Part IB — Electromagnetism
Based on lectures by D. Tong
Notes taken by Dexter Chua
Lent 2015
These notes are not endorsed by the lecturers, and I have modified them (often
significantly) after lectures. They are nowhere near accurate representations of what
was actually lectured, and in particular, all errors are almost surely mine.
Electromagnetism and Relativity
Review of Special Relativity; tensors and index notation. Lorentz force law. Electro
magnetic tensor. Lorentz transformations of electric and magnetic fields. Currents
and the conservation of charge. Maxwell equations in relativistic and nonrelativistic
forms. [5]
Electrostatics
Gauss’s law. Application to spherically symmetric and cylindrically symmetric charge
distributions. Point, line and surface charges. Electrostatic potentials; general charge
distributions, dipoles. Electrostatic energy. Conductors. [3]
Magnetostatics
Magnetic fields due to steady currents. Ampre’s law. Simple examples. Vector
potentials and the BiotSavart law for general current distributions. Magnetic dipoles.
Lorentz force on current distributions and force between currentcarrying wires. Ohm’s
law. [3]
Electrodynamics
Faraday’s law of induction for fixed and moving circuits. Electromagnetic energy and
Poynting vector. 4vector potential, gauge transformations. Plane electromagnetic
waves in vacuum, polarization. [5]
Contents
0 Introduction
1 Preliminaries
1.1 Charge and Current
1.2 Forces and Fields
2 Electrostatics
2.1 Gauss’ Law
2.2 Electrostatic potential
2.2.1 Point charge
2.2.2 Dipole
2.2.3 General charge distribution
2.2.4 Field lines and equipotentials
2.3 Electrostatic energy
2.4 Conductors
3 Magnetostatics
3.1 Ampere’s Law
3.2 Vector potential
3.3 Magnetic dipoles
3.4 Magnetic forces
4 Electrodynamics
4.1 Induction
4.2 Magnetostatic energy
4.3 Resistance
4.4 Displacement currents
4.5 Electromagnetic waves
4.6 Poynting vector
5 Electromagnetism and relativity
5.1 A review of special relativity
5.1.1 A geometric interlude on (co)vectors
5.1.2 Transformation rules
5.1.3 Vectors and covectors in SR
5.2 Conserved currents
5.3 Gauge potentials and electromagnetic fields
5.4 Maxwell Equations
5.5 The Lorentz force law
0 Introduction
Electromagnetism is one of the four fundamental forces of the universe. Apart
from gravity, most daily phenomena can be explained by electromagnetism. The
very existence of atoms requires the electric force (plus weird quantum effects)
to hold electrons to the nucleus, and molecules are formed again due to electric
forces between atoms. More macroscopically, electricity powers all our electrical
appliances (by definition), while the magnetic force makes things stick on our
fridge. Finally, it is the force that gives rise to light, and allows us to see things.
While it seems to be responsible for so many effects, the modern classical
description of electromagnetism is rather simple. It is captured by merely four
short and concise equations known as Maxwell’s equations. In fact, in the
majority of this course, we would simply be exploring different solutions to these
equations.
Historically, electromagnetism has another significance — it led to Einstein’s
discovery of special relativity. At the end of the course, we will look at how
Maxwell’s equations naturally fit in nicely in the framework of relativity. Written
relativistically, Maxwell’s equation look much simpler and more elegant. We will
discover that magnetism is entirely a relativistic effect, and makes sense only in
the context of relativity.
As we move to the world of special relativity, not only do we get a better
understanding of Maxwell’s equation. As a takeaway, we also get to understand
relativity itself better, as we provide a more formal treatment of special relativity,
which becomes a powerful tool to develop theories consistent with relativity.
1 Preliminaries
1.1 Charge and Current
The strength of the electromagnetic force experienced by a particle is determined
by its (electric) charge. The SI unit of charge is the Coulomb. In this course, we
assume that the charge can be any real number. However, at the fundamental
level, charge is quantised. All particles carry charge
q
=
ne
for some integer
n
,
and the basic unit
e ≈ 1.6 × 10
−19
C
. For example, the electron has
n
=
−
1,
proton has n = +1, neutron has n = 0.
Often, it will be more useful to talk about charge density ρ(x, t).
Definition
(Charge density)
.
The charge density is the charge per unit volume.
The total charge in a region V is
Q =
Z
V
ρ(x, t) dV
When we study charged sheets or lines, the charge density would be charge
per unit area or length instead, but this would be clear from context.
The motion of charge is described by the current density J(x, t).
Definition (Current and current density). For any surface S, the integral
I =
Z
S
J · dS
counts the charge per unit time passing through
S
.
I
is the current, and
J
is
the current density, “current per unit area”.
Intuitively, if the charge distribution
ρ
(
x, t
) has velocity
v
(
x, t
), then (ne
glecting relativistic effects), we have
J = ρv.
Example.
A wire is a cylinder of crosssectional area
A
. Suppose there are
n
electrons per unit volume. Then
ρ = nq = −ne
J = nqv
I = nqvA.
It is well known that charge is conserved — we cannot create or destroy
charge. However, the conservation of charge does not simply say that “the total
charge in the universe does not change”. We want to rule out scenarios where a
charge on Earth disappears, and instantaneously appears on the Moon. So what
we really want to say is that charge is conserved locally: if it disappears here, it
must have moved to somewhere nearby. Alternatively, charge density can only
change due to continuous currents. This is captured by the continuity equation:
Law (Continuity equation).
∂ρ
∂t
+ ∇ ·J = 0.
We can write this into a more intuitive integral form via the divergence
theorem.
The charge Q in some region V is defined to be
Q =
Z
V
ρ dV.
So
dQ
dt
=
Z
V
∂ρ
∂t
dV = −
Z
V
∇ · J dV = −
Z
S
J · dS.
Hence the continuity equation states that the change in total charge in a volume
is given by the total current passing through its boundary.
In particular, we can take
V
=
R
3
, the whole of space. If there are no
currents at infinity, then
dQ
dt
= 0
So the continuity equation implies the conservation of charge.
1.2 Forces and Fields
In modern physics, we believe that all forces are mediated by fields (not to
be confused with “fields” in algebra, or agriculture). A field is a dynamical
quantity (i.e. a function) that assigns a value to every point in space and time.
In electromagnetism, we have two fields:
– the electric field E(x, t);
– the magnetic field B(x, t).
Each of these fields is a vector, i.e. it assigns a vector to every point in space
and time, instead of a single number.
The fields interact with particles in two ways. On the one hand, fields cause
particles to move. On the other hand, particles create fields. The first aspect is
governed by the Lorentz force law:
Law (Lorentz force law).
F = q(E + v × B)
while the second aspect is governed by Maxwell’s equations.
Law (Maxwell’s Equations).
∇ · E =
ρ
ε
0
∇ · B = 0
∇ × E +
∂B
∂t
= 0
∇ × B −µ
0
ε
0
∂E
∂t
= µ
0
J,
where we have two constants of nature:
– ε
0
= 8.85 × 10
−12
m
−3
kg
−1
s
2
C
2
is the electric constant;
– µ
0
= 4π × 10
−6
m kg C
−2
is the magnetic constant.
Some prefer to call these constants the “permittivity of free space” and “per
meability of free space” instead. But why bother with these complicated and
easilyconfused names when we can just call them “electric constant” and “mag
netic constant”?
We’ve just completed the description of all of (classical, nonrelativistic) elec
tromagnetism. The remaining of the course would be to study the consequences
of and solutions to these equations.
2 Electrostatics
Electrostatics is the study of stationary charges in the absence of magnetic
fields. We take
ρ
=
ρ
(
x
),
J
= 0 and
B
= 0. We then look for timeindependent
solutions. In this case, the only relevant equations are
∇ · E =
ρ
ε
0
∇ × E = 0,
and the other two equations just give 0 = 0.
In this chapter, our goal is to find
E
for any
ρ
. Of course, we first start with
simple, symmetric cases, and then tackle the more general cases later.
2.1 Gauss’ Law
Here we transform the first Maxwell’s equation into an integral form, known as
Gauss’ Law.
Consider a region
V ⊆ R
3
with boundary
S
=
∂V
. Then integrating the first
equation over the volume V gives
Z
V
∇ · E dV =
1
ε
0
Z
V
ρ dV.
The divergence theorem gives
R
V
∇ · E
d
V
=
R
S
E ·
d
S
, and by definition,
Q =
R
V
ρ dV . So we end up with
Law (Gauss’ law).
Z
S
E · dS =
Q
ε
0
,
where Q is the total charge inside V .
Definition
(Flux through surface)
.
The flux of
E
through the surface
S
is
defined to be
Z
S
E · dS.
Gauss’ law tells us that the flux depends only on the total charge contained
inside the surface. In particular any external charge does not contribute to the
total flux. While external charges do create fields that pass through the surface,
the fields have to enter the volume through one side of the surface and leave
through the other. Gauss’ law tells us that these two cancel each other out
exactly, and the total flux caused by external charges is zero.
From this, we can prove Coulomb’s law:
Example
(Coulomb’s law)
.
We consider a spherically symmetric charge density
ρ
(
r
) with
ρ
(
r
) = 0 for
r > R
, i.e. all the charge is contained in a ball of radius
R
.
R
E
S
By symmetry, the force is the same in all directions and point outward radially.
So
E = E(r)
ˆ
r.
This immediately ensures that ∇ × E = 0.
Put S to be a sphere of radius r > R. Then the total flux is
Z
S
E · dS =
Z
S
E(r)
ˆ
r · dS
= E(r)
Z
S
ˆ
r · dS
= E(r) · 4πr
2
By Gauss’ law, we know that this is equal to
Q
ε
0
. Therefore
E(r) =
Q
4πε
0
r
2
and
E(r) =
Q
4πε
0
r
2
ˆ
r.
By the Lorentz force law, the force experienced by a second charge is
F(r) =
Qq
4πε
0
r
2
ˆ
r,
which is Coulomb’s law.
Strictly speaking, this only holds when the charges are not moving. However,
for most practical purposes, we can still use this because the corrections required
when they are moving are tiny.
Example. Consider a uniform sphere with
ρ(r) =
(
ρ r < R
0 r > R
.
Outside, we know that
E(r) =
Q
4πε
0
r
2
ˆ
r
Now suppose we are inside the sphere.
R
E
r
Then
Z
S
E · dS = E(r)4πr
2
=
Q
ε
0
r
3
R
3
So
E(r) =
Qr
4πε
0
R
3
ˆ
r,
and the field increases with radius.
Example (Line charge). Consider an infinite line with uniform charge density
per unit length η.
We use cylindrical polar coordinates:
z
r =
p
x
2
+ y
2
E
By symmetry, the field is radial, i.e.
E(r) = E(r)
ˆ
r.
Pick
S
to be a cylinder of length
L
and radius
r
. We know that the end caps do
not contribute to the flux since the field lines are perpendicular to the normal.
Also, the curved surface has area 2πrL. Then
Z
S
E · dS = E(r)2πrL =
ηL
ε
0
.
So
E(r) =
η
2πε
0
r
ˆ
r.
Note that the field varies as 1
/r
, not 1
/r
2
. Intuitively, this is because we have
one more dimension of “stuff” compared to the point charge, so the field does
not drop as fast.
Example
(Surface charge)
.
Consider an infinite plane
z
= 0, with uniform
charge per unit area σ.
By symmetry, the field points vertically, and the field bottom is the opposite of
that on top. we must have
E = E(z)
ˆ
z
with
E(z) = −E(−z).
Consider a vertical cylinder of height 2
z
and crosssectional area
A
. Now only
the end caps contribute. So
Z
S
E · dS = E(z)A − E(−z)A =
σA
ε
0
.
So
E(z) =
σ
2ε
0
and is constant.
Note that the electric field is discontinuous across the surface. We have
E(z → 0+) − E(z → 0−) =
σ
ε
0
.
This is a general result that is true for any arbitrary surfaces and
σ
. We can prove
this by considering a cylinder across the surface and then shrink it indefinitely.
Then we find that
ˆ
n · E
+
−
ˆ
n · E
−
=
σ
ε
0
.
However, the components of E tangential to the surface are continuous.
2.2 Electrostatic potential
In the most general case, we will have to solve both
∇·E
=
ρ/ε
0
and
∇×E
=
0
.
However, we already know that the general form of the solution to the second
equation is E = −∇φ for some scalar field φ.
Definition
(Electrostatic potential)
.
If
E
=
−∇φ
, then
φ
is the electrostatic
potential.
Substituting this into the first equation, we obtain
∇
2
φ =
ρ
ε
0
.
This is the Poisson equation, which we have studied in other courses. If we are
in the middle of nowhere and ρ = 0, then we get the Laplace equation.
There are a few interesting things to note about our result:
– φ
is only defined up to a constant. We usually fix this by insisting
φ
(
r
)
→
0
as
r → ∞
. This statement seems trivial, but this property of
φ
is actually
very important and gives rise to a lot of interesting properties. However,
we will not have the opportunity to explore this in this course.
–
The Poisson equation is linear. So if we have two charges
ρ
1
and
ρ
2
, then
the potential is simply
φ
1
+
φ
2
and the field is
E
1
+
E
2
. This is the
principle of superposition. Among the four fundamental forces of nature,
electromagnetism is the only force with this property.
2.2.1 Point charge
Consider a point particle with charge Q at the origin. Then
ρ(r) = Qδ
3
(r).
Here δ
3
is the generalization of the usual delta function for (3D) vectors.
The equation we have to solve is
∇
2
φ = −
Q
ε
0
δ
3
(r).
Away from the origin
r
=
0
,
δ
3
(
r
) = 0, and we have the Laplace equation. From
the IA Vector Calculus course, the general solution is
φ =
α
r
for some constant α.
The constant
α
is determined by the delta function. We integrate the equation
over a sphere of radius r centered at the origin. The left hand side gives
Z
V
∇
2
φ dV =
Z
S
∇φ · dS =
Z
S
−
α
r
2
ˆ
r · dS = −4πα
The right hand side gives
−
Q
ε
0
Z
V
δ
3
(r) dV = −
Q
ε
0
.
So
α =
Q
4πε
0
and
E = −∇φ =
Q
4πε
0
r
2
ˆ
r.
This is just what we get from Coulomb’s law.
2.2.2 Dipole
Definition
(Dipole)
.
A dipole consists of two point charges, +
Q
and
−Q
at
r = 0 and r = −d respectively.
To find the potential of a dipole, we simply apply the principle of superposition
and obtain
φ =
1
4πε
0
Q
r
−
Q
r + d
.
This is not a very helpful result, but we can consider the case when we are far,
far away, i.e.
r d
. To do so, we Taylor expand the second term. For a general
f(r), we have
f(r + d) = f(r) + d · ∇f(r) +
1
2
(d · ∇)
2
f(r) + ··· .
Applying to the term we are interested in gives
1
r + d
=
1
r
− d ·∇
1
r
+
1
2
(d · ∇)
2
1
r
+ ···
=
1
r
−
d · r
r
3
−
1
2
d · d
r
3
−
3(d · r)
2
r
5
+ ··· .
Plugging this into our equation gives
φ =
Q
4πε
0
1
r
−
1
r
+
d · r
r
3
+ ···
∼
Q
4πε
0
d · r
r
3
.
Definition
(Electric dipole moment)
.
We define the electric dipole moment to
be
p = Qd.
By convention, it points from ve to +ve.
Then
φ =
p ·
ˆ
r
4πε
0
r
2
,
and
E = −∇φ =
1
4πε
0
3(p ·
ˆ
r)
ˆ
r − p
r
3
.
2.2.3 General charge distribution
To find
φ
for a general charge distribution
ρ
, we use the Green’s function for the
Laplacian. The Green’s function is defined to be the solution to
∇
2
G(r, r
0
) = δ
3
(r − r
0
),
In the section about point charges, We have shown that
G(r, r
0
) = −
1
4π
1
r − r
0

.
We assume all charge is contained in some compact region V . Then
φ(r) = −
1
ε
0
Z
V
ρ(r
0
)G(r, r
0
) d
3
r
0
=
1
4πε
0
Z
V
ρ(r
0
)
r − r
0

d
3
r
0
Then
E(r) = −∇φ(r)
= −
1
4πε
0
Z
V
ρ(r
0
)∇
1
r − r
0

d
3
r
0
=
1
4πε
0
Z
V
ρ(r
0
)
(r − r
0
)
r − r
0

3
d
3
r
0
So if we plug in a very complicated ρ, we get a very complicated E!
However, we can ask what φ and E look like very far from V , i.e. r r
0
.
We again use the Taylor expansion.
1
r − r
0

=
1
r
+ r
0
· ∇
1
r
+ ···
=
1
r
+
r · r
0
r
3
+ ··· .
Then we get
φ(r) =
1
4πε
0
Z
V
ρ(r
0
)
1
r
+
r · r
0
r
3
+ ···
d
3
r
0
=
1
4πε
0
Q
r
+
p ·
ˆ
r
r
2
+ ···
,
where
Q =
Z
V
ρ(r
0
) dV
0
p =
Z
V
r
0
ρ(r
0
) dV
0
ˆ
r =
r
krk
.
So if we have a huge lump of charge, we can consider it to be a point charge
Q
,
plus some dipole correction terms.
2.2.4 Field lines and equipotentials
Vectors are usually visualized using arrows, where longer arrows represent larger
vectors. However, this is not a practical approach when it comes to visualizing
fields, since a field assigns a vector to every single point in space, and we don’t
want to draw infinitely many arrows. Instead, we use field lines.
Definition
(Field line)
.
A field line is a continuous line tangent to the electric
field E. The density of lines is proportional to E.
They begin and end only at charges (and infinity), and never cross.
We can also draw the equipotentials.
Definition
(Equipotentials)
.
Equipotentials are surfaces of constant
φ
. Because
E = −∇φ, they are always perpendicular to field lines.
Example.
The field lines for a positive and a negative charge are, respectively,
+

We can also draw field lines for dipoles:
+

2.3 Electrostatic energy
We want to calculate how much energy is stored in the electric field. Recall from
IA Dynamics and Relativity that a particle of charge
q
in a field
E
=
−∇φ
has
potential energy U(r) = qφ(r).
U
(
r
) can be thought of as the work done in bringing the particle from infinity,
as illustrated below:
work done = −
Z
r
∞
F · dr
= −
Z
r
∞
E · dr
= q
Z
r
∞
∇φ · dr
= q[φ(r) − φ(∞)]
= U(r)
where we set φ(∞) = 0.
Now consider
N
charges
q
i
at positions
r
i
. The total potential energy stored
is the work done to assemble these particles. Let’s put them in one by one.
(i) The first charge is free. The work done is W
1
= 0.
(ii) To place the second charge at position r
2
takes work. The work is
W
2
=
q
1
q
2
4πε
0
1
r
1
− r
2

.
(iii) To place the third charge at position r
3
, we do
W
3
=
q
3
4πε
0
q
1
r
1
− r
3

+
q
2
r
2
− r
3

(iv) etc.
The total work done is
U =
N
X
i=1
W
i
=
1
4πε
0
X
i<j
q
i
q
j
r
i
− r
j

.
Equivalently,
U =
1
4πε
0
1
2
X
i6=j
q
i
q
j
r
i
− r
j

.
We can write this in an alternative form. The potential at point
r
i
due to all
other particles is
φ(r
i
) =
1
4πε
0
X
j6=i
q
j
r
i
− r
j

.
So we can write
U =
1
2
N
X
i=1
q
i
φ(r
i
).
There is an obvious generalization to continuous charge distributions:
U =
1
2
Z
ρ(r)φ(r) d
3
r.
Hence we obtain
U =
ε
0
2
Z
(∇ · E)φ d
3
r
=
ε
0
2
Z
[∇ · (Eφ) −E · ∇φ] d
3
r.
The first term is a total derivative and vanishes. In the second term, we use the
definition E = −∇φ and obtain
Proposition.
U =
ε
0
2
Z
E · E d
3
r.
This derivation of potential energy is not satisfactory. The final result shows
that the potential energy depends only on the field itself, and not the charges.
However, the result was derived using charges and electric potentials — there
should be a way to derive this result directly with the field, and indeed there is.
However, this derivation belongs to a different course.
Also, we have waved our hands a lot when generalizing to continuous dis
tributions, which was not entirely correct. If we have a single point particle,
the original discrete formula implies that there is no potential energy. However,
since the associated field is nonzero, our continuous formula gives a nonzero
potential.
This does not mean that the final result is wrong. It is correct, but it
describes a more sophisticated (and preferred) conception of “potential energy”.
Again, we shall not go into the details in this course.
2.4 Conductors
Definition
(Conductor)
.
A conductor is a region of space which contains lots
of charges that are free to move.
In electrostatic situations, we must have
E
= 0 inside a conductor. Otherwise,
the charges inside the conductor would move till equilibrium. This almost
describes the whole of the section: if you apply an electric field onto a conductor,
the charges inside the conductor move until the external field is canceled out.
From this, we can derive a lot of results. Since
E
= 0,
φ
is constant inside
the conductor. Since
∇ · E
=
ρ/ε
0
, inside the conductor, we must have
ρ
= 0.
Hence any net charge must live on the surface.
Also, since
φ
is constant inside the conductor, the surface of the conductor is
an equipotential. Hence the electric field is perpendicular to the surface. This
makes sense since any electric field with components parallel to the surface would
cause the charges to move.
Recall also from our previous discussion that across a surface, we have
ˆ
n · E
outside
−
ˆ
n · E
inside
=
σ
ε
0
,
where σ is the surface charge density. Since E
inside
= 0, we obtain
E
outside
=
σ
ε
0
ˆ
n
This allows us to compute the surface charge given the field, and vice versa.
Example.
Consider a spherical conductor with
Q
= 0. We put a positive plate
on the left and a negative plate on the right. This creates a field from the left to
the right.
With the conductor in place, since the electric field lines must be perpendicular
to the surface, they have to bend towards the conductor. Since field lines end
and start at charges, there must be negative charges at the left and positive
charges at the right.
− +
We get an induced surface charge.
Example.
Suppose we have a conductor that fills all space
x <
0. We ground
it such that
φ
= 0 throughout the conductor. Then we place a charge
q
at
x = d > 0.
φ = 0 +
We are looking for a potential that corresponds to a source at
x
=
d
and satisfies
φ
= 0 for
x <
0. Since the solution to the Poisson equation is unique, we can
use the method of images to guess a solution and see if it works — if it does, we
are done.
To guess a solution, we pretend that we don’t have a conductor. Instead, we
have a charge
−q
and
x
=
−d
. Then by symmetry we will get
φ
= 0 when
x
= 0.
The potential of this pair is
φ =
1
4πε
0
"
q
p
(x − d)
2
+ y
2
+ z
2
−
q
p
(x + d)
2
+ y
2
+ z
2
#
.
To get the solution we want, we “steal” part of this potential and declare our
potential to be
φ =
1
4πε
0
q
√
(x−d)
2
+y
2
+z
2
−
q
√
(x+d)
2
+y
2
+z
2
if x > 0
0 if x ≤ 0
Using this solution, we can immediately see that it satisfies Poisson’s equations
both outside and inside the conductor. To complete our solution, we need to find
the surface charge required such that the equations are satisfied on the surface
as well.
To do so, we can calculate the electric field near the surface, and use the
relation
σ
=
ε
0
E
outside
·
ˆ
n
. To find
σ
, we only need the component of
E
in the
x
direction:
E
x
= −
∂φ
∂x
=
q
4πε
0
x − d
r − d
3/2
−
x + d
r + d
3/2
for x > 0. Then induced surface charge density is given by E
x
at x = 0:
σ = E
x
ε
0
= −
q
2π
d
(d
2
+ y
2
+ z
2
)
3/2
.
The total surface charge is then given by
Z
σ dy dz = −q.
3 Magnetostatics
Charges give rise to electric fields, currents give rise to magnetic fields. In this
section, we study the magnetic fields induced by steady currents, i.e. in situations
where J 6= 0 and ρ = 0. Again, we look for timeindependent solutions.
Since there is no charge, we obtain
E
= 0. The remaining Maxwell’s equations
are
∇ × B = µJ
∇ · B = 0
The objective is, given a J, find the resultant B.
Before we start, what does the condition
ρ
= 0 mean? It does not mean that
there are no charges around. We want charges to be moving around to create
current. What it means is that the positive and negative charges balance out
exactly. More importantly, it stays that way. We don’t have a net charge flow
from one place to another. At any specific point in space, the amount of charge
entering is the same as the amount of charge leaving. This is the case in many
applications. For example, in a wire, all electrons move together at the same
rate, and we don’t have charge building up at parts of the circuit.
Mathematically, we can obtain the interpretation from the continuity equa
tion:
∂ρ
∂t
+ ∇ ·J = 0.
In the case of steady currents, we have ρ = 0. So
∇ · J = 0,
which says that there is no net flow in or out of a point.
3.1 Ampere’s Law
Consider a surface
S
with boundary
C
. Current
J
flows through
S
. We now
integrate the first equation over the surface S to obtain
Z
S
(∇ × B) ·dS =
I
C
B · dr = µ
0
Z
S
J · dS.
So
Law (Ampere’s law).
I
C
B · dr = µ
0
I,
where I is the current through the surface.
Example
(A long straight wire)
.
A wire is a cylinder with current
I
flowing
through it.
We use cylindrical polar coordinates (
r, ϕ, z
), where
z
is along the direction
of the current, and r points in the radial direction.
I
S
z
r
By symmetry, the magnetic field can only depend on the radius, and must lie
in the
x, y
plane. Since we require that
∇ · B
= 0, we cannot have a radial
component. So the general form is
B(r) = B(r)
ˆ
ϕ.
To find
B
(
r
), we integrate over a disc that cuts through the wire horizontally.
We have
I
C
B · dr = B(r)
Z
2π
0
r dϕ = 2πrB(r)
By Ampere’s law, we have
2πrB(r) = µ
0
I.
So
B(r) =
µ
0
I
2πr
ˆ
ϕ.
Example
(Surface current)
.
Consider the plane
z
= 0 with surface current
density k (i.e. current per unit length).
Take the xdirection to be the direction of the current, and the
z
direction to be
the normal to the plane.
We imagine this situation by considering this to be infinitely many copies of
the above wire situation. Then the magnetic fields must point in the
y
direction.
By symmetry, we must have
B = −B(z)
ˆ
y,
with B(z) = −B(−z).
Consider a vertical rectangular loop of length L through the surface
L
Then
I
C
B · dr = LB(z) − LB(−z) = µ
0
kL
So
B(z) =
µ
0
k
2
for z > 0.
Similar to the electrostatic case, the magnetic field is constant, and the part
parallel to the surface is discontinuous across the plane. This is a general result,
i.e. across any surface,
ˆ
n × B
+
−
ˆ
n × B
−
= µ
0
k.
Example
(Solenoid)
.
A solenoid is a cylindrical surface current, usually made
by wrapping a wire around a cylinder.
z
r
C
We use cylindrical polar coordinates with
z
in the direction of the extension of
the cylinder. By symmetry, B = B(r)
ˆ
z.
Away from the cylinder,
∇ × B
= 0. So
∂B
∂r
= 0, which means that
B
(
r
)
is constant outside. Since we know that
B
=
0
at infinity,
B
=
0
everywhere
outside the cylinder.
To compute
B
inside, use Ampere’s law with a curve
C
. Note that only the
vertical part (say of length
L
) inside the cylinder contributes to the integral.
Then
I
C
B · dr = BL = µ
o
INL.
where
N
is the number of wires per unit length and
I
is the current in each wire
(so INL is the total amount of current through the wires).
So
B = µ
0
IN.
3.2 Vector potential
For general current distributions J, we also need to solve ∇ ·B = 0.
Recall that for the electric case, the equation is
∇ · E
=
ρ/ε
0
. For the
B
field, we have 0 on the right hand side instead of
ρ/ε
0
. This is telling us that
there are no magnetic monopoles, i.e. magnetic charges.
The general solution to this equation is B = ∇ × A for some A.
Definition (Vector potential). If B = ∇ × A, then A is the vector potential.
The other Maxwell equation then says
∇ × B = −∇
2
A + ∇(∇·A) = µ
0
J. (∗)
This is rather difficult to solve, but it can be made easier by noting that
A
is
not unique. If A is a vector potential, then for any function χ(x),
A
0
= A + ∇χ
is also a vector potential of B, since ∇× (A + ∇χ) = ∇× A.
The transformation A 7→ A + ∇χ is called a gauge transformation.
Definition
((Coulomb) gauge)
.
Each choice of
A
is called a gauge. An
A
such
that ∇ ·A = 0 is called a Coulomb gauge.
Proposition. We can always pick χ such that ∇ ·A
0
= 0.
Proof.
Suppose that
B
=
∇×A
with
∇·A
=
ψ
(
x
). Then for any
A
0
=
A
+
∇χ
,
we have
∇ · A
0
= ∇A + ∇
2
χ = ψ + ∇
2
χ.
So we need a
χ
such that
∇
2
χ
=
−ψ
. This is the Poisson equation which we
know that there is always a solution by, say, the Green’s function. Hence we can
find a χ that works.
If B = ∇ ×A and ∇ · A = 0, then the Maxwell equation (∗) becomes
∇
2
A = −µ
0
J.
Or, in Cartesian components,
∇
2
A
i
= −µ
0
J
i
.
This is 3 copies of the Poisson equation, which we know how to solve using
Green’s functions. The solution is
A
i
(r) =
µ
0
4π
Z
J
i
(r
0
)
r − r
0

dV
0
,
or
A(r) =
µ
0
4π
Z
J(r
0
)
r − r
0

dV
0
,
both integrating over r
0
.
We have randomly written down a solution of
∇
2
A
=
−µ
0
J
. However, this
is a solution to Maxwell’s equations only if it is a Coulomb gauge. Fortunately,
it is:
∇ · A(r) =
µ
0
4π
Z
J(r
0
) · ∇
1
r − r
0

dV
0
= −
µ
0
4π
Z
J(r
0
) · ∇
0
1
r − r
0

dV
0
Here we employed a clever trick — differentiating 1
/r−r
0

with respect to
r
is the
negative of differentiating it with respect to
r
0
. Now that we are differentiating
against r
0
, we can integrate by parts to obtain
= −
µ
0
4π
Z
∇
0
·
J(r
0
)
r − r
0

−
∇
0
· J(r
0
)
r − r
0

dV
0
.
Here both terms vanish. We assume that the current is localized in some region
of space so that
J
=
0
on the boundary. Then the first term vanishes since it is
a total derivative. The second term vanishes since we assumed that the current
is steady (∇ · J = 0). Hence we have Coulomb gauge.
Law (BiotSavart law). The magnetic field is
B(r) = ∇ × A =
µ
0
4π
Z
J(r
0
) ×
r − r
0
r − r
0

3
dV
0
.
If the current is localized on a curve, this becomes
B =
µ
0
I
4π
I
C
dr
0
×
r − r
0
r − r
0

3
,
since J(r
0
) is nonzero only on the curve.
3.3 Magnetic dipoles
According to Maxwell’s equations, magnetic monopoles don’t exist. However, it
turns out that a localized current looks like a dipole from far far away.
Example
(Current loop)
.
Take a current loop of wire
C
, radius
R
and current
I.
Based on the fields generated by a straight wire, we can guess that
B
looks like
this, but we want to calculate it.
By the BiotSavart law, we know that
A(r) =
µ
0
4π
Z
J(r
0
)
r − r
0

dV =
µ
0
I
4π
I
C
dr
0
r − r
0

.
Far from the loop, r − r
0
 is small and we can use the Taylor expansion.
1
r − r
0

=
1
r
+
r · r
0
r
3
+ ···
Then
A(r) =
µ
0
I
4π
I
C
1
r
+
r · r
0
r
3
+ ···
dr
0
.
Note that
r
is a constant of the integral, and we can take it out. The first
1
r
term vanishes because it is a constant, and when we integrate along a closed
loop, we get 0. So we only consider the second term.
We claim that for any constant vector g,
I
C
g ·r
0
dr
0
= S × g,
where
S
=
R
d
S
is the vector area of the surface bounded by
C
. It follows from
Green’s theorem for the function f(r
0
):
I
C
f(r
0
)dr
0
=
Z
S
∇f × dS,
taking f (r
0
) = g ·r
0
. Then
I
C
g ·r
0
dr
0
=
Z
S
g ×dS = S ×g.
Using this, we have
A(r) ≈
µ
0
4π
m × r
r
3
,
where
Definition (Magnetic dipole moment). The magnetic dipole moment is
m = IS.
Then
B = ∇ ×A =
µ
0
4π
3(m ·
ˆ
r)
ˆ
r − m
r
3
.
This is the same form as
E
for an electric dipole, except that there is no 1
/r
2
leading term here, because we have no magnetic monopoles.
After doing it for a loop, we can do it for a general current distribution:
Example. We have
A
i
(r) =
µ
0
4π
Z
J
i
(r
0
)
r − r
0

dV
0
=
µ
0
4π
Z
J
i
(r
0
)
r
−
J
i
(r
0
)(r · r
0
)
r
3
+ ···
dV
0
.
We will show that the first term vanishes by showing that it is a total derivative.
We have
∂
0
j
(J
j
r
0
i
) = (∂
0
j
J
j
)
 {z }
=∇·J=0
r
0
i
+ (∂
0
j
r
0
i
)
{z}
=δ
ij
J
j
= J
i
.
For the second term, we look at
∂
0
j
(J
j
r
0
i
r
0
k
) = (∂
0
j
J
j
)r
0
i
r
0
k
+ J
i
r
0
k
+ J
k
r
0
i
= J
k
r
0
i
+ J
i
r
0
k
.
Apply this trick to
Z
J
i
r
j
r
0
j
dV =
Z
r
j
2
(J
i
r
0
j
− J
j
r
0
i
) dV
0
,
where we discarded a total derivative
∂
0
j
(
J
j
r
0
i
r
0
u
). Putting it back in vector
notation,
Z
J
i
r · r
0
dV =
Z
1
2
(J
i
(r · r
0
) − r
i
(J · r
0
)) dV
=
Z
1
2
[r × (J ×r
0
)]
i
dV.
So the longdistance vector potential is again
A(r) =
µ
0
4π
m × r
r
3
,
with
Definition (Magnetic dipole moment).
m =
1
2
Z
r
0
× J(r
0
) dV
0
.
3.4 Magnetic forces
We’ve seen that moving charge produces currents which generates a magnetic
field. But we also know that a charge moving in a magnetic field experiences a
force F = qr × B. So two currents will exert a force on each other.
Example (Two parallel wires).
d
z
x
y
We know that the field produced by each current is
B
1
=
µ
0
I
2πr
ˆϕ.
The particles on the second wire will feel a force
F = qv × B
1
= qv ×
µ
0
I
1
2πd
ˆ
y.
But
J
2
=
nqv
and
I
2
=
J
2
A
, where
n
is the density of particles and
A
is the
crosssectional area of the wire. So the number of particles per unit length is
nA, and the force per unit length is
nAF =
µ
0
I
1
I
2
2πd
ˆ
z ×
ˆ
y = −µ
0
I
1
I
2
2πd
ˆ
x.
So if
I
1
I
2
>
0, i.e. the currents are in the same direction, the force is attractive.
Otherwise the force is repulsive.
Example
(General force)
.
A current
J
1
, localized on some closed curve
C
1
, sets
up a magnetic field
B(r) =
µ
0
I
1
4π
I
C
1
dr
1
×
r − r
1
r − r
1

3
.
A second current J
2
on C
2
experiences the Lorentz force
F =
Z
J
2
(r) × B(r) dV.
While we are integrating over all of space, the current is localized at a curve
C
2
.
So
F = I
2
I
C
2
dr
2
× B(r
2
).
Hence
F =
µ
0
4π
I
1
I
2
I
C
1
I
C
2
dr
2
×
dr
1
×
r
2
− r
1
r
2
− r
1

3
.
For wellseparated currents, approximated by
m
1
and
m
2
, we claim that the
force can be written as
F =
µ
0
4π
∇
3(m
1
·
ˆ
r)(m
2
·
ˆ
r) − (m
1
· m
2
)
r
3
,
whose proof is too complicated to be included.
Note that we’ve spent a whole chapter discussing “magnets”, but they are
nothing like what we stick on our fridges. It turns out that these “real” magnets
are composed of many tiny aligned microscopic dipoles arising from electron
spin. However, obviously we do not care about magnets in real life.
4 Electrodynamics
So far, we have only looked at fields that do not change with time. However, in
real life, fields do change with time. We will now look at timedependent
E
and
B fields.
4.1 Induction
We’ll explore the Maxwell equation
∇ × E +
∂B
∂t
= 0.
In short, if the magnetic field changes in time, i.e.
∂B
∂t
6
= 0, this creates an
E
that accelerates charges, which creates a current in a wire. This process is called
induction. Consider a wire, which is a closed curve C, with a surface S.
We integrate over the surface S to obtain
Z
S
(∇ × E) ·dS = −
Z
S
∂B
∂t
· dS.
By Stokes’ theorem and commutativity of integration and differentiation (assum
ing S and C do not change in time), we have
Z
C
E · dr = −
d
dt
Z
S
B · dS.
Definition (Electromotive force (emf)). The electromotive force (emf) is
E =
Z
C
E · dr.
Despite the name, this is not a force! We can think of it as the work done on a
unit charge moving around the curve, or the “voltage” of the system.
For convenience we define the quantity
Definition (Magnetic flux). The magnetic flux is
Φ =
Z
S
B · dS.
Then we have
Law (Faraday’s law of induction).
E = −
dΦ
dt
.
This says that when we change the magnetic flux through
S
, then a current
is induced. In practice, there are many ways we can change the magnetic flux,
such as by moving bar magnets or using an electromagnet and turning it on and
off.
The minus sign has a significance. When we change a magnetic field, an emf
is created. This induces a current around the wire. However, we also know that
currents produce magnetic fields. The minus sign indicates that the induced
magnetic field opposes the initial change in magnetic field. If it didn’t and the
induced magnetic field reinforces the change, we will get runaway behaviour and
the world will explode. This is known as Lenz’s law.
Example. Consider a circular wire with a magnetic field perpendicular to it.
If we decrease
B
such that
˙
Φ <
0, then
E >
0. So the current flows anticlockwise
(viewed from above). The current generates its own
B
. This acts to increase
B
inside, which counteracts the initial decrease.
This means you don’t get runaway behaviour.
There is a related way to induce a current: keep B fixed and move wire.
Example.
d
Slide the bar to the left with speed
v
. Each charge
q
will experience a Lorentz
force
F = qvB,
in the counterclockwise direction.
The emf, defined as the work done per unit charge, is
E = vBd,
because work is only done for particles on the bar.
Meanwhile, the change of flux is
dΦ
dt
= −vBd,
since the area decreases at a rate of −vd.
We again have
E = −
dΦ
dt
.
Note that we obtain the same formula but different physics — we used Lorentz
force law, not Maxwell’s equation.
Now we consider the general case: a moving loop
C
(
t
) bounding a surface
S(t). As the curve moves, the curve sweeps out a cylinder S
c
.
S(t + δt)
S(t)
S
c
The change in flux
Φ(t + δt) −Φ(t) =
Z
S(t+δt)
B(t + δt) ·dS −
Z
S(t)
B(t) · dS
=
Z
S(t+δt)
B(t) +
∂B
∂t
· dS −
Z
S(t)
B(t) · dS + O(δt
2
)
= δt
Z
S(t)
∂B
∂t
· dS +
"
Z
S(t+δt)
−
Z
S(t)
#
B(t) · dS + O(δt
2
)
We know that
S
(
t
+
δt
),
S
(
t
) and
S
c
together form a closed surface. Since
∇ · B = 0, the integral of B over a closed surface is 0. So we obtain
"
Z
S(t+δt)
−
Z
S(t)
#
B(t) · dS +
Z
S
c
B(t) · dS = 0.
Hence we have
Φ(t + δt) −Φ(t) = δt
Z
S(t)
∂B
∂t
· dS −
Z
S
c
B(t) · dS = 0.
We can simplify the integral over S
c
by writing the surface element as
dS = (dr ×v) δt.
Then B ·dS = δt(v × B) · dr. So
dΦ
dt
= lim
δ→0
δΦ
δt
=
Z
S(t)
∂B
∂t
· dS −
Z
C(t)
(v ×B) · dr.
From Maxwell’s equation, we know that
∂B
∂t
= −∇ × E. So we have
dΦ
dt
= −
Z
C
(E + v × B) dr.
Defining the emf as
E =
Z
C
(E + v × B) dr,
we obtain the equation
E = −
∂Φ
∂t
for the most general case where the curve itself can change.
4.2 Magnetostatic energy
Suppose that a current
I
flows along a wire
C
. From magnetostatics, we know
that this gives rise to a magnetic field B, and hence a flux Φ given by
Φ =
Z
S
B · dS,
where S is the surface bounded by C.
Definition (Inductance). The inductance of a curve C, defined as
L =
Φ
I
,
is the amount of flux it generates per unit current passing through
C
. This is a
property only of the curve C.
Inductance is something engineers care a lot about, as they need to create
real electric circuits and make things happen. However, us mathematicians find
these applications completely pointless and don’t actually care about inductance.
The only role it will play is in the proof we perform below.
Example
(The solenoid)
.
Consider a solenoid of length
`
and crosssectional
area A (with `
√
A so we can ignore end effects). We know that
B = µ
0
IN,
where
N
is the number of turns of wire per unit length and
I
is the current. The
flux through a single turn (pretending it is closed) is
Φ
0
= µ
0
INA.
So the total flux is
Φ = Φ
0
N` = µ
0
IN
2
V,
where V is the volume, A`. So
L = µ
0
N
2
V.
We can use the idea of inductance to compute the energy stored in magnetic
fields. The idea is to compute the work done in building up a current.
As we build the current, the change in current results in a change in magnetic
field. This produces an induced emf that we need work to oppose. The emf is
given by
E = −
dΦ
dt
= −L
dI
dt
.
This opposes the change in current by Lenz’s law. In time
δt
, a charge
Iδt
flows
around C. The work done is
δW = EIδt = −LI
dI
dt
δt.
So
dW
dt
= −LI
dI
dt
= −
1
2
L
dI
2
dt
.
So the work done to build up a current is
W =
1
2
LI
2
=
1
2
IΦ.
Note that we dropped the minus sign because we switched from talking about
the work done by the emf to the work done to oppose the emf.
This work done is identified with the energy stored in the system. Recall
that the vector potential A is given by B = ∇ ×A. So
U =
1
2
I
Z
S
B · dS
=
1
2
I
Z
S
(∇ × A) ·dS
=
1
2
I
I
C
A · dr
=
1
2
Z
R
3
J · A dV
Using Maxwell’s equation ∇ × B = µ
0
J, we obtain
=
1
2µ
0
Z
(∇ × B) ·A dV
=
1
2µ
0
Z
[∇ · (B ×A) + B · (∇ × A)] dV
Assuming that
B × A
vanishes sufficiently fast at infinity, the integral of the
first term vanishes. So we are left with
=
1
2µ
0
Z
B · B dV.
So
Proposition. The energy stored in a magnetic field is
U =
1
2µ
0
Z
B · B dV.
In general, the energy stored in E and B is
U =
Z
ε
0
2
E · E +
1
2µ
0
B · B
dV.
Note that while this is true, it does not follow directly from our results for pure
magnetic and pure electric fields. It is entirely plausible that when both are
present, they interact in weird ways that increases the energy stored. However,
it turns out that this does not happen, and this formula is right.
4.3 Resistance
The story so far is that we change the flux, an emf is produced, and charges
are accelerated. In principle, we should be able to compute the current. But
accelerating charges are complicated (they emit light). Instead, we invoke a new
effect, friction.
In a wire, this is called resistance. In most materials, the effect of resistance
is that
E
is proportional to the speed of the charged particles, rather than the
acceleration.
We can think of the particles as accelerating for a very short period of time,
and then reaching a terminal velocity. So
Law (Ohm’s law).
E = IR,
Definition (Resistance). The resistance is the R in Ohm’s law.
Note that
E
=
R
E ·
d
r
and
E
=
−∇φ
. So
E
=
V
, the potential difference.
So Ohm’s law can also be written as V = IR.
Definition
(Resistivity and conductivity)
.
For the wire of length
L
and cross
sectional area A, we define the resistivity to be
ρ =
AR
L
,
and the conductivity is
σ =
1
ρ
.
These are properties only of the substance and not the actual shape of the
wire. Then Ohm’s law reads
Law (Ohm’s law).
J = σE.
We can formally derive Ohm’s law by considering the field and interactions
between the electron and the atoms, but we’re not going to do that.
Example.
d
z
x
y
Suppose the bar moves to the left with speed
v
. Suppose that the sliding bar
has resistance
R
, and the remaining parts of the circuit are superconductors
with no resistance.
There are two dynamical variables, the position of the bar
x
(
t
), and the
current I(t).
If a current I flows, the force on a small segment of the bar is
F = IB
ˆ
y ×
ˆ
z
So the total force on a bar is
F = IB`
ˆ
x.
So
m¨x = IB`.
We can compute the emf as
E = −
dΦ
dt
= −B` ˙x.
So Ohm’s law gives
IR = −B` ˙x.
Hence
m¨x = −
B
2
`
2
R
˙x.
Integrating once gives
˙x(t) = −ve
−B
2
`
2
t/mR
.
With resistance, we need to do work to keep a constant current. In time
δt
,
the work needed is
δW = EIδt = I
2
Rδt
using Ohm’s law. So
Definition
(Joule heating)
.
Joule heating is the energy lost in a circuit due to
friction. It is given by
dW
dt
= I
2
R.
4.4 Displacement currents
Recall that the Maxwell equations are
∇ · E =
ρ
ε
0
∇ · B = 0
∇ × E = −
∂B
∂t
∇ × B = µ
0
J + ε
0
∂E
∂t
So far we have studied all the equations apart from the
µ
0
ε
0
∂E
∂t
term. Historically
this term is called the displacement current.
The need of this term was discovered by purely mathematically, since people
discovered that Maxwell’s equations would be inconsistent with charge conserva
tion without the term.
Without the term, the last equation is
∇ × B = µ
0
J.
Take the divergence of the equation to obtain
µ
0
∇ · J = ∇ · (∇ ×B) = 0.
But charge conservation says that
˙ρ + ∇·J = 0.
These can both hold iff
˙ρ
= 0. But we clearly can change the charge density —
pick up a charge and move it elsewhere! Contradiction.
With the new term, taking the divergence yields
µ
0
∇ · J + ε
0
∇ ·
∂E
∂t
= 0.
Since partial derivatives commute, we have
ε
0
∇ ·
∂E
∂t
= ε
0
∂
∂t
(∇ · E) = ˙ρ
by the first Maxwell’s equation. So it gives
∇ · J + ˙ρ = 0.
So with the new term, not only is Maxwell’s equation consistent with charge
conservation — it actually implies charge conservation.
4.5 Electromagnetic waves
We now look for solutions to Maxwell’s equation in the case where
ρ
= 0 and
J = 0, i.e. in nothing/vacuum.
Differentiating the fourth equation with respect to time,
µ
0
ε
0
∂
2
E
∂t
2
=
∂
∂t
(∇ × B)
= ∇ ×
∂B
∂t
= ∇( ∇ · E

{z}
=ρ/ε
0
=0
) + ∇
2
E by vector identities
= ∇
2
E.
So each component of E obeys the wave equation
1
c
2
∂
2
E
∂t
2
− ∇
2
E = 0.
We can do exactly the same thing to show that B obeys the same equation:
1
c
2
∂
2
B
∂t
2
− ∇
2
B = 0,
where the speed of the wave is
c =
1
√
µ
0
ε
0
Recall from the first lecture that
– ε
0
= 8.85 × 10
−12
m
−3
kg
−1
s
2
C
2
– µ
0
= 4π × 10
−6
m kg C
−2
So
c = 3 ×10
8
m s
−1
,
which is the speed of light!
We now look for plane wave solutions which propagate in the
x
direction,
and are independent of y and z. So we can write our electric field as
E(x) = (E
x
(x, t), E
y
(x, t), E
z
(x, t)).
Hence any derivatives wrt
y
and
z
are zero. Since we know that
∇ · E
= 0,
E
x
must be constant. We take
E
x
= 0. wlog, assume
E
z
= 0, ie the wave propagate
in the
x
direction and oscillates in the
y
direction. Then we look for solutions of
the form
E = (0, E(x, t), 0),
with
1
c
2
∂
2
E
∂t
2
−
∂
2
E
∂x
2
= 0.
The general solution is
E(x, t) = f(x − ct) + g(x + ct).
The most important solutions are the monochromatic waves
E = E
0
sin(kx − ωt).
Definition (Amplitude, wave number and frequency).
(i) E
0
is the amplitude
(ii) k is the wave number.
(iii) ω is the (angular) frequency.
The wave number is related to the wavelength by
λ =
2π
k
.
Since the wave has to travel at speed c, we must have
ω
2
= c
2
k
2
So the value of k determines the value of ω, vice versa.
To solve for B,we use
∇ × E = −
∂B
∂t
.
So B = (0, 0, B) for some B. Hence the equation gives.</