MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Physics Department
Physics 8.07: Electromagnetism II
November 15, 2012
Prof. Alan Guth
QUIZ 2
Reformatted to Remove Blank Pages∗
THE FORMULA SHEETS ARE AT THE END OF THE EXAM.
Your Name
∗
Recitation
Problem
Maximum
1
25
2
35
3
20
4
10
5
10
TOTAL
100
Score
A few clarifications that were posted on the blackboard during the quiz are incorporated here into the text.
8.07 QUIZ 2, FALL 2012
p. 2
PROBLEM 1: THE MAGNETIC FIELD OF A SPINNING, UNIFORMLY
CHARGED SPHERE (25 points)
This problem is based on Problem 1 of Problem Set 8.
A uniformly charged solid sphere of radius R carries a total charge Q, and is set
spinning with angular velocity ω about the z axis.
(a) (10 points) What is the magnetic dipole moment of the sphere?
r ) at large
(b) (5 points) Using the dipole approximation, what is the vector potential A(
is a vector, so it is not enough to merely specify its
distances? (Remember that A
magnitude.)
(c) (10 points) Find the exact vector potential INSIDE the sphere. You may, if you wish,
make use of the result of Example 5.11 from Griffiths’ book. There he considered a
spherical shell, of radius R, carrying a uniform surface charge σ, spinning at angular
velocity ω
directed along the z axis. He found the vector potential
 µ Rωσ
0

r sin θ φˆ , (if r ≤ R)

3
θ, φ) =
A(r,
4

 µ0 R ωσ sin θ φˆ , (if r ≥ R) .
3
r2
(1.1)
PROBLEM 2: SPHERE WITH VARIABLE DIELECTRIC CONSTANT (35
points)
A dielectric sphere of radius R has variable permittivity, so the permittivity throughout
space is described by
0 (R/r)2 if r < R
(r ) =
(2.1)
if r > R .
0 ,
There are no free charges anywhere in this problem. The sphere is embedded in a constant
= E0 ẑ, which means that V (r ) ≈ −E0 r cos θ for r R.
external electric field E
(a) (9 points) Show that V (r ) obeys the differential equation
∇2 V +
d ln ∂V
=0
dr ∂r
(2.2)
for all r , as a consequence of the laws of electrostatics.
(b) (4 points) Explain why the solution can be written as
V (r, θ) =
∞
=0
V (r){ ẑi1 . . . ẑi } r̂i1 . . . r̂i ,
(2.3a)
8.07 QUIZ 2, FALL 2012
p. 3
or equivalently (your choice)
V (r, θ) =
∞
V (r)P (cos θ) ,
(2.3b)
=0
where { . . . } denotes the traceless symmetric part of . . . , and P (cos θ) is the Legendre polynomial. (Your answer here should depend only on general mathematical
principles, and should not rely on the explicit solution that you will find in parts (c)
and (d).)
(c) (9 points) Derive the ordinary differential equation obeyed by V (r) (separately for
r < R and r > R) and give its two independent solutions in each region. Hint: they
are powers of r. You may want to know that
d
dP (cos θ)
sin θ
(2.4)
= −( + 1) sin θP (cos θ) .
dθ
dθ
The relevant formulas for the traceless symmetric tensor formalism are in the formula
sheets.
(d) (9 points) Using appropriate boundary conditions on V (r, θ) at r = 0, r = R, and
r → ∞, determine V (r, θ) for r < R and r > R.
(e) (4 points) What is the net dipole moment of the polarized sphere?
PROBLEM 3: PAIR OF MAGNETIC DIPOLES (20 points)
Suppose there are two magnetic dipoles. One has dipole moment m
1 = m0 ẑ and
1
is located at r 1 = + 2 a ẑ; the other has dipole moment m
2 = −m0 zˆ, and is located at
r 2 = − 12 a zˆ.
(a) (10 points) For a point on the z axis at large z, find the leading (in powers of 1/z)
0, z) and the magnetic field B(0,
0, z).
behavior for the vector potential A(0,
(b) (3 points) In the language of monopole ( = 0), dipole ( = 1), quadrupole ( = 2),
octupole ( = 3), etc., what type of field is produced at large distances by this
current configuration? In future parts, the answer to this question will be called a
whatapole.
(c) (3 points) We can construct an ideal whatapole — a whatapole of zero size — by
taking the limit as a → 0, keeping m0 an fixed, for some power n. What is the correct
value of n?
(d) (4 points) Given the formula for the current density of a dipole,
r δ 3 (r − r d ) ,
Jdip (r ) = −m
×∇
(3.1)
where r d is the position of the dipole, find an expression for the current density
of the whatapole constructed in part (c). Like the above equation, it should be
expressed in terms of δ-functions and/or derivatives of δ-functions, and maybe even
higher derivatives of δ-functions.
8.07 QUIZ 2, FALL 2012
p. 4
PROBLEM 4: UNIFORMLY MAGNETIZED INFINITE CYLINDER (10
points)
Consider a uniformly magnetized infinite circular cylinder, of radius R, with its axis
= M0 ẑ.
coinciding with the z axis. The magnetization inside the cylinder is M
r ) everywhere in space.
(a) (5 points) Find H(
r ) everywhere in space.
(b) (5 points) Find B(
PROBLEM 5: ELECTRIC AND MAGNETIC UNIFORMLY POLARIZED
SPHERES (10 points)
Compare the electric field of a uniformly polarized sphere with the magnetic field of
a uniformly magnetized sphere; in each case the dipole moment per unit volume points
along ẑ. Multiple choice: which of the following is true?
and B
field lines point in the same direction both inside and outside the
(a) The E
spheres.
and B
field lines point in the same direction inside the spheres but in opposite
(b) The E
directions outside.
and B
field lines point in opposite directions inside the spheres but in the
(c) The E
same direction outside.
and B
field lines point in opposite directions both inside and outside the
(d) The E
spheres.
No justification needed. (But if you give a justification, there is a chance that you might
get partial credit for an incorrect answer.)
8.07 FORMULA SHEET FOR QUIZ 2, V. 2, FALL 2012
p. 5
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Physics Department
Physics 8.07: Electromagnetism II
November 13, 2012
Prof. Alan Guth
FORMULA SHEET FOR QUIZ 2, V. 2
Exam Date: November 15, 2012
∗∗∗
Some sections below are marked with asterisks, as this section is. The asterisks
indicate that you won’t need this material for the quiz, and need not understand it. It is
included, however, for completeness, and because some people might want to make use
of it to solve problems by methods other than the intended ones.
Index Notation:
·B
= Ai Bi ,
A
×B
i = ijk Aj Bk ,
A
ijk pqk = δip δjq − δiq δjp
det A = i1 i2 ···in A1,i1 A2,i2 · · · An,in
Rotation of a Vector:
Ai = Rij Aj ,
Orthogonality: Rij Rik = δjk
j=1
Rotation about z-axis by φ: Rz (φ)ij

i=1  cos φ


= i=2 

 sin φ
i=3
0
(RT T = I)
j=2
j=3
− sin φ
cos φ
0
0
0
1







Rotation about axis n̂ by φ:∗∗∗
R(n̂, φ)ij = δij cos φ + n̂i n̂j (1 − cos φ) − ijk n̂k sin φ .
Vector Calculus:
∂
ϕ)i = ∂i ϕ ,
∂i ≡
Gradient:
(∇
∂xi
Divergence:
∇ · A ≡ ∂i A i
Curl:
× A)
i = ijk ∂j Ak
(∇
Laplacian:
· (∇
ϕ) =
∇ ϕ=∇
2
∂ 2ϕ
∂xi ∂xi
Fundamental Theorems of Vector Calculus:
b
Gradient:
− ϕ(a)
ϕ · d = ϕ(b)
∇
a
· da
A
·A
d x=
∇
3
Divergence:
V
Curl:
S
S
where S is the boundary of V
· d
(∇ × A) · da =
A
P
where P is the boundary of S
8.07 FORMULA SHEET FOR QUIZ 2, V. 2, FALL 2012
p. 6
Delta Functions:
ϕ(r )δ 3 (r − r ) d3 x = ϕ(r )
ϕ(x)δ(x − x ) dx = ϕ(x ) ,
d
dϕ ϕ(x) δ(x − x ) dx = −
dx x=x
dx
δ(x − xi )
, g(xi ) = 0
δ(g(x)) =
|g (xi )|
i
1
r
−
r
·
= 4πδ 3 (r − r )
= −∇2
∇
3
|r − r |
|r − r |
x 4π
1
δij − 3r̂i r̂j
r̂j
j
+
δij δ 3 (r)
≡ ∂i 3 = −∂i ∂j
=
∂i
2
3
r
r
r
r
3
ˆ − d
8π 3
· 3(d · r)r̂
∇
=
−
(d · ∇)δ (r )
3
r3
× 3(d · r̂)r̂ − d = − 4π d × ∇
δ 3 (r )
∇
3
r
3
Electrostatics:
, where
F = qE
(r − r )
1
1 (r − r ) qi
=
r ) d3 x
E(r ) =
3 ρ(
4π0
4π0 i |r − r |3
|r − r |
0 =permittivity of free space = 8.854 × 10−12 C2 /(N·m2 )
1
= 8.988 × 109 N·m2 /C2
4π0
r
1
ρ(r ) 3 d x
V (r ) = V (r 0 ) −
E(r ) · d =
4π0
|r − r |
r0
·E
= ρ ,
×E
= 0,
= −∇
V
∇
∇
E
0
ρ
(Poisson’s Eq.) ,
ρ = 0 =⇒ ∇2 V = 0 (Laplace’s Eq.)
∇2 V = −
0
Laplacian Mean Value Theorem (no generally accepted name): If ∇2 V = 0, then
the average value of V on a spherical surface equals its value at the center.
Energy:
1 1 qi qj
ρ(r )ρ(r )
1 1
d3 x d3 x
W =
=
2 4π0
rij
2 4π0
|r − r |
1
W =
2
ij
i=j
1
d xρ(r )V (r ) = 0
2
3
2 3
E
d x
8.07 FORMULA SHEET FOR QUIZ 2, V. 2, FALL 2012
p. 7
Conductors:
= σ n̂
Just outside, E
0
Pressure on surface:
1
2 σ |E|outside
Two-conductor system with charges Q and −Q: Q = CV , W = 12 CV 2
N isolated conductors:
Vi =
Pij Qj ,
Pij = elastance matrix, or reciprocal capacitance matrix
Cij Vj ,
Cij = capacitance matrix
j
Qi =
j
a
a2
Image charge in sphere of radius a: Image of Q at R is q = − Q, r =
R
R
Separation of Variables for Laplace’s Equation in Cartesian Coordinates:
V =
cos αx
sin αx
cos βy
sin βy
cosh γz
sinh γz
where γ 2 = α2 + β 2
Separation of Variables for Laplace’s Equation in Spherical Coordinates:
Traceless Symmetric Tensor expansion:
1
1 ∂
2 ∂ϕ
+ 2 ∇2θ ϕ = 0 ,
r
∇ ϕ(r, θ, φ) = 2
∂r
r
r ∂r
where the angular part is given by
1 ∂ 2ϕ
1 ∂
∂ϕ
2
sin θ
∇θ ϕ ≡
+
sin θ ∂θ
∂θ
sin2 θ ∂φ2
2
()
()
∇2θ Ci1 i2 ...i n̂i1 n̂i2 . . . n̂i = −( + 1)Ci1 i2 ...i n̂i1 n̂i2 . . . n̂i ,
()
where Ci1 i2 ...i is a symmetric traceless tensor and
n̂ = sin θ cos φ ê1 + sin θ sin φ ê2 + cos θ ê3 .
General solution to Laplace’s equation:
()
∞
C
()
i2 ...i
r̂i1 r̂i2 . . . r̂i ,
V (r ) =
Ci1 i2 ...i r + i1+1
r
=0
where r = rr̂
8.07 FORMULA SHEET FOR QUIZ 2, V. 2, FALL 2012
p. 8
Azimuthal Symmetry:
∞ B
V (r ) =
A r + +1 { ẑi1 . . . ẑi } r̂i1 . . . rˆi
r
=0
where { . . . } denotes the traceless symmetric part of . . . .
Special cases:
{1} = 1
{ ẑi } = ẑi
{ ẑi ẑj } = ẑi ẑj − 13 δij
ẑi δjk + ẑj δik + ẑk δij
{ ẑi ẑj ẑk ẑm } = ẑi ẑj ẑk ẑm − 17 ẑi ẑj δkm + ẑi zˆk δmj + ẑi ẑm δjk + ẑj ẑk δim
1
δij δkm + δik δjm + δim δjk
+ ẑj ẑm δik + ẑk ẑm δij + 35
{ ẑi ẑj ẑk } = ẑi ẑj ẑk −
1
5
Legendre Polynomial / Spherical Harmonic expansion:
General solution to Laplace’s equation:
∞ Bm
Am r + +1 Ym (θ, φ)
V (r ) =
r
=0 m=−
2π
Orthonormality:
π
dφ
0
0
sin θ dθ Y∗ m (θ, φ) Ym (θ, φ) = δ δm m
Azimuthal Symmetry:
∞ B
A r + +1 P (cos θ)
V (r ) =
r
=0
Electric Multipole Expansion:
First several terms:
1 Q p
· r̂ 1 r̂i rˆj
+
·
·
·
, where
+ 2 +
V (r ) =
Q
ij
r
2 r3
4π0 r
3
3
Q = d x ρ(r ) , pi = d x ρ(r ) xi Qij = d3 x ρ(r )(3xi xj −δij |r |2 ) ,
dip (r ) = − 1 ∇
E
4π0
×E
dip (r ) = 0 ,
∇
p · r̂
r2
=
1 3(p · r̂)r̂ − p
1
−
pi δ 3 (r )
3
r
30
4π0
·E
dip (r ) = 1 ρdip (r ) = − 1 p · ∇
δ 3 (r )
∇
0
0
8.07 FORMULA SHEET FOR QUIZ 2, V. 2, FALL 2012
p. 9
Traceless Symmetric Tensor version:
V (r ) =
∞
1 1
()
Ci1 ...i rˆi1 . . . rˆi ,
+1
4π0
r
=0
where
()
Ci1 ...i
(2 − 1)!!
=
!
ρ(r ) { xi1 . . . xi } d3 x
(r ≡ rrˆ ≡ xi eˆi )
∞
(2 − 1)!! r 1
=
{ r̂i1 . . . r̂i } r̂i1 . . . r̂i ,
!
r +1
|r − r |
for r < r
=0
(2 − 1)!! ≡ (2 − 1)(2 − 3)(2 − 5) . . . 1 =
(2)!
, with (−1)!! ≡ 1 .
2 !
Reminder: { . . . } denotes the traceless symmetric part of . . . .
Griffiths version:
∞
1 1
V (r ) =
r ρ(r )P (cos θ ) d3 x
+1
4π0
r
=0
where θ = angle between r and r .
∞
r
1
<
=
P (cos θ ) ,
+1 |r − r |
r>
=0
1
P (x) = 2 !
d
dx
∞
1
√
=
λ P (x)
2
1 − 2λx + λ
=0
(x2 − 1) ,
(Rodrigues’ formula)
P (1) = 1
P (−x) = (−1) P (x)
1
−1
dx P (x)P (x) =
2
δ 2 + 1
Spherical Harmonic version:∗∗∗
∞
1 4π qm
Ym (θ, φ)
V (r ) =
2 + 1 r +1
4π0
=0 m=−
where qm =
∗ Ym
r ρ(r ) d3 x
∞ 4π r ∗ 1
=
Y (θ , φ )Ym (θ, φ) ,
2 + 1 r +1 m
|r − r |
=0 m=−
for r < r
8.07 FORMULA SHEET FOR QUIZ 2, V. 2, FALL 2012
p. 10
Electric Fields in Matter:
Electric Dipoles:
p = d3 x ρ(r ) r
r δ 3 (r − r d ) , where r d = position of dipole
ρdip (r ) = −p · ∇
= (p · ∇)E
= ∇(p
· E)
F
(force on a dipole)
= p × E
(torque on a dipole)
U = −p · E
Electrically Polarizable Materials:
(r ) = polarization = electric dipole moment per unit volume
P
· n̂
ρbound = −∇ · P ,
σbound = P
≡ 0 E
+P
,
D
·D
= ρfree ,
∇
×E
= 0 (for statics)
∇
Boundary conditions:
⊥
⊥
Eab
ove − Ebelow =
σ
0
E
above − Ebelow = 0
⊥
⊥
Dabove
− Dbe
low = σfree
D
above − Dbelow = Pabove − Pbelow
Linear Dielectrics:
= 0 χe E,
P
χe = electric susceptibility
= E
≡ 0 (1 + χe ) = permittivity,
D
r =
= 1 + χe = relative permittivity, or dielectric constant
0
N α/0
, where N = number density of atoms
1 − Nα
30
or (nonpolar) molecules, α = atomic/molecular polarizability (P = αE)
1
·E
d3 x
(linear materials only)
Energy: W =
D
2
W (Even if one or more potential differences are
Force on a dielectric: F = −∇
held fixed, the force can be found by computing the gradient with the total
charge on each conductor fixed.)
Clausius-Mossotti equation: χe =
Magnetostatics:
Magnetic Force:
= q (E
+ v × B)
= dp ,
F
dt
where p = γm0v ,
1
γ= 1−
v2
c2
8.07 FORMULA SHEET FOR QUIZ 2, V. 2, FALL 2012
=
F
p. 11
=
Id × B
d3 x
J × B
Current Density:
J · da
Current through a surface S: IS =
S
Charge conservation:
∂ρ
· J
= −∇
∂t
Moving density of charge: J = ρv
Biot-Savart Law:
µ0
µ0
K(r ) × (r − r ) d × (r − r )
I
=
B(r ) =
da
4π
|r − r |3
|r − r |3
4π
µ0
J(r ) × (r − r ) 3
=
d x
4π
|r − r |3
where µ0 = permeability of free space ≡ 4π × 10−7 N/A2
Examples:
= µ0 I φ̂
Infinitely long straight wire: B
2πr
= µ0 nI0 ẑ , where n = turns per
Infintely long tightly wound solenoid: B
unit length
0, z) =
Loop of current on axis: B(0,
µ0 IR2
ẑ
2(z 2 + R2 )3/2
r ) = 1 µ0 K
× n̂ , n̂ = unit normal toward r
Infinite current sheet: B(
2
Vector Potential:
r )coul = µ0
A(
4π
r )
J(
d3 x ,
|r − r |
×A
,
=∇
B
·A
coul = 0
∇
·B
= 0 (Subject to modification if magnetic monopoles are discovered)
∇
(r ) = A(
r ) + ∇Λ(
r ) for any Λ(r ). B
=∇
×A
is
Gauge Transformations: A
unchanged.
Ampère’s Law:
· d = µ0 Ienc
B
×B
= µ0 J , or equivalently
∇
P
8.07 FORMULA SHEET FOR QUIZ 2, V. 2, FALL 2012
p. 12
Magnetic Multipole Expansion:
Traceless Symmetric Tensor version:
∞
{ r̂i1 . . . r̂i }
µ0 ()
Aj (r ) =
Mj;i1 i2 ...i
4π
r +1
=0
(2 − 1)!!
()
where Mj;i1 i2 ...i =
d3 xJj (r ){ xi1 . . . xi }
!
Current conservation restriction:
d3 x Sym(xi1 . . . xi−1 Ji ) = 0
i1 ...i
where Sym means to symmetrize — i.e. average over all
i1 ...i
orderings — in the indices i1 . . . i
Special cases: = 1:
d3 x Ji = 0
d3 x (Ji xj + Jj xi ) = 0
= 2:
× r̂
r ) = µ0 m
Leading term (dipole): A(
,
4π r 2
where
1
(1)
mi = − ijk Mj;k
2
1
1
m
= I
r × d =
d3 x r × J = Ia ,
2 P
2
where a =
da for any surface S spanning P
S
× r̂
µ0 3(m
· r̂)r̂ − m
2µ0
×m
dip (r ) = µ0 ∇
m
δ 3 (r )
=
B
+
2
3
4π
3
r
4π
r
·B
dip (r ) = 0 ,
×B
dip (r ) = µ0 Jdip (r ) = −µ0 m
δ 3 (r )
∇
∇
×∇
Griffiths version:
∞
µ0 I 1
(r ) P (cos θ )d
A(r ) =
4π
r +1
=0
Magnetic Fields in Matter:
Magnetic Dipoles:
1
1
m
= I
r × d =
d3 x r × J = Ia
2 P
2
8.07 FORMULA SHEET FOR QUIZ 2, V. 2, FALL 2012
p. 13
r δ 3 (r − r d ), where r d = position of dipole
×∇
Jdip (r ) = −m
=∇
(m
F
· B)
(force on a dipole)
=m
×B
U = −m
·B
(torque on a dipole)
Magnetically Polarizable Materials:
(r ) = magnetization = magnetic dipole moment per unit volume
M
×M
bound = M
,
× n̂
K
Jbound = ∇
≡ 1B
−M
,
×H
= Jfree ,
·B
=0
∇
H
∇
µ0
Boundary conditions:
⊥
⊥
⊥
⊥
⊥
⊥
Hab
Bab
ove − Bbelow = 0
ove − Hbelow = −(Mabove − Mbelow )
H
B
above − Bbelow = µ0 (K × n̂)
above − Hbelow = Kfree × n̂
Linear Magnetic Materials:
= χm H,
χm = magnetic susceptibility
M
= µH
B
µ = µ0 (1 + χm ) = permeability,
Magnetic Monopoles:
r ) = µ0 qm r̂ ;
B(
Force on a static monopole: F = qm B
4π r 2
= µ0 qe qm r̂ , where r̂ points
Angular momentum of monopole/charge system: L
4π
from qe to qm
µ0 qe qm
1
Dirac quantization condition:
= h̄ × integer
4π
2
Connection Between Traceless Symmetric Tensors and Legendre Polynomials
or Spherical Harmonics:
(2)!
P (cos θ) = { ẑi1 . . . ẑi } n̂i1 . . . n̂i
2 (!)2
For m ≥ 0,
(,m)
Ym (θ, φ) = Ci1 ...i n̂i1 . . . n̂i ,
(,m)
+
where Ci1 i2 ...i = dm { û+
ˆim+1 . . . ẑi } ,
i1 . . . ûim z
2m (2 + 1)
(−1)m (2)!
,
with dm =
2 !
4π ( + m)! ( − m)!
1
and û+ = √ (êx + iêy )
2
∗
Form m < 0, Y,−m (θ, φ) = (−1)m Ym
(θ, φ)
8.07 FORMULA SHEET FOR QUIZ 2, V. 2, FALL 2012
p. 14
More Information about Spherical Harmonics:∗∗∗
2 + 1 ( − m)! m
P (cos θ)eimφ
Ym (θ, φ) =
4π ( + m)! where Pm (cos θ) is the associated Legendre function, which can be defined by
Pm (x)
+m
(−1)m
2 m/2 d
=
(1 − x )
(x2 − 1)
+m
2 !
dx
Legendre Polynomials:
SPHERICAL HARMONICS Ylm(θ , φ)
l=0
1
Y00 =
4π
3
sin θeiφ
8π
Y11 = l=1
3
cos θ
4π
Y10 =
Y22 =
l=2
1
4
15
sin2 θe2iφ
2π
15
sin θ cosθeiφ
8π
Y21 = -
Y20 =
5
( 32 cos2θ
4π
1
)
2
35
sin3 θe3iφ
4π
Y33 = -
1
4
Y32 =
1
4
105
sin2 θ cos θe2iφ
2π
Y31 = -
1
4
21
sinθ (5cos2θ -1)eiφ
4π
l=3
Y30 =
7
( 5 cos3θ
4π 2
3
2
cos θ)
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8.07 Electromagnetism II
Fall 2012
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