PHYS 110B - HW #4Spring 2004, Solutions by David PaceAny referenced equations 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 magneticfields 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 powerflowing into the gap by integrating the Poynting vector over the relevant surface. Verifythat the input power is equivalent to the rate of increasing energy in the gap. (GriffithsHint: This verification amounts to proving the validity of equation 8.9 in the case whereW = 0.)

Solution

(a) The electric field between the plates of a parallel plate capacitor is known to be (seeexample 2.5 in Griffiths),

~E =σ

εo

z (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.The resultant charge density on each “plate” is then time-dependent because the flowingcurrent causes charge to pile up. At time zero there is no charge on the plates, so we knowthat the charge density increases 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.

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 (4)

where the enclosed current may be due to the displacement term; as it is in this problem.

The displacement current through the gap must be in the z direction (the only proper cur-rent flow is through the wire). This, coupled with using the displacement current leadsto,

Bφ(2πs) = µoId,enc (5)

1

Solving for the displacement current gives,

Id,enc =

∫~Jd · d~a (6)

=

∫ (εo

∂ ~E

∂t

)· d~a (7)

=

∫ s

0

∫ 2π

0

(εo

I

εoπa2z

)· s ds dφ z (8)

=Is2

a2(9)

Now the magnetic field is found from (5),

~B =µoIs

2πa2φ (10)

(b) The energy density is given in terms of the fields in the gap,

uem =1

2

(εoE

2 +B2

µo

)Eq. 8.13 (11)

=1

2

(εo

I2t2

ε2oπ

2a4+

1

µo

µ2oI

2s2

4π2a4

)(12)

=1

2

(I2

π2a4

)[t2

εo

+µos

2

4

](13)

noting that E2 = ~E · ~E.

The Poynting vector in the gap is,

~S =1

µo

~E × ~B Eq. 8.10 (14)

=1

µo

(It

εoπa2z × µoIs

2πa2φ

)(15)

= − I2ts

2εoπ2a4s (16)

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

Verification of equation 8.14 follows (umech = 0),

∂

∂t(umech + uem) = −~∇ · ~S (17)

∂

∂t

(I2

2π2a4

)[t2

εo

+µos

2

4

]= −1

s

∂

∂s

(s−I2ts

2εoπ2a4

)(18)

2

I2t

εoπ2a4= −1

s

(− I2ts

εoπ2a4

)(19)

=I2t

εoπ2a4(20)

(c) Since we have the energy density in the gap we are ready to determine the total energy.

Uem =1

2

∫ (εoE

2 +B2

µo

)dτ Eq. 8.5 (21)

=1

2

∫ a

0

∫ 2π

0

∫ w

0

(I2

π2a4

)[t2

εo

+µos

2

4

]s ds dφ dz (22)

=I2

2π2a4(2πw)

∫ a

0

[t2s

εo

+µos

3

4

]ds (23)

=I2w

πa4

[t2a2

2εo

+µoa

4

16

](24)

=I2w

2πa2

[t2

εo

+µoa

2

8

](25)

The problem tells us to determine the total power flowing into the gap by integrating thePoynting vector over the surface enclosing it. This is the cylindrical surface occurring ats = a. Technically, this also includes the circular surfaces at each of the plates, but for thesesurfaces 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 (26)

=

∫ 2π

0

∫ w

0

− I2ts

2εoπ2a4s dφ dz (27)

= − I2ta2

2εoπ2a4(2wπ) (28)

= − I2tw

εoπa2(29)

Finally, we essentially want to verify equation 8.9,

dW

dt= − d

dt

∫V

1

2

(εoE

2 +1

µo

B2

)dτ − 1

µo

∮S

(~E × ~B

)· d~a (30)

where this represents the rate at which work is done on a collection of charges in the vol-ume V , that is enclosed by the surface S.

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Since there are no charges in the gap, W = 0 and dW/dt = 0. The equation to verify is,

d

dt

∫V

1

2

(εoE

2 +1

µo

B2

)dτ = − 1

µo

∮S

(~E × ~B

)· d~a (31)

d

dtUem = −

∮S

~S · d~a (32)

we have solved for Uem in (25), and the integral on the right hand side of the above equa-tion is the power flowing into the gap, given by (29). Continuing with the verification ofequation 8.9,

d

dt

I2w

2πa2

[t2

εo

+µoa

2

8

]= −

(− I2tw

εoπa2

)(33)

I2tw

εoπa2=

I2tw

εoπa2(34)

this equation is verified.

[2.] Problem 8.4 from Griffiths

(a) Consider two point charges, of equal charge q, that are separated by a distance 2a.Integrate the Maxwell stress tensor for this system over the plane separating these chargesin order to determine the force of one on the other.

(b) Repeat (a) for charges of opposite sign.

Solution

(a) Generally, the force on charges in some volume V is,

~F =

∮S

↔T · d~a− εoµo

d

dt

∫V

~S dτ Eq. 8.22 (35)

In this problem there is no time dependence. The term involving the time derivative of thePoynting vector is zero.

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

↔T = εo

E2

x − E2

2ExEy ExEz

EyEx E2y − E2

2EyEz

EzEx EzEy E2z − E2

2

(36)

The expression in (36) was given directly in lecture on 5-5-2004, but it may also be derivedfrom the general form of the stress tensor.

Tij ≡ εo

(EiEj −

1

2δijE

2

)+

1

µo

(BiBj −

1

2δijB

2

)Eq. 8.19 (37)

4

Label the charges 1 and 2. I will draw the enclosed region around charge 2 and determinethe force that 1 exerts on it. Setting the origin of this system on charge 1, the electric fielddue to this charge is given by,

~E1 =q

4πεor2r (38)

Writing this in terms of Cartesian coordinates, and setting x = −a since we are only inter-ested in the electric field along the plane between the charges gives,

~E1 =q

4πεo

(1

(−a)2 + y2 + z2

)(−ax + yy + zz√(−a)2 + y2 + z2

)(39)

=q

4πεo

(1

a2 + y2 + z2

)(−ax + yy + zz√

a2 + y2 + z2

)(40)

using r =√

a2 + y2 + z2 as the distance from charge 1 to the plane. The −ax term comesfrom my setting the charge at the origin and placing the other charge at x = −2a.

The stress tensor depends on the total field in the system. As such, the effects of the secondcharge must be included. The symmetry in this problem allows the electric field due to thesecond charge to be written immediately after a translation in the x coordinate. The electricfield of charge 2 is the same as that due to charge 1, except that x → x + 2a.

~E2 =q

4πεo

(1

(−a + 2a)2 + y2 + z2

)((−a + 2a)x + yy + zz√

(−a + 2a)2 + y2 + z2

)(41)

=q

4πεo

(1

a2 + y2 + z2

)(ax + yy + zz√

a2 + y2 + z2

)(42)

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

~Etot = ~E1 + ~E2 (43)

=q

4πεo

(1

(a2 + y2 + z2)3/2

)[(−a + a)x + (y + y)y + (z + z)z] (44)

=q

2πεo

(1

(a2 + y2 + z2)3/2

)[yy + zz] (45)

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

Ex = 0 (46)

Ey =q

2πεo

(y

(a2 + y2 + z2)3/2

)(47)

Ez =q

2πεo

(z

(a2 + y2 + z2)3/2

)(48)

E2 =q2

4π2ε2o

(y2 + z2

(a2 + y2 + z2)3

)(49)

5

Referring back to (35), the force on charge 2 due to 1 may be written,

~F =

∫ ∞

−∞

∫ ∞

−∞

↔T ·dy dz x (50)

where the x represents the direction of the surface normal vector from the surface enclosingcharge 2.

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

↔T · x = εo

−E2

20 0

0 E2y − E2

2EyEz

0 EzEy E2z − E2

2

1

00

(51)

= εo

[−E2

2x + 0y + 0z

](52)

= −εo

2· q2

4π2ε2o

(y2 + z2

(a2 + y2 + z2)3

)x (53)

= − q2

8π2εo

(y2 + z2

(a2 + y2 + z2)3

)x (54)

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

~F =

∫ ∞

−∞

∫ ∞

−∞− q2

8π2εo

(y2 + z2

(a2 + y2 + z2)3

)dy dz x (55)

and it is seen that the final answer will put the force in either the positive or negative x. Ifthis were not the case, then we would already know our solution to be incorrect.

The integral in (55) is non-trivial. Since there is no x dependence in the expression for theforce along the plane we can use change of coordinate system to obtain a better integralwith which to work. Consider the yz plane to be in cylindrical coordinates. If we lets =

√y2 + z2 and then incorporate the φ direction into the integral we get,

~F =

∫ ∞

−∞

∫ ∞

−∞− q2

8π2εo

(y2 + z2

(a2 + y2 + z2)3

)dy dz x (56)

=

∫ ∞

0

∫ 2π

0

− q2

8π2εo

(s2

(a2 + s2)3

)s ds dφ x (57)

where the new integral is the result of starting over with the geometry considerations andnot a mathematical change of variable. Notice that there is an extra factor of s that comesfrom the da term in cylindrical coordinates. The x dependence is left alone because it camefrom the tensor work and the change to cylindrical coordinates was made after solving thispart.

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

6

∫ ∞

0

s3

(a2 + s2)3ds → Let u = s2 du = 2s ds (58)

=

∫ ∞

0

us

(a2 + u)3

du

2s(59)

=1

2

∫ ∞

0

u

(a2 + u)3du (60)

=1

2(−1)

[u

(u + a2)2+

a2

2(a2 + u)2

]∞0

(61)

= −1

2

[0 + 0− 0− a2

2a4

](62)

=1

4a2(63)

The integral over φ results in a factor of 2π and the final solution is,

~F = − q2

8π2εo

(2π)1

4a2x (64)

= − q2

4πεo(2a)2x (65)

where this is known to be the correct answer because it represents the force on charge 2due to the similarly charged particle 1 that is a distance 2a away. For the setup I used, the−x represents the direction of a repulsive force.

(b) Now we let the particles have opposite charge. We know what answer to expect, thesame as in (65) but with opposite sign since these particles will be attracted to each other.

Keep the same setup as in part (a) and use (40) as the electric field due to charge 1. Letcharge 2 change to a negative charge. This means the electric field of charge 2 (in thegeometry where charge 1 is placed at the origin) is given simply by changing the sign on(42),

~E2 = − q

4πεo

(1

a2 + y2 + z2

)(ax + yy + zz√

a2 + y2 + z2

)(66)

Now comes the big change. The total electric field is completely different from part (a),

~Etot = ~E1 + ~E2 (67)

=q

4πεo(a2 + y2 + z2)3/2(−2ax) (68)

because the y and z component terms cancel out due to the new ~E2.

7

Taking into account Ey = Ez = 0 and E2x = E2, the stress tensor is,

↔T = εo

E2

20 0

0 −E2

20

0 0 −E2

2

(69)

The normal vector to the surface enclosing charge 2 is still in the x direction because thatpoints outward from the enclosed volume. The dot product of the stress tensor and x is,

↔T · x =

εoE2

2x (70)

=(εo

2

) 4a2q2

16π2ε2o(a

2 + y2 + z2)3x (71)

=a2q2

8π2εo(a2 + y2 + z2)3x (72)

and we should be encouraged to see that so far our solution is at least in the correct direc-tion.

The force is,

~F =

∫ ∞

−∞

∫ ∞

−∞

a2q2

8π2εo(a2 + y2 + z2)3dy dz (73)

Moving to cylindrical coordinates provides,

~F =

∫ ∞

0

∫ 2π

0

a2q2

8π2εo(a2 + s2)3s ds dφ x (74)

=a2q2

4πεo

∫ ∞

0

s

(a2 + s2)3ds x (75)

I use, ∫x

(f + x2)3dx = − 1

4(f + x2)2(76)

∫ ∞

0

s

(a2 + s2)3ds = − 1

4(a2 + s2)2

∣∣∣∣∞0

(77)

=1

4a4(78)

Leading to the solution for the force on charge 2,

8

~F =a2q2

4πεo

· 1

4a4x (79)

=q2

16πεoa2x (80)

=q2

4πεo(2a)2x (81)

This is the force we expected all along.

[3.] Problem 8.6 from Griffiths

A parallel plate capacitor is shown in figure 8.6 in Griffiths. We may assume the electricfield between the plates of this capacitor is ~E = Ez, and that it is placed in a uniformbackground magnetic field of ~B = Bx.

(a) Find the EM momentum between the plates.

(b) Connect the plates with a resistive wire stretched along the z-axis. As the capacitordischarges the current in the wire will experience a force due to the background magneticfield. Find the total impulse delivered to the system during the discharge.

(c) Consider an alternative to the situation in part (a). Slowly turn off the magnetic fieldand let the induced electric field exert a force on the plates. Show that the total impulsedelivered to the system is equal to the momentum originally stored in the electromagneticfields.

Solution

(a) The momentum stored in the EM fields of a particular volume is,

~pem = µoεo

∫V

~Sdτ Eq. 8.29 (82)

The fields have been given to us and they are uniform. As such, the integral simplifies tothe total volume contained within the plates.

~pem = µoεoAd~S (83)

= εoAd (Ez ×Bx) (84)

= εoAEBd y (85)

where A is the given area of each plate and d is the distance between them (keep track of dso that you don’t mistakenly think it represents a differential element of something).

(b) The impulse delivered to a system is the force applied to it through time.

Imp =

∫~Fdt (86)

9

For a current flowing in a wire we know the force is ~F = ~I× ~B. The capacitor is dischargingso the current is related to the charge on the plates by I = −dQ/dt. Putting this togetherallows for the impulse to be calculated,

Imp =

∫ ∞

0

(~I × ~B)dt (87)

=

∫ ∞

0

I(dz ×Bx)dt (88)

= Bd

∫ ∞

0

(−dQ

dt

)dt y (89)

= −Bd(Q)∞0 y (90)

= BdQo y (91)

where Q = Q(t).

The final charge on the plates is zero because eventually the capacitor is completely dis-charged. The initial charge on the plates can be determined.

Qo = σ · A (92)

= εoEA (93)

The charge density on the plates for a parallel plate capacitor is known from the relationE = σ/εo.

The total impulse delivered to the capacitor is,

Imp = εoAEBd y (94)

which is equivalent to the total momentum originally stored in the fields.

(c) We need to determine the force for which to take our time integral. The problem state-ment tells us to solve for the induced electric fields so we expect the force to be,

~F = q ~E∗ = εoEA~E∗ (95)

where E is the field given in the problem and ~E∗ is the induced field as the magneticfield turns off. This is the expression for the force on one of the plates. One could simplymultiply this value by 2 to account for the total force on the capacitor (intuitively oneexpects that the electric fields will be directed oppositely on the plates, but their charge isalso opposite so they experience a force in the same direction). In the following I will leaveeverything general; skipping over this ability to simplify the problem.

Solve for the induced electric field using Faraday’s law. The closed loop is a rectangle oflength l and height d. This lengths of this loop lie on the plates and therefore will provide

10

the induced electric field that exerts a force on them.∮~E∗ · d~l = −dΦ

dt(96)

E∗(2l) = −lddB

dt(97)

E∗ =d

2

dB

dt(98)

The vector nature of E∗ is neglected here because the geometry shows that it will be di-rected in the ±y depending on which plate we consider.

In the Griffiths’ figure the bottom plate is shown to have the positive charge density (theelectric field points outward from this plate). The induced electric field is directed to the+y on this plate. Writing out the total force on this capacitor as the sum of the forces oneach plate,

~F = ~Fbot + ~Ftop (99)

= −εoEA

(d

2

dB

dty +

−d

2

dB

dt(−y)

)(100)

= −εoEAddB

dty (101)

where the initial negative sign comes from Faraday’s law.

Now the impulse may be calculated,

Imp =

∫ ∞

0

−εoEAddB

dtdt y (102)

= −εoEAd(−Bo) y (103)

= εoEABd y (104)

because the initial magnetic field is given in the problem. This impulse correctly agreeswith the original momentum stored in the fields, as given by (85).

[4.] Problem by Professor Carter

Consider a cylindrical capacitor of length L with charge +Q on the inner cylinder of radiusa and −Q on the outer cylinder of radius b. The capacitor is filled with a lossless dielectricwith dielectric constant equal to 1. The capacitor is located in a region with a uniformmagnetic field B, which points along the symmetry axis of the cylindrical capacitor. A flawdevelops in the dielectric insulator, and a current flow develops between the two plates ofthe capacitor. Because of the magnetic field, this current flow results in a torque on thecapacitor, which begins to rotate. After the capacitor is fully discharged (total charge onboth plates is no zero), what is the magnitude and direction of the angular velocity of thecapacitor? The moment of inertia of the capacitor (about the axis of symmetry) is I , andyou may ignore fringing fields in the calculation.

11

Reference the diagram provided on the original homework assignment sheet.

Solution

Considering only the conservation of angular momentum this problem is solved quickly.Before the current flow there is a certain amount of angular momentum stored in the EMfields. After the dielectric breaks down the capacitor discharges until there is no longer anelectric field between its plates. Since there is no longer an electric field there is no longerany angular momentum stored in the fields. All of this angular momentum must exist inthe physical rotation of the system.

One could solve this problem considering the force on the current, but that would requireconsiderably more work than the method shown here. Furthermore, the problem state-ment doesn’t say much about the current; is it uniform or localized?

The total angular momentum in the fields is,

~Lem =

∫εo

[~r × ( ~E × ~B)

]dτ (105)

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 isknown to be zero outside the plates. Inside the capacitor we have,

~E =λ

2πεss (106)

where λ is the charge per unit length of the inner cylinder and ε is the dielectric constant ofthe material 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 vol-ume integral in (105) are then decided. Continuing with the solution for the total angularmomentum stored in the fields gives,

~Lem =

∫ b

a

∫ 2π

0

∫ L

0

[~s×

(λ

2πss×Bz

)]s ds dφ dz (107)

=

∫ b

a

∫ 2π

0

∫ L

0

[s s× λB

2πs(−φ)

]s ds dφ dz (108)

=

∫ b

a

∫ 2π

0

∫ L

0

−λBs

2πds dφ dz z (109)

= −λB

2π(2π)(L)

∫ b

a

s ds z (110)

= −λB

2(b2 − a2)L z (111)

= −QB

2(b2 − a2) z (112)

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

~ω =~L

I= −QB

2I(b2 − a2) z (113)

where this provides both the direction and magnitude of the angular velocity.

[5.] Problem 8.9 from Griffiths

Consider a very long solenoid. This solenoid has radius a, turns per unit length n, and acurrent 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 currentin 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 Poynt-ing vector just outside the solenoid and verify that it is directed toward the loop. (GriffithsHint: Use the electric field due to the changing flux in the solenoid and the magnetic fielddue to the current in the wire loop.) Integrate the Poynting vector over the entire surfaceof the solenoid to verify that the total energy “emitted” by the solenoid is equal to thatdissipated 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 througha resistive wire is Ir = E/R, where E is the emf (voltage) across the wire. The emf iscalculated according to,

E = −dΦ

dt(114)

= − d

dtµonIsπa2 (115)

= −µonπa2dIs

dt(116)

where this takes advantage of the properties of solenoid fields (i.e. the field outside is zeroand the field inside is uniform).

The current in the loop is positive, though the direction depends on how we orient thesolenoid, let ~B → z.

Ir =µonπa2

R

∣∣∣∣dIs

dt

∣∣∣∣ (117)

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

(b) Begin by solving for the fields Griffiths tells us to use. The electric field at the surface ofthe solenoid is found using Faraday’s law, (96). Since we care about the field at the surface

13

of the solenoid we set s = a.

Eφ(2πa) = −dΦ

dt(118)

~E = −µonπa2dIs

dt

1

2πaφ (119)

=µona

2

∣∣∣∣dIs

dt

∣∣∣∣ φ (120)

Notice that the final direction of the electric field is in the positive φ. There is a negativesign in Faraday’s law, but we also know that the time derivative of the solenoid current isnegative since it is being slowly decreased.

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

~B(z) =µoIr

2

b2

(b2 + z2)3/2z Eq. 5.38 (121)

where the direction is set by the orientation of the solenoid.

The Poynting vector may now be calculated,

~S =1

µo

~E × ~B =1

µo

µona

2

∣∣∣∣dIs

dt

∣∣∣∣ φ× µoIr

2

b2

(b2 + z2)3/2z (122)

=µonab2Ir

4(b2 + z2)3/2

∣∣∣∣dIs

dt

∣∣∣∣ s (123)

The Poynting vector is directed toward the resistive loop, as expected. The next step is tointegrate this 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 (124)

=µona2b2Ir

4

∣∣∣∣dIs

dt

∣∣∣∣ (2π)

∫ ∞

−∞

dz

(b2 + z2)3/2(125)

From integral tables, ∫dx

(f + cx2)3/2=

x

f(f + gx2)1/2(126)

Power =µoπna2b2Ir

2

∣∣∣∣dIs

dt

∣∣∣∣ [ z

b2(b2 + z2)1/2

]∞−∞

(127)

=µoπna2b2Ir

2

∣∣∣∣dIs

dt

∣∣∣∣ ( 2

b2

)(128)

= µoπna2Ir

∣∣∣∣dIs

dt

∣∣∣∣ (129)

14

Using (117) to rewrite (129),Power = I2

r R (130)

and the power directed from the solenoid toward the resistive loop is equal to the energydissipated in this loop.

[6.] Problem 8.11 from Griffiths

Treat the electron as a uniformly charged spherical shell (charge e) of radius R, spinningwith 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 itsfields, Uem = mec

2. Furthermore, let the spin angular momentum of the electron be dueentirely to its fields, Lem = ~/2. Solve for the angular velocity and radius of the electronin this case. Calculate the value Rω and comment on whether this value makes senseclassically.

Solution

(a) The total energy in the fields is given by (21). This is simply the addition of the energyin the electric field plus that of the magnetic field, so I will determine Ue and Um indepen-dently. 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 uniformlycharged spherical shell is zero. Also, the field outside is simply that of a point charge.Therefore,

~Ein = 0 (131)

~Eout =e

4πεor2r (132)

The magnetic field inside a uniformly charges spinning spherical shell is given in example5.11 of Griffiths,

~Bin =2

3µoσRω z Eq. 5.68 (133)

This was a remarkable result when we first saw it because the field inside the shell is uni-form. This can be written in terms of the variables given in the problem because we cansolve for the surface charge density,

σ =Qtot

A=

e

4πR2(134)

The magnetic field inside the shell is,

~Bin =µoωe

6πRz (135)

The magnetic field outside of the shell is that of a dipole. This is known from a variety ofsources: HW #9 of PHYS 110A with Prof. Carter, reading problem 5.36 from Griffiths (that

15

was on HW #9), or using the given vector potential from example 5.11 in Griffiths to solvefor the field directly by way of ~B = ~∇× ~A. Regardless of your method, the magnetic fieldoutside the shell is,

~Bout =µom

4πr3

(2 cos θ r + sin θ θ

)(136)

=µoωeR2

12πr3

(2 cos θ r + sin θ θ

)(137)

The magnetic dipole moment was found when this spinning shell was considered in HW#9 mentioned previously. Below is a copy of one of the methods used to find m in thatassignment.

Copied from HW #9 Solutions, PHYS 110A Winter 2003

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

d~m = dI ~a (138)

The differential element on the right hand side of (138) must be for the current because thearea vector of any individual ring is,

~a = πl2z

= π(r sin θ)2z

~a = πr2 sin2 θz (139)

where l is the radius of the ring. The direction z is determined by the orientation of thesphere and its rotation and can therefore be set to whatever value we want.

To find dI we need to write out the current in 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

d~m = πσωR4 sin3 θ dθ (140)

where I have taken into account the fact that this is a shell so r = R.

16

The magnetic dipole moment is found through an integration of d~m,

~m = πσωR4

∫ π

0

sin3 θ dθ z

= πσωR4

[−1

3cos θ

(sin2 θ + 2

)]π

0

z

~m =4πσωR4

3z (141)

End copied section

Replace the charge density in (141) with that from (134) to get the final expression for ~Bout.

Now begins the calculation of the total energy in the fields. The electric field is zero insidethe shell so it contributes nothing to the total energy. The electric field outside the shellcontributes,

Ue,out =

∫εoE

2

2dτ (142)

=εo

2

∫ ∞

R

∫ π

0

∫ 2π

0

(e

4πεor2

)2

r2 sin θ dr dθ dφ (143)

=e2

32π2εo

(4π)

∫ ∞

R

dr

r2(144)

=e2

8πεo

(−1

r

)∞R

(145)

=e2

8πεoR(146)

On to the magnetic field energy inside the shell. Since the field inside the shell is uniformthe integral may be skipped. The energy inside the shell is simply the magnetic energydensity multiplied by the interior volume.

Um,in = um,in ·4

3πR3 (147)

=1

2µo

(µoωe

6πR

)2

· 4

3πR3 (148)

=µoω

2e2R

54π(149)

Calculating the energy stored in the magnetic field outside of the shell will illustrate why I

17

wrote the dipole field in terms of spherical coordinates. Once again, recall that B2 = ~B · ~B.

Um,out =

∫ ∞

R

∫ π

0

∫ 2π

0

1

2µo

(µ2

oω2e2R4

144π2r6

)(4 cos2 θ + sin2 θ)r2 sin θ dr dθ dφ (150)

=µoω

2e2R4

288π2(2π)

∫ ∞

R

∫ π

0

1

r4(4 cos2 θ + sin2 θ) sin θ dr dθ (151)

=µoω

2e2R4

144π

(− 1

3r3

)∞R

∫ π

0

(4 cos2 θ + sin2 θ) sin θ dθ (152)

=µoω

2e2R

432π

∫ π

0

(4 cos2 θ + sin2 θ) sin θ dθ (153)

This is another integral that you may solve in any manner. Here I make the substitution4 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 (154)

= 4(2)− 3

∫ π

0

sin3 θ dθ (155)

= 8− 3

[−1

3cos θ

(sin2 θ + 2

)]π

0

(156)

= 8− 4 = 4 (157)

Returning to the energy expression we have,

Um,out =µoω

2e2R

432π· 4 (158)

=µoω

2e2R

108π(159)

The total energy in the fields is the sum of (146), (149), and (159).

Utot =e2

8πεoR+

µoω2e2R

54π+

µoω2e2R

108π(160)

=e2

8πεoR+

µoω2e2R

36π(161)

(b) The angular momentum stored in the fields is found using (105). The zero electric fieldinside 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 finding the angular momentumdensity in the fields and then we’ll integrate that over the volume outside of the shell.

18

~lem = εo

[~r × ( ~Eout × ~Bout)

](162)

= εo

[~r ×

(e

4πεor2r × µoωeR2

12πr3

(2 cos θ r + sin θ θ

))](163)

= εo

[~r ×

(e

4πεor2· µoωeR2

12πr3sin θ φ

)](164)

=µoωe2R2

48π2r5sin θ (−r θ) (165)

= −µoωe2R2

48π2r4sin θ θ (166)

On to the total,

~Lem =

∫ ∞

R

∫ π

0

∫ 2π

0

−µoωe2R2

48π2r4sin θ θ r2 sin θ dr dθ dφ (167)

= −µoωe2R2

48π2(2π)

∫ ∞

R

∫ π

0

sin2 θ

r2dr dθ θ (168)

= −µoωe2R2

24π

(−1

r

)∞R

∫ π

0

sin2 θ dθ θ (169)

= −µoωe2R

24π

∫ π

0

sin2 θ dθ θ (170)

The integral in (170) is not as easy as it looks. The θ vector changes with the value of θ.Integrating over θ from 0 to π means the θ can be replaced with its z component. This wasshown graphically in Discussion on 7-5-2004, but it is a mathematical fact and you will beable to find it elsewhere (proven rigorously). The z component of θ is − sin θ.

−µoωe2R

24π

∫ π

0

sin2 θ dθ θ = −µoωe2R

24π

∫ π

0

sin2 θ dθ (− sin θ z) (171)

=µoωe2R

24π

∫ π

0

sin3 θ dθ z (172)

The solution to the integral is shown in (155) and (156).

~Lem =µoωe2R

18πz (173)

(c) We can use the angular momentum relation to solve for the product, ωR, immediately.

Lem =~2

(174)

19

µoωe2R

18π=

~2

(175)

ωR =18π~2µoe2

(176)

=9π(1.05× 10−34)

(4π × 10−7)(1.60× 10−19)2(177)

= 9.23× 1010 (178)

This product represents the physical speed of a point on the equator of the shell. It is con-siderably faster than the speed of light and therefore makes no sense physically, demon-strating the need for quantum mechanics in the explanation of various properties of theelectron.

To solve for the values independently use (178) and the mass relation given in the problemstatement. This provides two equations with which you can solve for the two unknowns.The numerical values are approximately:

R = 2.96× 10−11 m (179)

ω = 3.12× 1021 s−1 (180)

[7.] Problem 9.6 from Griffiths

(a) Write a revised boundary condition (replacing equation 9.27 from Griffiths) for the caseof 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 secondstring is 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 (181)

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 9.27 on page 365 of Griffiths(also presented in lecture on 10-5-2004),

∆F ∼= T

(∂f

∂z

∣∣∣∣+

− ∂f

∂z

∣∣∣∣−

)(182)

The above equation is considered at the point z = 0 in the case of a knot being there.Griffiths arrives at equation 9.27 by taking the left side of (182) to be zero because the massof the knot is set to zero. In part (a) of this problem we are asked to consider a knot of

20

some mass, m. This is equivalent to setting ∆F = ma, which for a one dimensional casebecomes,

m∂2f

∂t2= T

(∂f

∂z

∣∣∣∣+

− ∂f

∂z

∣∣∣∣−

)(183)

all of which is evaluated at z = 0.

(b) This part is a boundary value problem. We will use two equations to solve for theamplitudes and phases of the reflected and transmitted waves in terms of the incidentvalues (which may always be assumed to be given). The first boundary condition is solvedfor in (183). The second boundary condition comes from the fact that the rope itself iscontinuous and therefore requires that the functions to the left and right of the knot beequal at z = 0.

f(0−, t) = f(0+, t) Eq. 9.26 (184)

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

f− =∼AI ei(k1z−ωt)+

∼AR ei(−k1z−ωt) (185)

f+ =∼

AT ei(k2z−ωt) (186)

where∼AI ,

∼AR, and

∼AT refer to the complex amplitudes of the incident (coming from the

left), reflected, and transmitted waves respectively.

Condition (184) says that we can use either wave function for the time derivative term in(183). Using f+ we have,

m∂2f

∂t2= −mω2

∼AT e−iωt (187)

recalling that this is evaluated at z = 0. The shortcut method is to use ∂/∂t = −iω whendealing with waves of this sort.

Computing the right side of (183),

−mω2∼

AT e−iωt = iT (k2

∼AT e−iωt − k1

∼AI e−iωt + k1

∼AR e−iωt) (188)

−mω2∼

AT = iT (k2

∼AT −k1

∼AI +k1

∼AR) (189)

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

(iTk2 + mω2)∼

AT = iTk1(∼AI −

∼AR) (190)

Writing out (184) allows us to simplify it,

∼AI +

∼AR=

∼AT (191)

21

Multiply (191) by iTk1 and add this to (190),

iTk1 · (∼AI +

∼AR =

∼AT ) (192)

+

(iTk2 + mω2)∼

AT = iTk1(∼AI −

∼AR) (193)

Result :

2iTk1

∼AI =

[iT (k1 + k2) + mω2

] ∼AT (194)

∼AT =

2iTk1

[iT (k1 + k2) + mω2]

∼AI (195)

Putting this expression for∼

AT into (191) and solving for∼

AR gives,∼

AR =iT (k1 − k2)−mω2

iT (k1 + k2) + mω2

∼AI (196)

Now it is time to make use of the fact that the second string is massless. For waves onstrings the velocity is given by

v =

√T

µEq. 9.3 (197)

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 thevelocities by,

k2

k1

=v1

v2

Eq. 9.24 (198)

which leads to k2/k1 = 0 in this case.

Return to the expressions for the transmitted and reflected amplitudes in terms of the in-cident amplitude and factor k1 out of the denominators. This allows those expressions tosimplify to,

∼AT =

2

1− imω2

k1T

∼AI (199)

∼AR =

1 + imω2

k1T

1− imω2

k1T

∼AI (200)

Separate the real amplitude from the phase of the waves as follows,

AT eiδT =2

1− imω2

k1T

AIeiδI (201)

AReiδR =1 + imω2

k1T

1− imω2

k1T

AIeiδI (202)

22

Thus concludes the setup part of this problem. From this point forward it is all algebra.One method is the following,

AT eiδT

AIeiδI=

2

1− imω2

k1T

(203)

AT eiδT

AReiδR=

1 + imω2

k1T

1− imω2

k1T

(204)

If you square both sides of the above equations you will be able to separate out the ratioof the amplitudes from a relation between the phases. Coupling this with the followingidentity,

tan φ =ei2φ − 1

i (ei2φ + 1)(205)

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 (206)

AR = AI (207)

δT = δI + tan−1 mω2

k1T(208)

δR = δI + tan−12mω2

k1T

1− m2ω4

k21T 2

(209)

23