MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Physics Department
Physics 8.07: Electromagnetism II
November 21, 2012
Prof. Alan Guth
QUIZ 2 SOLUTIONS
QUIZ DATE: NOVEMBER 15, 2012
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
A(r, θ, φ) =
4

 µ0 R ωσ sin θ φˆ , (if r ≥ R) .
r2
3
(1.1)
PROBLEM 1 SOLUTION:
(a) A uniformly charged solid sphere of radius R carries a total charge Q, hence it has
charge density ρ = Q/( 43 πR3 ). To find the magnetic moment of sphere we can divide
the sphere into infinitesimal charges. Using spherical polar coordinates, we can take
dq = ρ dτ = ρ r 2 dr sin θ dθ dφ, with the contribution to the dipole moment given by
dm
= 12 r × J dτ . One method would be to write down the volume integral directly,
using J = ρv = ρω × r. We can, however, integrate over φ before we start, so we are
breaking the sphere into rings, where a given ring is indicated by its coordinates r
and θ, and its size dr and dθ. The volume of each ring is dτ = 2πr 2 dr sin θ dθ. The
current dI in the ring is given by dq/T , where T = 2π/ω is the period, so
dI =
dq
ωρdτ
=
= ωρr 2 dr sin θ dθ .
T
2π
(1.2)
8.07 QUIZ 2 SOLUTIONS, FALL 2012
p. 2
The magnetic dipole moment of each ring is then given by
1
1
dm
ring =
r × J dτ = dI
r × d = dI(πr 2 sin2 θ) ẑ .
2
2 ring
ring
(1.3)
The total magnetic dipole moment is then
m
= ωρr 2 sin θ (πr 2 sin2 θ) dr dθ ẑ
R
π
4
r dr
= πωρ
0
(1 − cos2 θ) sin θ dθ ẑ
0
Q R5 4
=
3 5 3
πR
3
= πω 4
1
QωR2 zˆ .
5
(1.4)
(b) The vector potential in dipole approximation is,
× r
µ0 |m
| sin θ
= µ0 m
A
=
φ̂ =
3
4π r
4π
r2
µ0 QωR2 sin θ ˆ
φ.
4π 5
r2
(1.5)
(c) To calculate the exact vector potential inside the sphere, we split the sphere into
shells. Let r be the integration variable and the radius of a shell, moreover let
dr denote the thickness of the shell. Then we can use the results of Example 5.11
(pp. 236-37) in Griffiths, if we replace σ by its value for this case. The value of σ is
found equating charges
Q
σ(4πr 2) = 4
(4πr 2 )dr (1.6)
3
3 πR
and therefore we must replace
σ→
Q
dr
4
3
3 πR
.
Making this replacement in Griffiths’ Eq. (5.67), quoted above as Eq. (1.1), we now
have

 r r if r < r ω
Q
µ
0
dAφ (r, θ, φ) = 4 3 dr sin θ r 4
(1.7)
 2 if r > r .
3
πR
3
r
Note that the R of Griffiths has been replaced by r , which is the radius of the
integration shell. Now we can calculate the vector potential inside the sphere at
8.07 QUIZ 2 SOLUTIONS, FALL 2012
p. 3
some radius r < R. The integration will require two pieces, a piece where 0 < r < r
and the other where r < r < R, thus using the two options in Eq. (1.7):
µ0 Qω
sin θ
Aφ (r, θ, φ) =
4π R3
0
r
r 4
dr 2 +
r
R
dr rr .
(1.8)
r
Doing the integrals one finds
3r 3
rR2 µ0 Qω
sin θ −
+
.
Aφ (r, θ, φ) =
4π R3
10
2
(1.9)
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)
(b) (4 points) Explain why the solution can be written as
V (r, θ) =
∞
V (r){ ẑi1 . . . ẑi } r̂i1 . . . r̂i ,
(2.3a)
=0
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).)
8.07 QUIZ 2 SOLUTIONS, FALL 2012
p. 4
(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
= −( + 1) sin θP (cos θ) .
(2.4)
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 2 SOLUTION:
(a) Since we don’t have free charges anywhere,
·D
=∇
· (E),
∇
· (∇
) + ∇
·E
=0.
=E
(2.5)
= d êr . Then putting this
The permittivity only depends on r, so we can write ∇
dr
result into Eq. (2.5) with E = −∇V , we find
V ) · êr d + ∇2 V
0 = (∇
dr
∂V d 1
=
+ ∇2 V
∂r dr =⇒
0=
∂V d ln + ∇2 V .
∂r dr
(2.6)
(b) With an external field along the z-axis, the problem has azimuthal symmetry, implying ∂V /∂φ = 0, so V = V (r, θ). The Legendre polynomials P (cos θ) are a complete
set of functions of the polar angle θ for 0 ≤ θ ≤ π, implying that at each value of
r, V (r, θ) can be expanded in a Legendre series. In general, the coefficients may be
functions of r, so we can write
V (r, θ) =
∞
=0
V (r)P (cos θ) .
(2.7)
8.07 QUIZ 2 SOLUTIONS, FALL 2012
p. 5
The same argument holds for an expansion in { ẑi1 . . . ẑi } r̂i1 . . . r̂i , since these are
in fact the same functions, up to a multiplicative constant. Note that if depended
on θ as well as r, then the completeness argument would still be valid, and it would
still be possible to write V (r, θ) as in Eqs. (2.3). In that case, however, the equations
for the functions V (r) would become coupled to each other, making them much more
difficult to solve.
d ln 2
(c) For r < R we have
= − . Using the hint, Eq. (2.4) in the problem statement,
dr
r
we write
∞
1 ∂
∂V d ln dV
2
( + 1)
2 ∂V
∇ V +
=
P (cos θ ) 2
r
+
−
V = 0 .
−
∂r dr
r2
∂r
dr
r
r ∂r
=0
(2.8)
For this equation to hold for all r < R and for all θ, the term inside the square
brackets should be zero. (To show this, one would multiply by P (cos θ) sin θ and
then integrate from θ = 0 to θ = 2π. By the orthonormality of the Legendre
polynomials, only the = term would survive, so it would have to vanish for every
.) Thus,
2
1 ∂
r 2 ∂r
r
2 ∂V
∂r
dV
+
dr
2
( + 1)
( + 1)
d2 V
−
−
V
=
−
V = 0 .
2
2
r
r
dr
r2
(2.9)
The general solution to Eq. (2.9) is
V (r) = A r +1 +
B
.
r
(2.10)
(This can be verified by inspection, but it can also be found by assuming a trial
function in the form of a power, V ∝ r p . Inserting the trial function into the
differential equation, one finds p(p − 1) = ( + 1) . One might see by inspection that
this is solved by p = + 1 or p = −, or one can solve it as a quadratic equation,
finding
1 ± (2 + 1)
p=
= + 1 or − .)
2
For r > R,
1 ∂ 2 ∂V ( + 1)
r
−
V = 0.
∂r
r2
r 2 ∂r
(2.11)
The general solution to Eq. (2.11) is,
V (r ) = C r +
D
.
r +1
(2.12)
8.07 QUIZ 2 SOLUTIONS, FALL 2012
p. 6
(d) The coefficients B are zero, B = 0, to avoid a singularity at r = 0. The potential
goes as V (r) = −E0 r cos θ for r R; this gives C = 0 except for C1 = −E0 . The
potential V (r, θ) is continuous at r = R, implying that

D

 A R+1 = +1
for = 1
R
(2.13)

 A R2 = −E R + D1
for = 1 .
1
0
R2
In addition, the normal component of the displacement vector is continuous on the
boundary of the sphere. Since is continuous at r = R, this means that Er =
−∂V /∂r is continuous, which one could also have deduced from Eq. (2.2), since any
discontinuity in ∂V /∂r would produce a δ-function in ∂ 2 V /∂r 2 . Setting ∂V /∂r at
r = R− equal to its value at r = R+ , we find

 ( + 1)A R = −( + 1) D

for = 1
R+2
(2.14)
D1


for = 1 .
2A1 R = −2 3 − E0
R
Solving Eq. (2.13) and Eq. (2.14) as two equations (for each ) for the two unknowns
A and D , we see that A = D = 0 for = 1, and that
3E0
,
4R
Then we find the potential as
A1 = −
C1 = −E0 ,
and D1 =

2
 − 3E0 r cos θ


4R V (r, θ) =
R3


 E0 cos θ
−r
4r2
E0 R 3
.
4
(2.15)
for r < R
(2.16)
for r < R .
(e) Eq. (2.16) tells us that for r > R, the potential is equal to that of the applied external
field, Vext = −E0 r cos θ, plus a term that we attribute to the sphere:
E0 R 3
cos θ .
4r 2
This has exactly the form of an electric dipole,
Vsphere (r, θ) =
Vdip =
1 p · r̂
,
4π0 r 2
(2.17)
(2.18)
if we identify
p = π0 R3 E0 zˆ .
(2.19)
8.07 QUIZ 2 SOLUTIONS, FALL 2012
p. 7
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 zˆ; the other has dipole moment m
2 = −m0 ẑ, 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.
PROBLEM 3 SOLUTION:
(a) For the vector potential, we have from the formula sheet that
× r̂
r ) = µ0 m
A(
,
4π r 2
(3.2)
which vanishes on axis, since m
= m0 ẑ, and r̂ = ẑ on axis. Thus,
0, z) = 0 .
A(0,
(3.3)
= 0, however, since B depends on derivatives of A
with
This does not mean that B
respect to x and y. From the formula sheet we have
· r̂)r̂ − m
dip (r ) = µ0 3(m
,
B
4π
r3
(3.4)
where we have dropped the δ-function because we are interested only in r =
0.
Evaluating this expression on the positive z axis, where r̂ = ẑ, we find
dip (0, 0, z) = µ0 2m0 ẑ = µ0 m0 ẑ .
B
2π r 3
4π r 3
(3.5)
8.07 QUIZ 2 SOLUTIONS, FALL 2012
p. 8
For 2 dipoles, we have
2 dip (0, 0, z) = µ0 m0
B
2π
µ0 m0
=
2πz 3
µ0 m0
≈
2πz 3
µ0 m0
2πz 3
µ0 m0
≈
2πz 3
≈
=
1
z − 12 a
1
1−
1
3 − 3
z + 12 a
1a 3
2z
−
1
1+
zˆ
1a 3
2z
ẑ
1
1
ẑ
−
1 − 32 az
1 + 32 az
3a
3a
1+
− 1−
ẑ
2z
2z
a
3
ẑ
z
3µ0 m0 a
zˆ .
4πz 4
(3.6)
(b) Since it falls off as 1/z 4 , it is undoubtedly a quadrupole ( = 2) . For either the E
fields, the monopole falls off as 1/r 2 , the dipole as 1/r 3 , and the quadrupole as
or B
1/r 4 .
(c) We wish to take the limit as a → 0 in such a way that the field at large z approaches
a constant, without blowing up or going to zero. From Eq. (3.6), we see that this
goal will be accomplished by keeping m0 a fixed, which means n = 1 .
(d) For the two-dipole system we add together the two contributions to the current
density, using the appropriate values of r d and m
:
J2
r)
dip (
r δ 3 r − a zˆ + m0 ẑ × ∇
r δ 3 r − a zˆ .
= −m0 ẑ × ∇
2
2
Rewriting,
J2
r)
dip (
r
= m0 aẑ × ∇
δ 3 (r + a2 ẑ) − δ 3 (r − a2 ẑ)
a
(3.7)
.
(3.8)
Now we can define Q ≡ m0 a, and if we take the limit a → 0 with Q fixed, the above
expression becomes
J2
r)
dip (
r ∂ δ 3 (r ) .
= Qẑ × ∇
∂z
(3.9)
8.07 QUIZ 2 SOLUTIONS, FALL 2012
p. 9
Since partial derivatives commute, this could alternatively be written as
J2
r)
dip (
= Qẑ ×
∂ 3
∇r δ (r ) .
∂z
(3.10)
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 4 SOLUTION:
r ) field is
= M0 ẑ. The curl of the H(
(a) The magnetization inside the cylinder is M
× H(
r ) = Jfree = 0 ,
∇
(4.1)
and the divergence is
·H
(r) = ∇
·
∇
r)
B(
(r)
−M
µ0
=
1 ∇·B−∇·M =0 .
µ0
(4.2)
Note that for a finite length cylinder, the divergence would be nonzero because of the
at the boundaries. Since H(
r ) is divergenceless and curl-free,
abrupt change in M
we can say
r) = 0
H(
everywhere in space.
(4.3)
r ) = 0 everywhere in space, we can find magnetic field as
(b) Having H(
(r ) = B(r ) − M
(r ) = 0
H
µ0
=⇒
r ) =
B(
µ0 M0 ẑ
0
for r < R ,
for r > R .
(4.4)
×M
= 0 and
In this question we could alternatively find the bound currents as Jb = ∇
ˆ
Kb = M × n̂ = M0 φ. Then, using Ampère’s law as we did for a solenoid, we could find
obtaining the same answers as above.
the magnetic field and then also H,
8.07 QUIZ 2 SOLUTIONS, FALL 2012
p. 10
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.
PROBLEM 5 SOLUTION:
field of a uniformly
E
polarized sphere
field of a uniformly
B
magnetized sphere
and B
field lines point in opposite directions inside the spheres but
The answer is (c), E
in the same direction outside, as shown in the diagrams, which were scanned from the
·E
=
first edition of Jackson. Note that the diagram on the left shows clearly that ∇
0
at the boundary of the sphere, so it could not possibly
be a picture of B. It is at least
×E
= 0, or equivalently E
· d = 0 for any closed loop, as it
visually consistent with ∇
must be to describe an electrostatic field. The diagram on the right, on the other hand,
×B
=
· d =
shows clearly that ∇
0, or equivalently B
0, so it could not possibly be a
·B
= 0, as
picture of an electrostatic field. It is at least qualitatively consistent with ∇
it must be.
8.07 FORMULA SHEET FOR QUIZ 2, V. 2, FALL 2012
p. 11
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. 12
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
dx x=x
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
≡ ∂i 3 = −∂i ∂j
2
3
r
3
r
r
r
· 3(d · r̂)r̂ − d = − 8π (d · ∇
)δ 3 (r )
∇
r3
3
r − d
4π
× 3(d · r̂)ˆ
δ 3 (r )
∇
= − d × ∇
3
3
r
Electrostatics:
, where
F = qE
1
(r − r )
1 (r − r ) qi
E(r ) =
=
) d3 x
3 ρ(r
4π0 i |r − r |3
4π0
|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 E(r ) · d =
d x
V (r ) = V (r 0 ) −
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
W =
=
d3 x d3 x
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. 13
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
r
+ 2 ∇2θ ϕ = 0 ,
∇ ϕ(r, θ, φ) = 2
∂r
r
r ∂r
where the angular part is given by
1 ∂
∂ϕ
1 ∂ 2ϕ
2
sin θ
+
∇θ ϕ ≡
sin θ ∂θ
∂θ
sin2 θ ∂φ2
2
()
()
∇2θ Ci1 i2 ...i n
ˆ i1 n
ˆ i2 . . . n
ˆ i = −( + 1)Ci1 i2 ...i n̂i1 n̂i2 . . . n
ˆ i ,
()
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
rˆi1 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. 14
Azimuthal Symmetry:
∞ B
A r + +1 { ẑi1 . . . ẑi } r̂i1 . . . r̂i
V (r ) =
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 − 71 ẑi ẑj δkm + ẑi ẑk δmj + ẑi ẑm δjk + ẑj zˆk 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
V (r ) =
Am r + +1 Ym (θ, φ)
r
=0 m=−
2π
π
dφ
Orthonormality:
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 r̂j
Q
+
·
·
·
, where
V (r ) =
+ 2 +
ij
2 r3
4π0 r
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
4π0
r
30
·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. 15
Traceless Symmetric Tensor version:
V (r ) =
∞
1 1
()
Ci1 ...i r̂i1 . . . r̂i ,
+1
r
4π0
=0
where
()
Ci1 ...i
(2 − 1)!!
=
!
ρ(r ) { xi1 . . . xi } d3 x
(r ≡ rr̂ ≡ xi êi )
∞
(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
1
P (−x) = (−1) P (x)
−1
dx P (x)P (x) =
2
δ 2 + 1
Spherical Harmonic version:∗∗∗
∞
1 4π qm
V (r ) =
Ym (θ, φ)
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. 16
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
⊥
⊥
Dab
ove − Dbelow = 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. 17
=
I d × 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
=
da
B(r ) =
4π
|r − r |3
4π
|r − r |3
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
J(r ) 3 d 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. 18
Magnetic Multipole Expansion:
Traceless Symmetric Tensor version:
∞
{ r̂i1 . . . r̂i }
µ0 ()
Aj (r ) =
Mj;i1 i2 ...i
r +1
4π
=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:
× rˆ
(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
2 P
where a =
da for any surface S spanning P
S
× r̂
µ0 3(m
· r̂)r̂ − m
2µ0
dip (r ) = µ0 ∇
×m
m
δ 3 (r )
B
=
+
2
3
4π
r
3
4π
r
·B
dip (r ) = 0 ,
×B
dip (r ) = µ0 Jdip (r ) = −µ0 m
3 (r )
∇
∇
× ∇δ
Griffiths version:
∞
µ0 I 1
(r ) P (cos θ )d
A(r ) =
r +1
4π
=0
Magnetic Fields in Matter:
Magnetic Dipoles:
1
1
r × d =
m
= I
d3 x r × J = Ia
2 P
2
8.07 FORMULA SHEET FOR QUIZ 2, V. 2, FALL 2012
p. 19
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 = ∇
·B
=0
≡ 1B
−M
,
×H
= Jfree ,
∇
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 rˆ ;
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)!
{ ẑi1 . . . ẑi } n̂i1 . . . n̂i
P (cos θ) = 2 (!)2
For m ≥ 0,
(,m)
Ym (θ, φ) = Ci1 ...i n̂i1 . . . n̂i ,
(,m)
+
ˆim+1 . . . ẑi } ,
where Ci1 i2 ...i = dm { û+
i1 . . . ûim z
2m (2 + 1)
(−1)m (2)!
,
with dm =
4π ( + m)! ( − m)!
2 !
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. 20
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)
Image by MIT OpenCourseWare.
MIT OpenCourseWare
http://ocw.mit.edu
8.07 Electromagnetism II
Fall 2012
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.