−d/2)x + yy + zz√(−d/2)2 + y2 + z2
)(44)
=σR2
εo
(1
(d2/4) + y2 + z2
)((−d/2)x + yy + zz√
(d2/4) + y2 + z2
)(45)
Add these to get the total electric field along the plane between them.
~Etot = ~E1 + ~E2 (46)
=σR2
εo
(1
(d2/4) + y2 + z2
)3/2
[((d/2)− (d/2))x + (y + y)y + (z + z)z] (47)
=2σR2
εo
(1
(d2/4) + y2 + z2
)3/2
[yy + zz] (48)
This provides all the information needed to completely write out the stress tensor. First, note thefollowing,
Ex = 0 (49)
Ey =2σR2
εo
(y
((d2/4) + y2 + z2)3/2
)(50)
Ez =2σR2
εo
(z
((d2/4) + y2 + z2)3/2
)(51)
E2 =4σ2R4
ε2o
(y2 + z2
((d2/4) + y2 + z2)3
)(52)
Referring back to (36), the force on sphere 2 due to 1 may be written,
~F =∫ ∞
−∞
∫ ∞
−∞
↔T ·dy dz (−x) (53)
where the−x represents the direction of the surface normal vector from the surface enclosing sphere2.
6
In the following steps I have included the fact that Ex = 0 in order to simplify.
↔T · (−x) = εo
−E2
2 0 0
0 E2y − E2
2 EyEz
0 EzEy E2z − E2
2
−1
00
(54)
= −εo
[−E2
2x + 0 y + 0 z
](55)
=εo
2· 4σ2R4
ε2o
(y2 + z2
((d2/4) + y2 + z2)3
)x (56)
=2σ2R4
εo
(y2 + z2
((d2/4) + y2 + z2)3
)x (57)
Putting this back into the integral expression for the force provides,
~F =∫ ∞
−∞
∫ ∞
−∞
2σ2R4
εo
(y2 + z2
((d2/4) + y2 + z2)3
)x dy dz (58)
and it is seen that the final answer will put the force in either the positive or negative x. If this werenot the case, then we would already know our solution to be incorrect. Consider the top part ofsphere 1. It exerts a force on the bottom of sphere 2 that is directed partially along the −z direction.The bottom part of sphere 1, however, exerts a force on the top part of 2 that is directed along the+z direction and cancels out the previously mentioned −z force. This symmetry is preserved forthe total force along the y and z directions, resulting in only a non-zero component along the xdirection. Furthermore, since the shells have the same surface charge density we know the forcebetween them should be repulsive. The force on 2 must be along the +x direction or we know theanswer is incorrect.
The integral in (58) is non-trivial. Since there is no x dependence in the expression for the force alongthe plane we can use change of coordinate system to obtain a better integral with which to work.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,
~F =∫ ∞
−∞
∫ ∞
−∞
2σ2R4
εo
(y2 + z2
((d2/4) + y2 + z2)3
)x dy dz (59)
=∫ ∞
0
∫ 2π
0
2σ2R4
εo
(s2
((d2/4) + s2)3
)x s ds dφ (60)
where the new integral is the result of starting over with the geometry considerations and not amathematical change of variable. Notice that there is an extra factor of s that comes from the daterm in cylindrical coordinates. The x dependence is left alone because it came from the tensorwork 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.∫ ∞
0
s3
((d2/4) + s2)3ds → Let u = s2 du = 2s ds (61)
=∫ ∞
0
us
((d2/4) + u)3du
2s(62)
=12
∫ ∞
0
u
((d2/4) + u)3du (63)
=12(−1)
[u
(u + (d2/4))2+
(d2/4)2((d2/4) + u)2
]∞0
(64)
= −12
[0 + 0− 0− (d2/4)
2(d2/4)2
](65)
=1
4(d2/4)(66)
=1d2
(67)
The integral over φ results in a factor of 2π and the final solution is,
~F =2σ2R4
εo(2π)
1d2
x (68)
=4πσ2R4
εod2x (69)
where you can use the expression q = 4πR2σ to prove to yourself that this force is equivalent to thatbetween 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 withdielectric 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 dielectricinsulator, and a current flow develops between the two plates of the capacitor. Because of the mag-netic field, this current flow results in a torque on the capacitor, which begins to rotate. After thecapacitor is fully discharged (total charge on both plates is now zero), what is the magnitude anddirection of the angular velocity of the capacitor? The moment of inertia of the capacitor (about theaxis 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. Beforethe current flow there is a certain amount of angular momentum stored in the EM fields. After thedielectric breaks down, the capacitor discharges until there is no longer an electric field between itsplates. Since there is no longer an electric field there is no longer any angular momentum stored inthe fields. All of this angular momentum must now exist in the physical rotation of the system.
8
+Q
-Q
B
It is still possible to solve this problem considering the force on the current, but that would requireconsiderably more work than the method shown here. Furthermore, the problem statement doesn’tsay much about the current; is it uniform or localized?
The total angular momentum in the fields is,
~Lem =∫
εo
[~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 tobe 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),
~E =λ
2πεss (71)
where λ is the charge per unit length of the inner cylinder and ε is the dielectric constant of thematerial between the plates. We are given that ε = 1.
The term ~E × ~B = 0 everywhere except the region between the plates. The limits of the volumeintegral in (70) are then decided. Continuing with the solution for the total angular momentumstored in the fields gives,
~Lem =∫ b
a
∫ 2π
0
∫ L
0
[~s×
(λ
2πss×Bz
)]s ds dφ dz (72)
=∫ b
a
∫ 2π
0
∫ L
0
[s s× λB
2πs(−φ)
]s ds dφ dz (73)
=∫ b
a
∫ 2π
0
∫ L
0−λBs
2πds dφ dz z (74)
= −λB
2π(2π)(L)
∫ b
as ds z (75)
9
= −λB
2(b2 − a2)L z (76)
= −QB
2(b2 − a2) z (77)
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 ofthe capacitor goes to zero this angular momentum will be entirely contained within the physicalrotation of the system. Angular momentum is related to angular velocity as ~L = I~ω, where I is themoment of inertia. The solution is,
~ω =~L
I= −QB
2I(b2 − a2) z (78)
where this provides both the direction and magnitude of the angular velocity. The cylindrical ca-pacitor will be spinning about its axis as electromagnetic angular momentum is converted to kineticangular momentum.
[4.] Problem 8.9 from Griffiths
Consider a very long solenoid. This solenoid has radius a, turns per unit length n, and a currentIs flowing through it. A loop of wire with resistance R is coaxial with the solenoid. The radius ofthis wire loop, b, is much greater than the radius of the solenoid. The current in the solenoid is thenslowly 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 vectorjust outside the solenoid and verify that it is directed toward the loop. (Griffiths Hint: Use theelectric field due to the changing flux in the solenoid and the magnetic field due to the current inthe wire loop.) Integrate the Poynting vector over the entire surface of the solenoid to verify that thetotal energy “emitted” by the solenoid is equal to that dissipated in the resistive wire, I2
r R.
Solution
(a) This part is a review of the topics covered in Ch. 7 of Griffiths. The current through a resistivewire is Ir = E/R, where E is the emf (voltage) across the wire. The emf is calculated according to,
E = −dΦdt
(79)
= − d
dtµonIsπa2 (80)
= −µonπa2 dIs
dt(81)
where this takes advantage of the properties of solenoid fields (i.e. the field outside is zero and thefield inside is uniform).
The current in the loop is positive, though the direction depends on how we orient the solenoid, let~B → z.
Ir =µonπa2
R
∣∣∣∣dIs
dt
∣∣∣∣ (82)
10
The absolute value of the time derivative term is taken because we know the current in the solenoidis decreasing.
(b) Begin by solving for the fields Griffiths tells us to use. The electric field at the surface of thesolenoid is found using Faraday’s law, ~∇× ~E = −d ~B/dt. Since we care about the field at the surfaceof the solenoid we set s = a. ∮
~E · d~l = − d
dt
∫~B · d~a (83)
Eφ(2πa) = −dΦdt
(84)
~E = −µonπa2 dIs
dt
12πa
φ (85)
=µona
2
∣∣∣∣dIs
dt
∣∣∣∣ φ (86)
Notice that the final direction of the electric field is in the positive φ. There is a negative sign inFaraday’s law, but we also know that the time derivative of the solenoid current is negative since itis being slowly decreased. We could also use Lenz’s law to determine this direction: the magneticfield of the solenoid is decreasing (and we said that ~B = Bz), so the current induced in the wirewill be such as to try and replace this decreasing field. A wire generates a magnetic field in the +zdirection with a current in the +φ direction.
The magnetic field at the surface of the solenoid is greatly simplified since b � a. This means wemay treat the entire surface as though it lies along the z-axis of the loop. The magnetic field alongthe z-axis of a current loop is given in example 5.6 of Griffiths,
~B(z) =µoIr
2b2
(b2 + z2)3/2z Eq. 5.38 (87)
where the direction is set by the orientation of the solenoid.
The Poynting vector may now be calculated,
~S =1µo
~E × ~B =1µo
µona
2
∣∣∣∣dIs
dt
∣∣∣∣ φ× µoIr
2b2
(b2 + z2)3/2z (88)
=µonab2Ir
4(b2 + z2)3/2
∣∣∣∣dIs
dt
∣∣∣∣ s (89)
The Poynting vector is directed toward the resistive loop, as expected. The next step is to integratethis vector over the entire surface of the solenoid (still using s = a).
Power =∫
~S · d~a =∫ 2π
0
∫ ∞
−∞
[µonab2Ir
4(b2 + z2)3/2
∣∣∣∣dIs
dt
∣∣∣∣ s] · a dφ dz s (90)
=µona2b2Ir
4
∣∣∣∣dIs
dt
∣∣∣∣ (2π)∫ ∞
−∞
dz
(b2 + z2)3/2(91)
From integral tables, ∫dx
(f + cx2)3/2=
x
f(f + gx2)1/2(92)
11
Power =µoπna2b2Ir
2
∣∣∣∣dIs
dt
∣∣∣∣ [ z
b2(b2 + z2)1/2
]∞−∞
(93)
=µoπna2b2Ir
2
∣∣∣∣dIs
dt
∣∣∣∣ ( 2b2
)(94)
= µoπna2Ir
∣∣∣∣dIs
dt
∣∣∣∣ (95)
Using (82) to rewrite (95),Power = I2
r R (96)
and the power directed from the solenoid toward the resistive loop is equal to the energy dissipatedin this loop.
[5.] Problem 8.11 from Griffiths
Treat the electron as a uniformly charged spherical shell (total charge e) of radius R, spinning withangular 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,Uem = mec
2. Furthermore, let the spin angular momentum of the electron be due entirely to itsfields, Lem = ~/2. Solve for the angular velocity and radius of the electron in this case. Calculatethe 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 fieldplus that of the magnetic field, so I will determine Ue and Um independently. Begin by determiningthe fields inside and outside the shell.
From electrostatics and Ch. 2 of Griffiths we know that the electric field inside a uniformly chargedspherical shell is zero. Also, the field outside is that of a point charge. Therefore,
~Ein = 0 (97)
~Eout =e
4πεor2r (98)
The magnetic field inside a uniformly charged spinning spherical shell is given in example 5.11 ofGriffiths,
~Bin =23µoσRω z Eq. 5.68 (99)
This can be rewritten in terms of the variables given in the problem after we solve for the surfacecharge density,
σ =Qtot
A=
e
4πR2(100)
12
The magnetic field inside the shell is,~Bin =
µoωe
6πRz (101)
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 potentialfrom example 5.11 in Griffiths to solve for the field directly by way of ~B = ~∇ × ~A. Regardless ofyour method, the magnetic field outside the shell is,
~Bout =µom
4πr3
(2 cos θ r + sin θ θ
)(102)
=µoωeR2
12πr3
(2 cos θ r + sin θ θ
)(103)
I have included the value of the dipole moment in (103) because you probably solved for it in 110Aand referenced that solution to provide its value here. If you do not remember this value, then youmight have solved for it in the following way.
Break the spinning shell into a series of infinitesimal rings. In this case the differential element ofthe 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 areavector of any individual ring is,
~a = πl2z (105)
= π(r sin θ)2z (106)
= πr2 sin2 θ z (107)
where l is the radius of the ring. The direction z is determined by the orientation of the sphere andits 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 usingthe surface current density.
dI = K dL where dL is the θ 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,
~m = πσωR4
∫ π
0sin3 θ dθ z (109)
= πσωR4
[−1
3cos θ
(sin2 θ + 2
)]π
0
z (110)
=4πσωR4
3z (111)
Replace the charge density in (111) with that from (100) to get the final expression for ~Bout.
Now begins the calculation of the total energy in the fields. The electric field is zero inside the shellso it contributes nothing to the total energy. The electric field outside the shell contributes,
Ue,out =∫
εoE2
2dτ (112)
=εo
2
∫ ∞
R
∫ π
0
∫ 2π
0
(e
4πεor2
)2
r2 sin θ dr dθ dφ (113)
=e2
32π2εo(4π)
∫ ∞
R
dr
r2(114)
=e2
8πεo
(−1
r
)∞R
(115)
=e2
8πεoR(116)
On to the magnetic field energy inside the shell. Since the field inside the shell is uniform the integralmay be skipped. The energy inside the shell is simply the magnetic energy density multiplied bythe interior volume.
Um,in = um,in ·43πR3 (117)
=1
2µo
(µoωe
6πR
)2· 43πR3 (118)
=µoω
2e2R
54π(119)
Calculating the energy stored in the magnetic field outside of the shell will illustrate why I wrotethe dipole field in terms of spherical coordinates. Once again, recall that B2 = ~B · ~B.
Um,out =∫ ∞
R
∫ π
0
∫ 2π
0
12µo
(µ2
oω2e2R4
144π2r6
)(4 cos2 θ + sin2 θ)r2 sin θ dr dθ dφ (120)
=µoω
2e2R4
288π2(2π)
∫ ∞
R
∫ π
0
1r4
(4 cos2 θ + sin2 θ) sin θ dr dθ (121)
=µoω
2e2R4
144π
(− 1
3r3
)∞R
∫ π
0(4 cos2 θ + sin2 θ) sin θ dθ (122)
=µoω
2e2R
432π
∫ π
0(4 cos2 θ + sin2 θ) sin θ dθ (123)
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,∫ π
0(4 cos2 θ + sin2 θ) sin θ dθ =
∫ π
0(4− 3 sin2 θ) sin θ dθ (124)
= 4(2)− 3∫ π
0sin3 θ dθ (125)
= 8− 3[−1
3cos θ
(sin2 θ + 2
)]π
0
(126)
= 8− 4 = 4 (127)
Returning to the energy expression we have,
Um,out =µoω
2e2R
432π· 4 (128)
=µoω
2e2R
108π(129)
The total energy in the fields is the sum of (116), (119), and (129).
Utot =e2
8πεoR+
µoω2e2R
54π+
µoω2e2R
108π(130)
=e2
8πεoR+
µoω2e2R
36π(131)
(b) The angular momentum stored in the fields is found using (70). The zero electric field inside theshell means that we only need be concerned with the region outside of the shell. All of the fieldshave already been found, so we begin by calculating the angular momentum density in the fieldsand then we’ll integrate that over the volume outside of the shell.
~lem = εo
[~r × ( ~Eout × ~Bout)
](132)
= εo
[~r ×
(e
4πεor2r × µoωeR2
12πr3
(2 cos θ r + sin θ θ
))](133)
= εo
[~r ×
(e
4πεor2· µoωeR2
12πr3sin θ φ
)](134)
=µoωe2R2
48π2r5sin θ (−r θ) (135)
= −µoωe2R2
48π2r4sin θ θ (136)
15
On to the total,
~Lem =∫ ∞
R
∫ π
0
∫ 2π
0−µoωe2R2
48π2r4sin θ θ r2 sin θ dr dθ dφ (137)
= −µoωe2R2
48π2(2π)
∫ ∞
R
∫ π
0
sin2 θ
r2dr dθ θ (138)
= −µoωe2R2
24π
(−1
r
)∞R
∫ π
0sin2 θ dθ θ (139)
= −µoωe2R
24π
∫ π
0sin2 θ dθ θ (140)
The integral in (140) is not as easy as it looks. The θ vector changes with the value of θ. Integratingover θ from 0 to π means the θ can be replaced with its z component. This is a mathematical factand you will be able to find it elsewhere (proven rigorously). The z component of θ is (− sin θ).
−µoωe2R
24π
∫ π
0sin2 θ dθ θ = −µoωe2R
24π
∫ π
0sin2 θ dθ (− sin θ z) (141)
=µoωe2R
24π
∫ π
0sin3 θ dθ z (142)
The solution to the integral is shown in (125) and (126).
~Lem =µoωe2R
18πz (143)
(c) We can use the angular momentum relation to solve for the product, ωR, immediately.
Lem =~2
(144)
µoωe2R
18π=
~2
(145)
ωR =18π~2µoe2
(146)
=9π(1.05× 10−34)
(4π × 10−7)(1.60× 10−19)2(147)
= 9.23× 1010 (148)
This product represents the physical speed (in m/s) of a point on the equator of the shell. It isconsiderably faster than the speed of light and therefore makes no sense physically, demonstratingthe 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-ment, Uem = mec
2. 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 atension 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 stringis massless. Find the amplitudes and phases of the reflected and transmitted waves.
Solution
(a) Begin with,∂f
∂z
∣∣∣∣0−
=∂f
∂z
∣∣∣∣0+
Eq. 9.27 (151)
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 ∼= T
(∂f
∂z
∣∣∣∣+
− ∂f
∂z
∣∣∣∣−
)(152)
The above equation is considered at the point z = 0, the location of the knot. Griffiths arrives atequation 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 tosetting ∆F = ma, which for a one dimensional case becomes,
m∂2f
∂t2= T
(∂f
∂z
∣∣∣∣+
− ∂f
∂z
∣∣∣∣−
)(153)
all of which is evaluated at z = 0. Recall that the function describing the string, f , represents itsposition. 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 andphases of the reflected and transmitted waves in terms of the incident values (which may always beassumed to be given). The first boundary condition is given in (153). The second boundary condi-tion comes from the fact that the rope itself is continuous and therefore requires that the functionsto 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.
f− =∼AI ei(k1z−ωt)+
∼AR ei(−k1z−ωt) (155)
f+ =∼
AT ei(k2z−ωt) (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,
m∂2f
∂t2= −mω2
∼AT e−iωt (157)
recalling that this is evaluated at z = 0. The shortcut method is to use ∂/∂t = −iω when dealingwith waves of this sort.
Computing the right side of (153),
−mω2∼
AT e−iωt = iT (k2
∼AT e−iωt − k1
∼AI e−iωt + k1
∼AR e−iωt) (158)
−mω2∼
AT = iT (k2
∼AT −k1
∼AI +k1
∼AR) (159)
This allows the transmitted amplitude to be written in terms of the other amplitudes as,
(iTk2 + mω2)∼
AT = iTk1(∼AI −
∼AR) (160)
Writing out (154) allows us to simplify it,
∼AI +
∼AR =
∼AT (161)
Multiply (161) by iTk1 and add this to (160),
iTk1 · (∼AI +
∼AR =
∼AT ) (162)
+
(iTk2 + mω2)∼
AT = iTk1(∼AI −
∼AR) (163)
Result :
2iTk1
∼AI =
[iT (k1 + k2) + mω2
] ∼AT (164)
∼AT =
2iTk1
[iT (k1 + k2) + mω2]
∼AI (165)
Putting this expression for∼
AT into (161) and solving for∼
AR gives,
∼AR =
iT (k1 − k2)−mω2
iT (k1 + k2) + mω2
∼AI (166)
Now it is time to make use of the fact that the second string is massless. For waves on strings thevelocity is given by
v =
√T
µEq. 9.3 (167)
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 velocitiesby,
k2
k1=
v1
v2Eq. 9.24 (168)
which leads to k2/k1 = 0 in this case.
Return to the expressions for the transmitted and reflected amplitudes in terms of the incident am-plitude and factor k1 out of the denominators. This allows those expressions to simplify to,
∼AT =
21− imω2
k1T
∼AI (169)
∼AR =
1 + imω2
k1T
1− imω2
k1T
∼AI (170)
Separate the real amplitude from the phase of the waves as follows,
AT eiδT =2
1− imω2
k1T
AIeiδI (171)
AReiδR =1 + imω2
k1T
1− imω2
k1T
AIeiδI (172)
Thus concludes the setup part of this problem. From this point forward it is all algebra. One methodis the following,
AT eiδT
AIeiδI=
21− imω2
k1T
(173)
AT eiδT
AReiδR=
1 + imω2
k1T
1− imω2
k1T
(174)
If you square both sides of the above equations you will be able to separate out the ratio of theamplitudes from a relation between the phases. Coupling this with the following identity,
tanφ =ei2φ − 1
i (ei2φ + 1)(175)
will allow you to solve for the desired values in terms of the incident parameters.
After some algebra, the real amplitudes and phases are,
AT =2√
1 + m2ω4
k21T 2
AI (176)
AR = AI (177)
δT = δI + tan−1 mω2
k1T(178)
δR = δI + tan−12mω2
k1T
1− m2ω4
k21T 2
(179)
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[7.] Problem by Professor Carter
A clever cub scout who has studied E&M decides he wants to cheat in the Pinewood derby raceby 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 figureout is his scheme will help him, calculate the length of time necessary for the laser to accelerate his0.1 kg car to a speed of 1 m/s. (Note that if the reflector isn’t perfect, then the car will be destroyedlong before reaching this speed.)
Solution
This is a problem in kinetics. The laser will exert a force on the car that is described according tothe radiation pressure of the laser. Since the laser is on continuously, the force on the car will beconstant. As such, the car’s acceleration due to the laser will also be constant. The time dependenceof 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 muststart from rest). The final velocity is v = 1.
Pressure, which is force (F ) per unit area (A), is given by,
P =F
A=
2I
cEq. 9.64 (181)
where I is the intensity of the incident light and we need to rewrite this to solve for the force on thecar due to the laser. The factor of 2, which is not actually written in Griffiths’ form of the equation,
F = PA (182)
=2IA
c(183)
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,
F =(
PL
A
)2A
c(184)
=2PL
c(185)
=2× 104
3× 108(186)
ma = 0.66× 10−4 (187)
a =0.66× 10−4
0.1(188)
= 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 unlikelythe 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 thez direction and polarized in the x 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 (193)
~B(z, t) =1cEo cos(kz − ωt + δ) y (194)
where the direction of propagation for this wave is given according to k = E × B = z
The term ~∇ · ~S is,
~∇ · ~S = ~∇ · 1µo
~E × ~B (195)
= ~∇ · E2o
µoccos2(kz − ωt + δ) z (196)
=E2
o
µoc
∂
∂zcos2(kz − ωt + δ) (197)
=−2kE2
o
µoccos(kz − ωt + δ) sin(kz − ωt + δ) (198)
The term ∂Uem is,
∂Uem
∂t=
∂
∂t
[εoE
2]
(199)
=∂
∂t
[εoE
2o cos2(kz − ωt + δ)
](200)
= εoE2o (2ω cos(kz − ωt + δ) sin(kz − ωt + δ)) (201)
= 2εoωE2o 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),
~∇ · ~S∂Uem
∂t
=−2kE2
oµoc cos(kz − ωt + δ) sin(kz − ωt + δ)
2εoωE2o cos(kz − ωt + δ) sin(kz − ωt + δ)
(203)
=−k
µoεoωc(204)
Now we must recall the following expressions relating k, ω, and the speed of light.
ω
k= c (205)
1µoεo
= c2 (206)
Continuing on,
~∇ · ~S∂Uem
∂t
=−c2
c2(207)
= −1 (208)
These terms are opposite in magnitude. As the energy in an EM wave decreases in time, the amountof energy flux away from the present location increases. This is another demonstration of energyconservation: whatever is lost within a system must have been radiated away. The topic of radiationarrives soon.
[9.] Problem 9.12 from Griffiths
Find the elements of the Maxwell stress tensor for a monochromatic plane wave (same one from theprevious problem, given by (193) and (194)). Is this the answer you expected? How is the Maxwellstress 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 ofTij . Most of the components can be immediately seen to be zero. The remaining terms are,
↔T = cos2(kz − ωt + δ)
εo
(E2
x − E2
2
)− B2
2µo0 0
0 − εoE2
2 + 1µo
(B2
y − B2
2
)0
0 0 − εoE2
2 − B2
2µo
(209)
where I have factored out the common cosine term.
In this problem E2x = E2 and B2
y = B2, which further simplifies this expression to,
↔T = cos2(kz − ωt + δ)
εoE2
2 − B2
2µo0 0
0 − εoE2
2 + B2
2µo0
0 0 − εoE2
2 − B2
2µo
(210)
22
Now recall from the previous work on electromagnetic energy that (see Eq. 9.54 for a reminder),
εoE2
2=
B2
2µo(211)
This makes two more of the terms in (210) zero,
↔T = cos2(kz − ωt + δ)
0 0 0
0 0 0
0 0 − εoE2
2 − B2
2µo
(212)
The Maxwell stress tensor is the negative of the energy density. For EM waves (light waves) theenergy density is completely manifested in the momentum density.
23