Lecture 01

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PX384 - Electrodynamics 1

PX384 Electrodynamics Erwin Verwichte

Electricity: electrum, Greek for amber or pale gold. Coined by William Gilbert (1540-1603) for physical force generated by rubbing amber.

Magnetism: magnetum, lodestone from Magnesia region in Thessaly (Greece) where magnetised ore was obtained.

Plasma: plasm, Greek for form, shape, something molded. Coined by Irving Langmuir (1927), connected high-speed electrons and ions in electrified fluid with blood cells in plasma of blood vessels.

PX384 - Electrodynamics 2

PX384 Electrodynamics Erwin Verwichte

Module webpage: http://go.warwick.ac.uk/physics/teach/syllabi/year3/px384 http://go.warwick.ac.uk/physics/teach/module_home/px384

Additional material: plasma formularium (handout) plasma calculator (online)

Leads from: Electromagnetic Theory and Optics (PX263) Leads to: Advanced Electrodynamics (PX472), Relativity & Electrodynamics (PX421)

and Physics of Fusion Power (PX438)

Recommended reading: Electromagnetism, I.S. Grant & W.R. Phillips, John Wiley & Sons, 2nd ed.(1990). Plasma Physics, R. O. Dendy, Oxford Science Publications (1990).

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PX384 - Electrodynamics 3

PX384: Electrodynamics

I. Introduction

II. Electromagnetic potentials

III. Single particle motion

IV. Plasma state of matter

V. Electromagnetic waves in media

Revision of Maxwells equations

Gauge theory

Motion of particle in constant and varying EM fields

Collective behaviour of charged particles

EM waves in media, in particular in a plasma

PX384 - Electrodynamics 4

I. Introduction

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PX384 - Electrodynamics 5

1. Maxwells equations in free space (microscopic)

.o

E

=

Gauss Law

. 0B =

Solenoidal condition

BEt

=

Faradays Law

o o oEB jt

= +

Ampres + Maxwells Law

PX384 - Electrodynamics 6

( , ) ( , )s s ss

r t r tj q n u=

With

E (r,t) [Vm-1] Electric field, force acting on unit charge B (r,t) [T] Magnetic induction, aka magnetic field

(r,t) [Cm-3] Electric charge density, charge per unit volume j (r,t) [Am-2] Electric current density, current per unit surface

n (r,t) [m-3] Number density, ne, ni for electron and ion resp. u (r,t) [ms-1] Bulk velocity q [C] Electric charge (qe= -e = -1.602 10-19 C)

o [Fm-1] Permittivity of free space = 8.854 10-12 Fm-1 o [Hm-1] Permeability of free space = 4 10-7 Hm-1 co [ms-1] Speed of light in free space = (o o) -1/2

( , ) ( , )s ss

r t r tq n =

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PX384 - Electrodynamics 7

Also

LorentzF qE qv B= +

Lorentz force

j E=

. 0jt + =

Charge conservation

Follows from conservation of particles per species

j E= Ohms Law

S

L I

I Uj ES LL I US

= = =

=

= Resistence

[C2N-1m2s-1] Conductivity, vacuum: 0, drinking water: 0.005 sea water: 5, copper: 60 106

PX384 - Electrodynamics 8

2. Maxwells equations in media (macroscopic)

free.D =

Gauss Law

. 0B =

Solenoidal condition

BEt

=

Faradays Law

freeDH jt

= +

Ampres + Maxwells Law

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PX384 - Electrodynamics 9

2.1. Dielectric materials

ro oD E P E = +

Electric displacement

r( 1) oP E =

Electric polarisation

free polarised. . .oP D E = = =

with free polarised = + , free.D =

. / oE =

and

The electric polarisation is linked with the atomic electric dipole moment

atomatom

( ) drP n p n r V= = E

+ - + - + -

+ - + - + -

P -p

PX384 - Electrodynamics 10

With

D(r,t) [Cm-2] Electric displacement, Mathematical construct P(r,t) [Cm-2] Electric polarisation p(r,t) [Cm] Electric dipole moment

r [1] Relative permittivity, vacuum: 1, air: 1.00059, paraffin wax: 2-2.5, porcelain: 6

E [1] Electric susceptibility, E = r - 1 = P/D

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PX384 - Electrodynamics 11

2.2. Magnetic materials

1 1 1ro oH B M B

=

Magnetic field

( )1 1r 1 oM B =

Magnetisation

with free magnj j j= +

, oB j =

freeH j =

and 1

free magnoM H B j j j = + = + =

The magnetisation is linked with the atomic magnetic dipole moment

atom ( )1 d2 V

rM n n r j V= =

B M

PX384 - Electrodynamics 12

With

H(r,t) [Cm-1s-1] Magnetic field, Mathematical construct M(r,t) [Cm-1s-1] Magnetisation (r,t) [Am2] Magnetic dipole moment

r [1] Relative permeability, vacuum: 1, water: 1.000008, copper: 1.000006, transformer iron at B=2mT: 0.00025

B [1] Magnetic susceptibility, B = r 1 = M/H

- diamagnetism: induced opposite to B, B0 - ferromagnetism: strong alignment of permanent

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PX384 - Electrodynamics 13

3. Inclusion of displacement current

Dt

freej

/D t

is the displacement current density, which was not included in Ampres original Law. Under what circumstances is this omission justified?

We neglect the displacement current for the moment and compare contribution of with in a solenoid.

Solenoid of N windings per unit length carries a time-varying current I(t) = I0sin(t).

1. Current Magnetic field

( ) sin( )o oB t NI t =

[Calculation I.1]

2. Time-varying magnetic field Electric field

01( ) cos( )2 o

E t NI r t =

I(t)

S C

B(t)

S C E(t)

PX384 - Electrodynamics 14

3. Electric field Displacement current density

20

1 sin( )2 o o

D NI r tt

=

21.d / .d2 oS S

D RS j St c

=

4. Compare free and displacement currents L

R

This ratio is small only if 2co/R

Example: A solenoid with R=5cm, using co 300000 km/s: 11 10

80.05 8 10 5 10

2 6 10o

Rc =

As long as the driving frequency is less than 1 GHz, the displacement current can be neglected.

Displacement current is important for high-frequency signals

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PX384 - Electrodynamics 15

Exercises Ex I.1: Write the dimensions of all quantities in terms of the basic SI units of mass (kg), distance (m), time (s) and electric charge (C). Ex I.2: Identify the scientists that contributed to EM theory on the money notes. Ex I.3: For a solenoid to have a meaningful displacement current with a current drive frequency that matches BBC Radio 4 FM transmission, what is the minimum radius of its coils? Ex I.4: Use the integral version of Gauss Law to find the electric field, E , electric displacement, D , and electric polarisation, P, for a slab of dielelectric of relative permittivity r which fills the space between z = a and contains a uniform density of electric charges free per unit volume. Use the integral version of to find the surface density of polarisation charge on the surface of the dielectric. Ex I.5: A steady beam of electrons has a radius R and a uniform charge density =-ene. Outside the beam there are no charges. Using Gauss Law, calculate the electric field for the system. For this use cylindrical coordinates and the radial part of the divergence operator . Match the solutions inside and outside the beam at r = R.

( ) ( )1. /rrE r rE r =