Chapter 1
Maxwell's Equations and Electromagnetic Fields
1.1 Introduction
1.1.1 Maxwell's Equations (1865)
The governing equations of electromagnetism
 

 
 

 
 

  (1.1) 
 

∇∧B
= μ_{0} j + 
1
c^{2}


∂E
∂t


 
 

 

E  electric field, describes the force felt by a (stationary)
charge q: F = q E 
B  magnetic field, describes the force felt by a current 
 i.e. a moving charge (velocity v): F = q v∧B 
Thus the Lorentz Force (on charge q) is
Figure 1.1:
Charge density is local charge per
unit volume. Current density is current per unit area.
ρ  electric charge density (Coulombs/m^{3}).
Total charge Q = ∫_{V} ρd^{3}x 
j  electric current density (Coulombs/s/m^{2}) 
 Current crossing area element dA is j . dA
Coulomb/s = Amps. 
[Note. Sometimes the first Maxwell equation is called Gauss's
rather than Coulomb's Law.]
1.1.2 Historical Note
Much scientific controversy in 2nd half of 19^{th} century concerned
question of whether E, B were `real' physical quantities
of science or else mere mathematical conveniences for expressing
the forces that charges exert on one another. English science
(Faraday, Maxwell) emphasized the fields; German mostly the
actatadistance. Since ∼ 1900 this question has been
regarded as settled in favor of the fields. And modern
physics, if anything, tends to regard the field as more
fundamental than the particle.
1.1.3 Auxiliary Fields and Electromagnetic Media
Electromagnetic texts often discuss two additional "auxiliary"
fields D the "electric displacement" and H
the "magnetic intensity" which account for dielectric and
magnetic properties of materials. These fields are not fundamental
and introduce unnecessary complication and possible confusion
for most of our topics. Therefore we will avoid them as much
as possible. For the vacuum, ϵ_{0} E
= D and
B
= μ_{0} H.
1.1.4 Units
Historically there were two (or more!) different
systems of units, one defining the quantity of charge in terms
of the force between two stationary charges (the
"Electrostatic" units) and one defining it in terms of forces
between (chargeless) currents (the "Electromagnetic"
system). Electrostatic units are based on Coulomb's law
∇. E
= ρ/ϵ_{0} and electromagnetic units on
the (steadystate version of) Ampere's law ∇∧B
=μ_{0} j. The quantities 1/ϵ_{0} and μ_{0} are
therefore fundamentally calibration factors that determine the
size of the unit charge. Choosing one or other of them to be 4π
amounts to choosing electrostatic or electromagnetic units.
However, with the unification of electromagnetism, and the
subsequent realization that the speed of light is a fundamental
constant, it became clear that the units of electromagnetism
ought to be defined in terms of only one of these laws and the speed
of light.
Therefore the "System Internationale" SI (or sometimes MKSA)
units adopts the electromagnetic definition because it can be
measured most easily, but with a different μ_{0}, as
follows.
"One Ampere is that current which, when flowing in two
infinitesimal parallel wires 1m apart produces a force of
2 ×10^{−7} Newtons per meter of their length."
An Amp is one Coulomb per second. So this defines the unit
of charge. We will show later that this definition amounts to
defining
μ_{0} = 4π×10^{−7} (Henry/meter) 
 (1.3) 
and that because the ratio of electromagnetic to electrostatic
units is c^{2}
ϵ_{0} = 
1
c^{2} μ_{0}

= 8.85 ×10^{−12} (Farad/meter) 
 (1.4) 
μ_{0} is called the "permeability of free space".
ϵ_{0} is called the "permittivity of free space".
[See J.D. Jackson 3rd Ed, Appendix for a detailed discussion.]
1.2 Vector Calculus and Notation
Electromagnetic quantities include vector fields E,B etc. and
so EM draws heavily on vector calculus.
∇ is shorthand for a vector operator (gradient)
∇ϕ =  ⎛ ⎝

∂ϕ
∂x

, 
∂ϕ
∂y

, 
∂ϕ
∂z
 ⎞ ⎠

= 
∂ϕ
∂x_{i}

(suffix notation) 
 (1.5) 
giving a vector gradient from a scalar field ϕ.
∇ can also operate on vector fields by scalar (.) or vector
(∧) multiplication.
1.2.1 Divergence
∇. E= 
∂E_{x}
∂x

+ 
∂E_{y}
∂y

+ 
∂E_{z}
∂z

= 
∂E_{i}
∂x_{i}


 (1.6) 
1.2.2 Curl
∇∧E=  ⎛ ⎝

∂E_{z}
∂y

− 
∂E_{y}
∂z

, 
∂E_{x}
∂z

− 
∂E_{z}
∂x

, 
∂E_{y}
∂x

− 
∂E_{x}
∂y
 ⎞ ⎠

= ϵ_{ijk} 
∂E_{k}
∂x_{j}


 (1.7) 
1.2.3 Volume Integration
d^{3}x is shorthand for dxdydz = dV, the volume element.
Figure 1.2:
Elements for surface and line integrals.
1.2.4 Surface Integration
The surface element dA or often dS is a vector normal to the element.
1.2.5 Line (Contour) Integration
Line element dl.
1.2.6 The Meaning of divergence: ∇.
Figure 1.3:
Cartesian volume element.
Consider a volume element.
Evaluate the total flux of a vector field v out across the
element's surface.
It is the sum v.dA over the six faces of the cuboid

[v_{x} (x + dx/2) − v_{x} (x−dx/2)] dydz+ 

 

[v_{y} (y + dy/2) − v_{y} (y−dy/2) ] dzdx+ 

 

[v_{z} (z + dz/2) − v_{z} (z−dz/2)] dxdy 


  (1.11) 

dxdydz  ⎡ ⎣

dv_{x}
dx

+ 
dv_{y}
dy

+ 
dv_{z}
dz
 ⎤ ⎦




 

So for this elemental volume:
 ⌠ ⌡

dS

v. dA=  ⌠ ⌡

dV

∇. v d^{3}x 
 (1.12) 
Figure 1.4:
Adjacent faces cancel out in
the sum of divergence from many elements.
But any arbitrary finite volume can be considered to be
the sum of many small cuboidal elements.
Adjacent internal face contribution cancel out
hence only the external surface contributions remain, so
 ⌠ ⌡

S

v. dA=  ⌠ ⌡

V

∇. v d^{3}x 
 (1.13) 
for any volume V with surface S, and arbitrary vector
field v.
This is Gauss's Theorem.
1.2.7 The Meaning of Curl: ∇∧
Figure 1.5:
Rectangular surface element with axes
chosen such that the normal is in the
zdirection.
Consider an arbitrary rectangular surface element
Choose axes such that normal is in zdirection and edges along
x and y. Arbitrary vector field v(x).
Evaluate the contour integral of v, clockwise around dC (relative
to the direction ∧z), round
boundary of element, regarding the center as x.
 

v(x,y−dy/2). dx + v(x+dx/2,y) . dy 
  (1.14) 
 

v(x,y+dy/2) . (−dx)+ v(x−dx/2,y) . (−dy) 
 
 

−dv_{x}dx + dv_{y}dy =  ⎛ ⎝

∂v_{y}
∂x

− 
∂v_{x}
∂y
 ⎞ ⎠

dxdy 
 
 

(∇∧v)_{z} dA = (∇∧v) . dA . 
  (1.15) 

So integral v.d l around element is equal to the curl
scalarproduct area element.
Figure 1.6:
Arbitrary surface may be divided into
the sum of many rectangular elements. Adjacent edge integral
contributions cancel.
Apply to arbitrary surface;
divide surface up into many elements dA. All internal
edge integrals cancel.
Hence
 ⌠ (⎜) ⌡

C

v.dl=  ⌠ ⌡

S

( ∇∧v) . dA 
 (1.16) 
This is Stokes' Theorem.
1.3 Electrostatics and Gauss' Theorem
Gauss's theorem is the key to understanding electrostatics in
terms of Coulomb's Law ∇.E
= ρ/ ϵ_{0}.
1.3.1 Point Charge q
Apply Gauss's Theorem to a sphere surrounding q
Figure 1.7:
Spherical volume, V over
which we perform an integral of Coulomb's law to deduce
E.
 ⌠ ⌡

S

E. dA=  ⌠ ⌡

V

∇. E d^{3}x =  ⌠ ⌡

V


ρ
ϵ_{0}

d^{3}x = 
q
ϵ_{0}

. 
 (1.17) 
But by spherical symmetry E must be in radial direction and
E_{r} has magnitude constant over the sphere.
Hence
∫_{S} E. dA
= ∫_{S} E_{r}dA = E_{r} ∫_{S} dA = E_{r} 4πr^{2}.
Thus
E_{r}4πr^{2} = 
q
ϵ_{0}

or 
 (1.18) 
E_{r} = 
q
4πϵ_{0} r^{2}

i.e. E
= 
q
4πϵ_{0}


r
r^{3}

. 
 (1.19) 
Consequently, force on a second charge at distance r is
F = 
q_{1}q_{2}
4πϵ_{0}r^{2}


 (1.20) 
Coulomb's inversesquarelaw of electrostatic force.
1.3.2 Spherically Symmetric Charge ρ(r)
Notice that pointcharge derivation depended only on symmetry.
So for a distributed chargedensity that is symmetric argument
works just the same i.e.
E= 
^
r

E_{r} E_{r} = 
q
4πϵ_{0}r^{2}


 (1.21) 
where now q =  ⌠ ⌡

V

ρ d^{3}x =  ⌠ ⌡

r
0

ρ(r) 4πr^{2}dr . 
 (1.22) 
Electric field due to a spherically symmetric charge density
is equal to that of a point charge of magnitude equal to the
total charge within the radius, placed at the spherical
center.
1.3.3 Arbitrary Charge Distribution
If there is no specific symmetry Gauss's Theorem still applies:
Figure 1.8:
Arbitrary volume for Gauss's Theorem.
 ⌠ ⌡

S

E. dA=  ⌠ ⌡

V

∇. E d^{3}x =  ⌠ ⌡

V


ρ
ϵ_{0}

d^{3}x = 
q
ϵ_{0}


 (1.23) 
q is the total charge (integral of charge density) over the volume.
∫_{S} E. dA is the total flux of electric field across
the surface S.
1.3.4 Intuitive Picture
Figure 1.9:
Intuitive picture of charges and
fieldlines.
Each (+ve) charge is the origination point of an electricfieldline.
[Each −ve charge is termination of a field line]. The total charge in
volume V determines the number of fieldlines that start in V.
Field lines only start/end on charges (Coulomb's Law) so all must
escape from the volume, crossing surface S (somewhere).
[Field lines that start and end in V contribute neither
to ∫E. dA nor to q, because of cancellation].
∫_{S} E. dA can be thought of as counting the
"number of field lines" crossing the surface.
[Of course it is an arbitrary choice how big we consider the
charge is that gives rise to one fieldline.]
Intuitive view of electric field "intensity":
Strength of E is proportional to the number of fieldlines
per unit area. (Fig 1.10).
Figure 1.10:
Spacing of fieldlines is
inversely proportional to fieldstrength.
All these intuitive views are conceptually helpful but are not
formally necessary. Electromagnetism is considered completely
described by Maxwell's equations without need for
these pictures.
1.3.5 Electric Potential (for static problems
[(∂)/(∂t)] → 0)
In the static situation there is no induction and Faraday's law
becomes ∇∧E
= 0.
By the way, this equation could also be derived from the inversesquarelaw
by noting that
Figure 1.11:
Each element contributes an
irrotational component to
E. Therefore the total
E is irrotational.
so by the linearity of the ∇∧ operator the sum (integral)
of all Electric field contributions from any charge distribution is
curlfree "irrotational":
∇∧  ⌠ ⌡


ρ(r′)
4πϵ_{0}


r−r′
r−r′^{3}

d^{3}r′ = 0 
 (1.25) 
[This shows that the spherical symmetry argument only works in the
absence of induction · B would define a preferred direction;
asymmetric!]
For any vector field E, ∇∧E
= 0 is a necessary
and sufficient condition that E can be written as the gradient
of a scalar E
= − ∇ϕ.
Necessary
( ∇∧∇ϕ)_{z} = 
∂
∂x


∂ϕ
∂y

− 
∂
∂y


∂ϕ
∂x

= 0 (et sim x, y) . 
 (1.26) 
Curl of a gradient is zero.
Sufficient (prove by construction)
Figure 1.12:
Two different paths from 0
to
x construct a closed contour when one is reversed.
Apply Stokes' theorem to a closed contour consisting of any 2
paths between points 0 and x.
 ⌠ (⎜) ⌡

C

E. dl= 
Path 2

− 
Path 1

=  ⌠ ⌡

S

∇∧E. dS 
= 0 by hypothesis.


 (1.27) 
So ∇∧E
= 0 ⇒ ∫_{o}^{x}E. dl is
independent of chosen path, i.e. it defines a
unique^{3}
quantity.
Call it −ϕ(x).
Consider ∇ϕ defined as the limit of δϕ between
adjacent points.
− ∇ϕ = ∇  ⎛ ⎝
 ⌠ ⌡

x

E. dl  ⎞ ⎠

= E . 
 (1.28) 
Many electrostatic problems are most easily solved in terms of
the electric potential ϕ because it is a scalar (so easier).
Governing equation:
 

− ∇. ∇ϕ = − ∇^{2} ϕ = 
ρ
ϵ_{0}


  (1.29) 
 


∂^{2}
∂x^{2}

+ 
∂^{2}
∂y^{2}

+ 
∂^{2}
∂x^{2}

is the "Laplacian" operator 
  (1.30) 
 

∇^{2} ϕ = 
−ρ
ϵ_{0}

"Poisson′s Equation". 
  (1.31) 

1.3.6 Potential of a Point Charge [General Potential Solution]
One can show by direct differentiation that
So by our previous expression E
= [q/(4 πϵ_{0})] [(r)/(r^{3})] we can identify
as the potential of a charge q (at the origin x
=0).
1.3.7 Green Function for the Laplacian
For a linear differential operator, L, mathematicians define
something called "Green's function" symbolically by the equation
If we can solve this equation in general, then solutions to
can be constructed for arbitrary ρ as
ϕ(x) =  ⌠ ⌡

G(x, x′) ρ(x′) d^{3}x′ 
 (1.36) 
because of the (defining) property of the δfunction
 ⌠ ⌡

f(x′) δ(x−x′) d^{3}x′ = f(x) . 
 (1.37) 
When L is the Laplacian, ∇^{2}, the Green function is
This fact may be derived directly from the solution for the
potential for a point charge. Indeed, a point charge is exactly
the deltafunction situation whose solution is the Green function.
In other words, the charge density for a point charge of magnitude
q at position x′ is
so the pointcharge potential,
namely,
ϕ = 
q
4 πϵ_{0}


1
x−x′

, 
 (1.40) 
is the solution of the equation:
∇^{2} ϕ = − 
q
ϵ_{0}

δ(x−x′) . 
 (1.41) 
Consequently, solution of Poisson's equation can be written as
the integral of the Green function:
ϕ(x) =  ⌠ ⌡

 ⎛ ⎝

−ρ( x′)
ϵ_{0}
 ⎞ ⎠

 ⎛ ⎝

−1
4πx−x′
 ⎞ ⎠

d^{3} x′ =  ⌠ ⌡


ρ(x′)
4 πϵ_{0}


d^{3}x′
x− x′

. 
 (1.42) 
Informally, the smooth charge distribution ρ can be
approximated as the sum (→ ∫) of many point charges
ρ(x′) d^{3}x′, and the potential is the sum of their
contributions.
1.3.8 Boundary Conditions
Strictly speaking, our solution of Poisson's equation is not unique.
We can always add to ϕ a solution of the homogeneous
(Laplace) equation ∇^{2}ϕ = 0.
The solution only becomes unique when boundary conditions are
specified. The solution
ϕ(x) =  ⌠ ⌡


ρ( x′)
4 πϵ_{0}


d^{3}x′
x− x′


 (1.43) 
is correct when the boundary conditions are that
no applied external field.
In practice most interesting electrostatic calculations involve specific
boundaries. A big fraction of the work is solving Laplace's equation
with appropriate boundary conditions. These are frequently the specification
of ϕ on (conducting) surfaces. The charge density on the conductors
is rarely specified initially.
1.3.9 Parallel Plate Capacitor
Figure 1.13:
The parallelplate capacitor.
Idealize as 1dimensional by ignoring the edge effects.
1d Laplace equation in the vacuum gap (where ρ = 0) is
Solution ϕ = a − Ez a,E const.
Hence electric field is
E
= E ∧z.
Notice how this arises purely from the
translational invariance of ϕ( [d/dx] = [d/dy] = 0 ).
Choose z=0 as one plate of capacitor. Other at z = d.
Choose ϕ(z=0) = 0: reference potential, making a = 0.
Potential of other plate: V = ϕ(d) = −Ed.
The question: how
much charge per unit area is there on the plates when the field is
E?
Figure 1.14:
Elemental volume for
calculating charge/field relationship.
Answer by considering a flat
elemental volume, with area A surrounding the +ve plate. Apply
Gauss's law
 

 
 

 ⌠ ⌡

V


ρ
ϵ_{0}

d^{3}x = 
Q
ϵ_{0}

= 
1
ϵ_{0}

σ A 
  (1.46) 

Hence
σ = − ϵ_{0} E . = ϵ_{0} 
V
d


 (1.47) 
Therefore if the total area is A, the total charge
Q, and the
voltage V between plates are related by
And the coefficient [(ϵ_{0}A)/d] is called capacitance, C.
Notice our approach:
 Solve Laplace's equation by choosing coordinates consistent
with problem symmetry.
 Obtain charge using Gauss's law to an appropriate
trial volume.
1.3.10 Charge on an arbitrary conductor
Consider a conductor, electrostatically charged.
Figure 1.15:
Arbitraryshaped conductor possesses
only surface charges related to the local normal field.
Current is zero.
So E is zero, anywhere inside because of conductivity.
Choose any volume internally: E
= 0 ⇒ ∇. E = 0 ⇒ρ = 0. There is no internal charge. It all resides
on surface.
At the surface there is an E just outside.
E is perpendicular to surface ds because surface is
an equipotential (& E
= − ∇ϕ).
Hence applying Gauss's law to a pill box

 ⌠ ⌡

V

∇. E d^{3}x =  ⌠ ⌡

S

E. dS 


 

=  ⌠ ⌡


ρ
ϵ_{0}

d^{3} x =  ⌠ ⌡


σ
ϵ_{0}

ds 


  (1.49) 

where σ = surface charge density
Hence σ = ϵ_{0} E.
Of course, in this general case E (= E_{normal}) is
not uniform on the surface but varies from place to place.
Again procedure would be: solve ϕ externally from ∇^{2}ϕ = 0;
then deduce σ; rather than the other way around.
1.3.11 Visualizing Electric Potential and Field
Figure 1.16:
Contours of potential and
corresponding fieldlines (marked with arrows). Only the fieldlines
emanating from the larger elliptical conductor are drawn.
Consider a (2D) contour plot of ϕ.
The value of ϕ can be thought
of as the potential energy of a charge of 1 Coulomb. Thus
there is a perfect analogy to gravitational potential energy
and height contours.
The force at any point on the hill is downward (on a +ve charge),
which is perpendicular to the
contours of constant ϕ. The strength of the force (E) is
proportional to the
steepness of the hill: i.e. how close
together the contours are (of ϕ).
When plotting fieldlines, i.e. lines following the electric field
direction, we generally also consider the electric field intensity
to be the number of fieldlines per unit area. So also the closeness
of fieldlines indicates field strength.
In chargefree regions ∇. E
= 0 implies fieldlines
have no beginning or end. However if ρ ≠ 0 then electric
field lines do possibly have ends (on the charges).
The potential contours never have ends.
1.3.12 Complex Potential Representation 2D
In chargefree region, ∇. E
= 0 ⇒ ∇^{2} ϕ = 0.
This causes there to be an intimate relationship between fieldlines and
ϕcontours. In 2 dimensions this relationship allows complex
analysis to be used to do powerful analysis of potential problems.
Consider a complex function f(z) = ϕ(z) + i ψ(z)
where z = x + iy is the complex argument with real and imaginary
parts x & y; and f has real and imaginary parts ϕ & ψ.
f is "analytic" if there exists a well defined complex derivative
[df/dz] (which is also analytic), defined in the usual
way as lim_{z′→ z} ( [(f( z′) − f ( z ))/(z′− z )] ) .
In order for this limit to be the same no matter what direction (x,y)
it is taken in, f must satisfy the "CauchyRiemann relations"

∂ϕ
∂x

= 
∂ψ
∂y

; 
∂ϕ
∂y

= − 
∂ψ
∂x


 (1.51) 
Which, by substitution imply ∇^{2}ϕ = 0, ∇^{2}ψ = 0,
and also
regarding x,y as 2d coordinates. This shows that
 The real part of an analytic function solves ∇^{2}ϕ = 0.
 The contours of the corresponding imaginary part, ψ,
then coincide
with the electric fieldlines.
Finding complex representations of potential problems is one of the most
powerful analytic solution techniques. However, for practical
calculations, numerical solution techniques are now predominant.
1.4 Electric Current in Distributed Media
Ohms law, V = IR, relates voltage current and resistance for
a circuit or discrete element. However we often care not just about
the total current but about the current density
in finitesized conductors (e.g. electromagnets).
This requires a local Ohm's law which is
where η is the medium's electric resistivity.
Often the conductivity σ = 1/η is used.
j = σE, (but I'll try to avoid confusion with
surface charge density σ).
Such a linear relationship applies in most metals.
1.4.1 Steady State Conduction
Conservation of charge can be written
so, in steady state, ∇. j = 0, i.e.
∇.  ⎛ ⎝

1
η

E  ⎞ ⎠

= ( E. ∇) 
1
η

+ 
1
η

∇. E= 0 
 (1.55) 
If conductivity
is uniform (∇[1/(η)] = 0) or invariant
along E, we therefore have ∇. E
= 0 ⇒ ρ = 0.
"Uniform conductivity conductors acquire zero volume charge
density in steady state".
1.4.2 Conductor Boundary Conditions (Steady Currents)
Figure 1.17:
A distributed conductor of finite
conductivity, carrying current.
If currents are flowing so that E ≠ 0 in the conductor
then conductors are not equipotential surfaces for
solutions of Laplace's equation outside.
Surface charges (only) are present on the conductor
(uniform η).
No current flows through the conductor surface (except at
contacts) so
inside conductor, while outside we have
surface charge density.
Normal components
Figure 1.18:
Boundary conditions across a
conductor/vacuum (or insulator) interface.
E_{n}_{inside} = 0 [ E_{n} ]_{inside}^{outside} = σ/ϵ_{0} 
 (1.58) 
Tangential components
[ E_{t} ]^{outside}_{inside} = 0 
 (1.59) 
In particular, for solving ∇^{2} ϕ = 0 inside uniform η
conductor, at conductor boundary:
unlike the usual electrostatic B.C. ϕ = given.
At electrical contacts ϕ given might be appropriate.
A general approach to solving a distributed steadycurrent
problem with uniformη media:
 Solve Laplace's equation ∇^{2}ϕ = 0 inside
conductors using Dirichlet (ϕgiven) or possibly inhomogeneous
Neumann (∇ϕ_{n} = given) BC's at contacts and
∇ϕ. n = 0 at insulating boundaries.
 Solve Laplace's equation ∇^{2} ϕ = 0 outside
conductors using ϕ = given (Dirichlet) B.C. with the ϕ
taken from the internal solution.
1.5 Magnetic Potential
Magnetic field has zero divergence ∇. B
= 0.
For any vector field^{4}
B, ∇. B
= 0 is a necessary and sufficient
condition that B can be written as the curl of a vector potential
B
= ∇∧A.
1.5.1 ∇.B
=0 Necessary
 


∂
∂x

 ⎛ ⎝

∂A_{z}
∂y

(z)

− 
∂A_{y}
∂z

(y)
 ⎞ ⎠


  (1.61) 
 


∂
∂y

 ⎛ ⎝

∂A_{x}
∂z

(x)

− 
∂A_{z}
∂x

(z)
 ⎞ ⎠


  (1.62) 
 


∂
∂z

 ⎛ ⎝

∂A_{y}
∂x

(y)

− 
∂A_{x}
∂y

(x)
 ⎞ ⎠

= 0 
  (1.63) 

So only divergenceless fields can be represented.
1.5.2 ∇.B
=0 Sufficient (outline proof by construction)
Consider the quantity
K(x) =  ⌠ ⌡


B( x′)
4 πx− x′

d^{3} x′ , 
 (1.64) 
a vector constructed from the integral of each Cartesian component
of B. Applying our knowledge of the Green function solution
of Poisson's equation, we know:
Vector operator theorem (for any v):
∇∧( ∇∧v ) = ∇( ∇. v ) − ∇^{2} v . 
 (1.66) 
Hence
B= − ∇^{2} K = ∇∧( ∇∧K) − ∇( ∇. K ) 
 (1.67) 
We have proved Helmholtz's theorem that any vector field can be
represented as the sum of grad + curl.]
When ∇. B
= 0 and B → 0 (fast enough) as
x → ∞, one can show that ∇. K = 0
and so we have constructed the required vector potential
A= ∇∧K = ∇∧  ⌠ ⌡


B( x′)
4 π


d^{3}x′
x− x′

. 
 (1.68) 
Notice that we have constructed A such that
∇. A
= 0. However A is undetermined from B because
we can add to it the gradient of an arbitrary scalar without
changing B, since ∇∧∇χ = 0. So in
effect we can make ∇. A equal any desired quantity
ψ(x) by adding to A ∇χ such that
∇^{2}χ = ψ.
Choosing ∇. A is known as choosing a "Gauge"
∇. A
= 0 is the "Coulomb Gauge".
1.5.3 General Vector Potential Solution (Magnetostatic)
Static Ampere's law ∇∧B
= μ_{0} j.
Now
 

 
 

∇( ∇. A) − ∇^{2} A= − ∇^{2} A (Coulomb Gauge). 
  (1.69) 

Hence Cartesian components of A are solutions of Poisson equation
∇^{2} A_{i} = − μ_{0} j_{i} 
 (1.70) 
Using our general solution of Poisson's equation (see eq 1.38):
A(x) = 
μ_{0}
4π

 ⌠ ⌡


j ( x′)
x− x′

d^{3}x′ 
 (1.71) 
Resulting B:
 

∇∧A= 
μ_{0}
4π

 ⌠ ⌡

∇∧ 
j ( x′)
x− x′

d^{3}x′ 
 
 


μ_{0}
4π

 ⌠ ⌡

− j (x′) ∧∇ 
1
x− x′

d^{3}x′ = 
μ_{0}
4π

 ⌠ ⌡


j( x′) ∧( x− x′)
x− x′^{3}

d^{3}x′ 
  (1.72) 

This is the distributedcurrent version of the law of
Biot and Savart
(dating from ∼ 1820).
For a wire carrying current I the integral over volume j is
replaced by the integral I dl i.e.
B= 
μ_{0}
4π

 ⌠ ⌡

− 
( x− x′)
x− x′^{3}

∧j d^{3} x = 
μ_{0}
4π

 ⌠ ⌡

− 
( x−x′)
x− x′^{3}

∧I dl . 
 (1.73) 
The BiotSavart law gives us a direct means to calculate B
by integrating over j(x′), numerically if necessary.
However this integration bruteforce method is excessively
computationally intensive and if symmetries are present in the problem
we can use them to simplify.
1.5.4 Cartesian Translational Symmetry (2d x,y)
Figure 1.19:
(a) The coordinates with respect
to an infinite straight filament carrying current I, and (b) the contour
and surface for use with Ampere's law.
If we consider a situation where ∂/∂z = 0,
corresponding to infinite straight parallel currents in zdirection
j = j(x,y) ∧z. Our general vector potential solution
shows us immediately that A
= A ∧z, A_{x} = A_{y} = 0.
(Assuming A, B→ 0 at ∞, i.e. no `external'
sources.) That fact tells us that B_{z} = (∇∧A)_{z} = 0.
We can consider the elementary building block of this problem to be
the single infinitesimal filament. Formally j = I δ(x) δ(y).
We could calculate B(x)
by integrating over this filament. Far easier to use
Ampere's Law directly
 ⌠ ⌡

S

( ∇∧B) . ds =  ⌠ ⌡

C

B. dl =  ⌠ ⌡

S

μ_{0} j . ds = μ_{0} I 
 (1.74) 
By symmetry (∫)_{C} B. dl = 2πr B_{θ}
So
Also B_{r} = 0 by applying Gauss's Theorem to a volume (of unit
length in zdir)
0 =  ⌠ ⌡

V

∇. B d^{3}x =  ⌠ ⌡

S

B. dS = 2πr B_{r} 
 (1.76) 
by symmetry.
Thus Maxwell's equations immediately show us what the 2d Green
function solving ∇∧B
= μ_{0} I ∧z δ(x−x′) is
B= 
^
θ


μ_{0} I
2 π


1
 √ 
( x−x′)^{2} + ( y−y′)^{2}



 (1.77) 
Any general j(x,y) can be handled by 2d integration using this
function.
1.5.5 Cylindrical Symmetry (Circular Loops with common axis)
If there exist cylindrical coordinates (r, θ, z) such that
∂/∂θ = 0, j = j ∧θ.
Then by symmetry A
= A_{θ} ∧θ, B_{θ} = 0.
This situation turns out to be soluble analytically but only in terms
of the special functions known as Elliptic Integrals.
Figure 1.20:
Cylindrical Coordinates near a circular
currentcarrying filament.
If
j_{θ} = I δ( r − a ) δ( z ) 
 (1.78) 
then
A_{θ} (r,z) = 
μ_{0}
4 π

2 
⎛ √


 ⎡ ⎣

( 2 − k^{2} ) K(k) − 2 E(k)
k
 ⎤ ⎦


 (1.79) 
where
k^{2} ≡ 
4 ra
( r + a )^{2} + z^{2}


 (1.80) 
and K, E are the complete elliptical integrals of the first
and second kind.
This general form is so cumbersome that it does not make general
analytic calculations tractable but it makes numerical evaluation
easier by using canned routines for K(k) & E(k).
On axis (r=0) the field is much simpler
B= B 
^
z

= 
μ_{0} I
4 π


2 πa^{2}
( z^{2} + a^{2} )^{3/2}


^
z

. 
 (1.81) 
1.5.6 General Property of Symmetry Situations: Flux Function
When there is a symmetry direction, the component of B
perpendicular to that direction can be expressed in terms of a
"flux function".
The magnetic flux between two positions is defined as the Bfield
flux crossing a surface spanning the gap (per unit length if
translational).
Figure 1.21:
Path from a reference point to a
field point defines a surface to which Stokes' theorem is applied, in
a situation of translational symmetry.
Since ∇. B
= 0 it does not matter how the surface
gets from the refpoint to P (provided it stays symmetric).
So the function
is well defined.
For translational (∧z) symmetry, a consequence is
This arises because
ψ =  ⌠ ⌡

S

B. dS =  ⌠ ⌡

∇∧A. dS = A_{z}(P) − A_{z}(0) 
 (1.84) 
So really ψ is identical to the zcomponent of the
vector potential and
 

B_{z} 
^
z

+ ∇∧  ⎛ ⎝

A_{z} 
^
z
 ⎞ ⎠

= B_{z} 
^
z

+ B_{⊥} 
 
 

A_{z} ∇∧ 
^
z

+ ( ∇A_{z} )∧ 
^
z


  (1.85) 
 

− 
^
z

∧∇A_{z} = − 
^
z

∧∇ψ 
 

B_{⊥} is the part of the field perpendicular to ∧z.
There could also be B_{z}.
For cylindrical symmetry some more variations arise from
curvilinear coordinate system. There are even other symmetries, for
example helical!
1.6 Electromagnetism and Magnets
1.6.1 Simple Solenoid
Figure 1.22:
Idealized long solenoid magnet coil.
A `Long' solenoid has a translational symmetry so B is
independent of z, as well as of θ. (Except near
ends).
So
0 = ∇. B= 
1
r


∂
∂r

r B_{r} + 
=0

B_{θ} + 
=0

B_{z} 
 (1.86) 
So
rB_{r} = const. and hence B_{r} = 0 . 
 (1.87) 
Also
∇∧B=  ⎛ ⎝

1
r


∂
∂r

r B_{θ}  ⎞ ⎠


^
z

+  ⎛ ⎝

− 
∂
∂r

B_{z}  ⎞ ⎠


^
θ

= μ_{0} j (steady) 
 (1.88) 
Inside the bore of the magnet, j = 0
so
rB_{θ} = const. and hence B_{θ} = 0 . 
 (1.89) 
(actually if j_{z} = 0 everywhere then B_{θ} = 0 everywhere, as
may be seen immediately from the BiotSavart law). Also

∂B_{z}
∂r

= 0 and hence B_{z} = const . 
 (1.90) 
Use the surface and bounding curve shown and write
 ⌠ ⌡

S

μ_{0} j . dS =  ⌠ ⌡

S

∇∧B. ds =  ⌠ (⎜) ⌡

C

B. dl 
 (1.91) 
So μ_{0} × current per unit length (denoted
J_{θ}) gives
μ_{0} J_{θ} = B_{z inside} − B_{z outside} 
 (1.92) 
But (by same approach) if B = 0 at infinity B_{z outside} = 0.
So, inside
Profile of field in coil:
Figure 1.23:
The field profile within the
conductor region of the coil depends on the currentdensity profile.
is determined by the current density in the coil:
B_{zb} − B_{za} = −  ⌠ ⌡

b
a

μ_{0} j_{θ} dr = − μ_{0} J_{θ} . 
 (1.95) 
(as before).
Notice that all this is independent of coil thickness (b−a).
Coils are usually multiturn so
where n is turns per unit length, I is current in each turn.
1.6.2 Solenoid of Arbitrary CrossSection
Figure 1.24:
Solenoid of arbitrary crosssection.
Consider BiotSavart Law, expressed as vector potential:
A(r) = 
μ_{0}
4π

 ⌠ ⌡


j ( r′)
r− r′

d^{3} r′ . 
 (1.99) 
If currents all flow in azimuthal direction, i.e. j_{z} = 0,
then A_{z} = 0.
⇒ B_{x} = B_{y} = 0 (everywhere.) 
 (1.100) 
Then integral form of Ampere's law is still
B_{z} (inside) = μ_{0} J_{p} 
 (1.101) 
where J_{p} is total current in azimuthal direction per unit length.
1.6.3 Coil Types
(a) Wire (Filament):
Figure 1.25:
Section through a wirewound
magnet coil.
Multiple layers wound on a former.
Usually only for lowfield lowcurrent work.
(b) Tape wound:
Figure 1.26:
Tapewound coils are stacked to
produce a solenoid.
Each coil consists of a spiralwound tape, n_{t} turns.
Many coils stacked to form a solenoid. Say n_{c} coils per unit length
n = n_{t} . n_{c}.
(c) Pancake:
Similar to tape but using square or rectangular conductor.
(Fewer turns/coil).
(d) Plate Coils:
Figure 1.27:
A pictureframe type plate coil
and the configuration of a solenoid.
Each turn is made of plate. Whole is a single
helix (topologically). Plates may be spaced by air or solid
insulator gap. n = n_{c}.
There are many other configurations of electromagnet, designed for a
tremendous variety of applications. Most require numerical
computation to determine the field and its spatial variation.
1.6.4 Magnetic Dipole
Figure 1.28:
Currents localized to a small
region close to the origin, with the field point far away.
The magnetic field from a "localized" current distribution.
Suppose we want the field at a point x which is far from the
currents, in the sense that for all points x′ where
j(x′) is nonnegligible, x′ << x,
(relative to an origin near the currents).
The general formula for A:
A(x) = 
μ_{0}
4π

 ⌠ ⌡


j ( x′)
x− x′

d^{3}x′ 
 (1.102) 
can be approximated by writing

1
x− x′

= 
1
(x^{2} − 2 x. x′+ x^{′2} )^{1/2}

≈ 
1
x

 ⎛ ⎝

1 + 
x. x′
x^{2}

+ ...  ⎞ ⎠


 (1.103) 
so
A ≅ 
μ_{0}
4π


1
x

 ⎡ ⎣
 ⌠ ⌡

j ( x′) d^{3}x′+ 
1
x^{2}

 ⌠ ⌡

x. x′ j ( x′) d^{3}x′  ⎤ ⎦

. 
 (1.104) 
Now we convert these integrals into more convenient expressions
using ∇. j = 0. Actually the first one is zero.
This follows immediately from the identity
∇. ( j x) = x( ∇. j ) +( j . ∇) x= j 
 (1.105) 
(which uses ∇x
= I i.e. ∂x_{i}/∂x_{j} = δ_{ij}, and ∇. j = 0).
So
 ⌠ ⌡

j d^{3}x′ =  ⌠ ⌡

∇′. ( j x′) d^{3}x′ =  ⌠ ⌡

S

x′j . dS = 0 , 
 (1.106) 
for any surface S that encloses all currents so that j = 0 on
S. An intuitive way to see this is that ∫jd^{3}x is the
average velocity of charges, and must be zero.
The second term is simplified using the same identity but being
careful to distinguish between x and x′,
and using notation ∇′ to denote the gradient operator
that operates on x′, j(x′), not
on x.

 ⌠ ⌡

(x. x′) j ( x′) d^{3}x′ 


 ⌠ ⌡

( x. x′) ∇′. ( j x′) d^{3}x′ 
 
 

 ⌠ ⌡

∇′. ( j x′( x. x′)) − x′j . ∇′( x′. x) d^{3}x′ 
 
 

−  ⌠ ⌡

x′j. ( ∇′x′) . x d^{3}x′ 
 
 

−  ⌠ ⌡

x′j . I . x d^{3}x′ = −  ⌠ ⌡

x′( j . x) d^{3} x′ 
  (1.107) 

But
x∧( x′∧j ) = ( j . x) x′− ( x. x′) j . 
 (1.108) 
So
 ⌠ ⌡

x∧( x′∧j ) d^{3}x′ = −2  ⌠ ⌡

x. x′j d^{3} x′ , 
 (1.109) 
by the integral relation just proved. [This identity is true for
any x].
Therefore our approximation for A is
A( x) = − 
μ_{0}
4π


x
x^{3}

∧  ⎛ ⎝

1
2

 ⌠ ⌡

x′∧j ( x′) d^{3} x′  ⎞ ⎠


 (1.110) 
or
A= 
μ_{0}
4π


m ∧x
x^{3}

, 
 (1.111) 
where the Magnetic Dipole Moment of the localized
current distribution is
m ≡ 
1
2

 ⌠ ⌡

x′∧j d^{3}x′ . 
 (1.112) 
We have derived this expression for an arbitrary j distribution;
but if the localized current is a loop current filament,
Figure 1.29:
Currentcarrying loop
integration to give dipole moment.
m = 
1
2

 ⌠ ⌡

x′∧j d^{3}x′ = 
1
2

 ⌠ ⌡

x∧I dl . 
 (1.113) 
If the loop is planar,
where ds
is the element of surface. So m is (current ×
area) for a planar filament.
The magnetic field is obtained from B
= ∇∧A
B= 
μ_{0}
4 π

 ⎡ ⎣

3 
x
x

 ⎛ ⎝

x
x

. m  ⎞ ⎠

− m  ⎤ ⎦


1
x^{3}


 (1.115) 
1.6.5 Revisionist History of Electromagnetic Induction
Michael Faraday first showed the effect of induction: a transient
current can be induced in one circuit by changes in another. This
was ∼ 1830. [Faraday knew no mathematics beyond the idea of
proportionality EMF ∝ rate of change of Bflux].
Suppose history had been different and we knew only the Lorentz
force law:
we could have "proved" the necessity of induction by "pure
thought".
Assume Galilean Invariance: physical laws must be invariant
under changes to moving coordinate systems x′ = x −vt,
t′ = t.
[Universally assumed in Faraday's time. Einstein doesn't come
till 1905!]
Consider a rigid (wire) circuit moved past a magnet:
Figure 1.30:
A rigid coil moving past a
steady magnet.
Each electron in the circuit (revisionist!) feels a Lorentz
force
as it is dragged through the magnetic field. The electric field in the
rest frame of the magnet is zero. And the total
electromotive force (integrated force per unit charge) round the
entire circuit is

1
q

 ⌠ (⎜) ⌡

C

F . dl =  ⌠ (⎜) ⌡

C

v∧B. dl 
 (1.118) 
This is generally a nonzero quantity.
In fact, this quantity can be transformed on the basis of purely geometrical
considerations.
Figure 1.31:
Surface elements in the
application of Gauss's theorem to succeeding instants of time.
Let's calculate the rate of change of total magnetic
flux due to circuit motion, in a static Bfield. Apply
Gauss' theorem to volume shown in Fig 1.31.
 

 ⌠ ⌡

V

∇. B d^{3} x =  ⌠ ⌡

S_{total}

B. dS =  ⌠ ⌡

S′

−  ⌠ ⌡

S

+  ⌠ ⌡

ribbon

B. ds 
 
 

 ⌠ ⌡

S′

B. dS −  ⌠ ⌡

S

B. dS +  ⌠ ⌡

B. ( dl∧vdt ) 
 
 

  (1.119) 

[where dΦ is change in flux].
So

d Φ
dt

= −  ⌠ (⎜) ⌡

C

( v∧B) . dl . 
 (1.120) 
(pure geometry when ∂B/∂t = 0).
This equation can alternatively be obtained algebraically by writing

d Φ
dt

=  ⌠ ⌡


dB
dt

. dS =  ⌠ ⌡

( v. ∇) B. dS 
 (1.121) 
and using
∇∧( B∧v) = ( v. ∇) B+ (∇.v) B− (B.∇) v− ( ∇. B) v = ( v. ∇) B . 
 (1.122) 
So

dΦ
dt

=  ⌠ ⌡

∇∧( B∧v) .dS =  ⌠ (⎜) ⌡

C

( B∧v) . dS 
 (1.123) 
Anyway EMF is

1
q

 ⌠ (⎜) ⌡

C

F . dl =  ⌠ (⎜) ⌡

C

(v∧B) . dS = − 
dΦ
dt


 (1.124) 
Now we consider the whole situation when the frame of reference
is changed to one in which the circuit is stationary and the magnet
is moving.
By Galilean invariance the total EMF is the same, and

1
q

 ⌠ (⎜) ⌡
 F . dl = − 
dΦ
dt


 (1.125) 
But now v
= 0, and instead B is changing so
In this case also the Lorentz force on the charges is
F = q ( E + v∧B) = q E (since v
=0) 
 (1.127) 
There has to be an electric field in this frame of reference.
And also

1
q

 ⌠ (⎜) ⌡
 F . dl =  ⌠ (⎜) ⌡
 E . dl = − 
dΦ
dt

= −  ⌠ ⌡


∂B
∂t

. dS 
 (1.128) 
Apply Stokes' theorem to the E . dl integral:
 ⌠ ⌡

S

 ⎡ ⎣

∇∧E + 
∂B
∂t
 ⎤ ⎦

. dS = 0 . 
 (1.129) 
But this integral has to be zero for all S (and C) which can be true only
if its integrand is everywhere zero:
"Faraday's" Law (expressed in differential form) (which Faraday
understood intuitively but could not have formulated in math)
1.6.6 Inductance
Suppose we have a set of circuits with currents I_{i} (i = 1 ... N).
These are inductively coupled if the current in one gives rise to flux
linking the others. Because Ampere's law is linear (B ∝ j),
the flux linking circuit j from current I_{i} is proportional to
I_{i}. Consequently, the total flux linking circuit j can be written
(Summation over I_{i}) different currents.
M is a matrix. The element M_{ij} is an inductance
between currents i and j..
Its units are

flux
current

↔ 
Wb
A

↔ Henrys . 
 (1.132) 
The electromotive force or voltage V_{j} induced in the j'th
circuit is then:
V_{j} = 
d
dt

Φ_{j} = 
∑
i

M_{ji} 
⋅
I

i

. 
 (1.133) 
For the simplest case N=1 circuit. M_{ii} → L
the self inductance
It can be shown from Maxwell's equations that M_{ij} is
symmetric.