3Forces

IA Dynamics and Relativity

3.5 Electromagnetism
Next we will study the laws of electromagnetism. We will only provide a
very rudimentary introduction to electromagnetism. Electromagnetism will be
examined more in-depth in the IB Electromagnetism and II Electrodynamics
courses.
As the name suggests, electromagnetism comprises two parts electricity
and magnetism. Similar to gravity, we generally imagine electromagnetism
working as follows: charges generate fields, and fields cause charges to move.
A common charged particle is the electron, which is currently believed to
be a fundamental particle. It has charge
q
e
=
1.6 × 10
19
C
. Other particles’
charges are always integer multiples of q
e
(unless you are a quark).
In electromagnetism, there are two fields the electric field
E
(
r, t
) and the
magnetic field
B
(
r, t
). Their effects on charged particles is described by the
Lorentz force law.
Law
(Lorentz force law)
.
The electromagnetic force experienced by a particle
with electric charge q is
F = q(E + v × B).
This is the first time where we see a force that depends on the velocity of
the particle. In all other forces we’ve seen, the force depends only on the field
which implicitly depends on the position only. This is weird, and seems to
violate Galilean relativity, since velocity is a relative concept that depends on
the frame of reference. It turns out that weird things happen to the
B
and
E
fields when you change the frame of reference. You will learn about these in the
IB Electromagnetism course (if you take it).
As a result, the magnetic force is not a conservative force, and it cannot be
given a (regular) potential. On the other hand, assuming that the fields are
time-independent, the electric field is conservative. We write the potential as
Φ
e
(r), and E = −∇Φ
e
.
Definition
(Electrostatic potential)
.
The electrostatic potential is a function
Φ
e
(r) such that
E = −∇Φ
e
.
While the magnetic force is not conservative in the traditional sense, it always
acts perpendicularly to the velocity. Hence it does no work. So overall, energy
is conserved under the action of the electromagnetic force.
Proposition. For time independent E(r) and B(r), the energy
E = T + V =
1
2
m|v|
2
+ qΦ
e
is conserved.
Proof.
dE
dt
= m
¨
r ·
˙
r + q(Φ
e
) ·
˙
r
= (m
¨
r qE) ·
˙
r
= (q
˙
r × B) ·
˙
r
= 0
Motion in a uniform magnetic field
Consider the particular case where there is no electric field, i.e.
E
=
0
, and that
the magnetic field is uniform throughout space. We choose our axes such that
B = (0, 0, B) is constant.
According to the Lorentz force law,
F
=
q
(
E
+
v × B
) =
qv × B
. Since
the force is always perpendicular to the velocity, we expect this to act as a
centripetal force to make the particle travel in circles.
Indeed, writing out the components of the equation of motion, we obtain
m¨x = qB ˙y (1)
m¨y = qB ˙x (2)
m¨z = 0 (3)
From (3), we see that there is uniform motion parallel to
B
, which is not
interesting. We will look at the x and y components.
There are many ways to solve this system of equations. Here we solve it
using complex numbers.
Let ζ = x + iy. Then (1) + (2)i gives
m
¨
ζ = iqB
˙
ζ.
Then the solution is
ζ = αe
t
+ β,
where
ω
=
qB/m
is the gyrofrequency, and
α
and
β
are complex integration
constants. We see that the particle goes in circles, with center β and radius α.
We can choose coordinates such that, at
t
= 0,
r
= 0 and
˙
r
= (0
, v, w
),
i.e. ζ = 0 and
˙
ζ = iv, and z = 0 and ˙z = w.
The solution is then
ζ = R(1 e
t
).
with R = v = (mv)/(qB) is the gyroradius or Larmor radius. Alternatively,
x = R(1 cos ωt)
y = R sin ωt
z = wt.
This is circular motion in the plane perpendicular to B:
(x R)
2
+ y
2
= R
2
,
combined with uniform motion parallel to B, i.e. a helix.
m
¨
r = q
˙
r × B
Let B = Bn with |n| = 1. Then
¨
r = ω
˙
r × n,
with our gyrofrequency
ω
=
qB/m
. We integrate once, assuming
r
(0) =
0
and
˙
r(0) = v
0
.
˙
r = ωr × n + v
0
. ()
Now we project () parallel to and perpendicular to B.
First we dot () with n:
˙
r · n = v
0
· n = w = const.
We integrate again to obtain
r · n = wt.
This is the part parallel to B.
To resolve perpendicularly, write r = (r · n)n + r
, with r
· n = 0.
The perpendicular component of () gives
˙
r
= wr
× n + v
0
(v
0
· n)n.
We solve this by differentiating again to obtain
¨
r
= ω
˙
r
× n = ω
2
r
+ ωv
0
× n,
which we can solve using particular integrals and complementary functions.
Point charges
So far we’ve discussed the effects of the fields on particles. But how can we
create fields in the first place? We’ll only look at the simplest case, where a
point charge generates an electric field.
Law
(Columb’s law)
.
A particle of charge
Q
, fixed at the origin, produces an
electrostatic potential
Φ
e
=
Q
4πε
0
r
,
where ε
0
8.85 × 10
12
m
3
kg
1
s
2
C
2
.
The corresponding electric field is
E = −∇Φ
e
=
Q
4πε
0
ˆ
r
r
2
.
The resulting force on a particle of charge q is
F = qE =
Qq
4πε
0
ˆ
r
r
2
.
Definition
(Electric constant)
. ε
0
is the electric constant or vacuum permittivity
or permittivity of free space.
The form of equations here are closely analogous to those of gravity. However,
there is an important difference: charges can be positive or negative. Thus
electrostatic forces can be either attractive or repulsive, whereas gravity is always
attractive.