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

−iωt

+ 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

−iωt

).

with R = v/ω = (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.

Alternatively, we can solve this with vector operations. Start with

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.