Electromagnetism
Objectives

The goal of this course is to provide an understanding of the behaviour of em walves in real materials and in artificial structures (such as waveguides and antennae).

Practically all applications of em thoery in the real world involve materials and structures.

The course:

  1. Revision of vacuum theory.
  2. The effect of materials on electric fields (here we shall introduce the D-field).
  3. The effect of materials on magnetic fields (introducing the H-field).
  4. Rederive Maxwell's equations in presence of materials.
  5. Propagation of em waves in idealised and in real materials.
  6. Behaviour of em waves at interfaces (e.g. reflection).
  7. Propagation of em waves in man made structures (e.g. cables, waveguides).
  8. Transmission and reception of em waves (aerials).
Text book

Elements of Electromagnetism by M N O Sadiku Oxford University Press

Revision Electrostatics
  • Electric Field E
  • Electrostatic Potential V
  • Electric Dipole p
  • Electric flux and the Gauss theorem in electrostatics
Current electricity
  • Electromotive force (EMF)
  • Current I
  • Current density J
Magnetic effects
  • Magnetic dipoles (m)
  • The magnetic B-field
  • Magnetic flux and the Gauss theorem in magnestatics
  • Magnetic fields set up by electrical currents
  • The Biôt-Savart law
  • The Ampère Law
Electromagnetic induction
  • The Faraday-Lenz law
Time-dependent effects
  • The displacement current
  • Maxwell's equations
  • The em wave equation (radio, light, ...)
Vector operators in Mathematics
gradscalar field∇V
divvector field∇.E
curlvector field∇×E
2scalar2V

Throughout this section, we will be assuming that everything is in a vacuum.

Electric field E ----o----> E

We define the E-field as the force on a unit charge at the point concerned.

Coulomb's Law r q o- - - - - - - - -o Q X

Force on a charge q at X is (Qq)/(4πε0r2).

ε0 = permittivity of free space.

Hence, the E-field at X due to Q is:

E = (Q)/(4πε0r2)
Lines of E Fields

The lines of an electric field E are given by:

dy/dx = Ey/Ex
Electrostatic Potential V

The work done in moving test charge q in the presence of the charge Q from infinity to an arbitrary point is the potential V of this arbitary point.

A o __ F |'.. /| : / ' | ''. / o <---- q | / '. rA / ,->\ ,''. | / | '..' '. / | : | / dl : PATH / '.. |/ ''... o- - - - - - - - - - - - - - -'-o Q rB B _ F /| / q /)θ o----o / dl / L Fext

Fext is the force an external agent requires to hold charge q in position against force F.

When q moves by dl, the work done by the external agent force Fext is:

dWext = Fext × (-dl cos θ)

But Fext = F = qE, so:

dWext = -q E dl cos θ dWext = -q El dl

Hence, in qoing from A to B, the work done by the external agent is (by adding together the work done for each little bit dl):

B Wext = - ∫ q E cos θ dl A B = - ∫ q E dr A

But E = Q/(4πε0r2), so:

B qQ.dr Wext = - ∫ ----- A 4πε0r2 [ qQ.dr ]B qQ qQ = [ ----- ] = ------ - ------ [ 4πε0r2]A 4πε0rB 4πε0rA

...which is independent of the route taken (and is thus a conservative force).

If q is a unit charge,

Q Q Wext = ------ - ------ 4πε0rB 4πε0rA

...which we write as:

Wext = vB - vA

...where

Q V(r) = ------ 4πε0rr

...which is the potential at the point concerned, and is the work required to bring a unit test charge (q=1) from ∞ (rA=∞) to the point concerned (rB).

The relation between E and V A->B B B Wext = - ∫ Eldl = - ∫ E.dl A A

for unit charge

B VB-VA = - ∫ E.dl A

Suppose A and B coincide (close loop). Then,

VB = VA

Hence,

E.dl = 0 closed loop

This is only correct in electrostatics.

From earlier:

dWext = -q El dl

If q=1, dWext = dV

Hence,

dV = -Eldl

i.e.

-∂V El = -- ∂l -∂V Ex = -- etc. ∂x

Since E is a vector, we have

E = -grad(v) = -∇V ∂V ∂V ∂V E = - i -- - j -- - k -- ∂x ∂y ∂z
Electric Dipole p

Two point charges +q and -q separated by a distance a constitute a dipole.

a o---------o -q +q --------> p

Usually we assume a is small.

The dipole moment p has magnitude

p = qa
Potential due to a dipole o X / / / r / / /) θ --->

The potential at X due to a dipole is given by:

p cos θ v = ------- 4πε0r2
Derivation of the potential due to a dipole X ,o r- .'/: ,' / :r+ ,' r/)θ: o---+---o -q a +q

From this diagram one sees that:

q q V = ------ - ------ 4πε0r+ 4πε0r- q ( r- - r+ ) = ----- ( ------- ) 4πε0 ( r+r- )

But, if you drop a perpendiculous from +q to r-, and assume that r>>a, you can approximate that r- - r+ ≅ a cos θ and r-r+ ≅ r2, so:

qa cos θ p cos θ V = -------- = ------- 4πε0r2 4πε0r2
The electric field caused by a dipole . /|\ Eθ |_ \ r | | \-----------------+------> Er \'θ X dipole -∂V Radial component: Er = -- ∂r -∂V -1 ∂V Tangential component: Eθ = -- = - -- ∂s r ∂θ

...where ds = rdθ is a deplacement in the tangential direction.

Hence:

2p cos θ ER = -------- 4πε0r3 p cos θ Eθ = ------- 4πε0r3

To find the electric field E, we use geometry:

. _ Eθ /|\ ,| E | ,' \ r |,')) φ \-----------------+------> Er p \'θ X

Hence:

E = (Er2 + Eθ2)½ p = ----- ( 3 cos2 θ + 1 )½ 4πε0r3 Eθ tan φ = -- = ½ tan θ Er
The electric field caused by a dipole _____________________\ +q / E o--->qE /| ____________ /_|_____\ / | / a/ |a sin α / | ________ /_____|_____\ / _| / -q/)α | | qE<---o--------+ _____________________\ /

Couple (Torque) acting on a dipole is denoted by the greek letter Γ and is:

Γ = (Eq)×(a sin α) Γ = pE sin α

In vector notation:

Γ = p × E
Electric flux and the Gauss theorem in electrostatics | _ normal-->| α /| E |,^./ | / ___|_/ / r|/\ \_____/<-- arbitrary area dS / / /

Electric Flux passing through dS is defined as: (normal Component of E) × (area) = En dS = E cos α dS = E.dS (for small dSE same throughout dS)

For a large surface, the flux passing through a surface is:

∬ En dS surface

...were E may be a function of the position on dS.

For a closed surface we can show that:

charged enclosed ∯ En dS = ---------------- whole ε0 surface

i.e. Electrical Flux leaving a closed surface is equal to the charge enclosed divided by ε0

This is the Gauss Theorem.

Charge density

ρ(v) = charge per unit volume at the point concerned.

_____ ( ,. ) C []' ) (_ _) ,. (_____) []' = volume dv

Clearly, the charge enclosed is given by:

∭ ρ(v) dv volume
Gauss Theorem

The Gauss Theorem becomes:

1 ∯ En dS = --- ∭ ρ(v) dv surface ε0 volume
Gauss Theorem in differential form

The divergence div(F) of a vector quantity F is the scalar which represents the net flux per unit volume of F leaving a small volume.

Let's consider the divergence of the electrical field E:

______________ ___ /| LHFACE /' /| /|\ : / | / | | : / / | |' / | / |.'|dz +-------------+ ..'| | . | | ...''|' | | ''|'....''' | | \|/ \ non uniform | + - - - -|- - + __...'''/ electric field E | | ../'' /\ | / ......''|' / / ..'|''''' | / /dy |/ |/ / +-------------+ \/_ RHFACE <-------------> x x+dx

Flux of E leaving LHFACE = -Ex dydz (negative because leaving in -x direction)

At RHFACE, the field is (Ex+dEx), so the flux leaving RHFACE = (Ex+dEx)dydz

Adding these two expressions, we get the net flux leaving in x-direction:

dEx dydz

...where dEx is:

∂Ex dEx = -- dx ∂x

Substituting:

∂Ex ∂Ex -- dxdydz = -- dv ∂x ∂x

Similarly for other directions, giving:

∂Ex ∂Ey ∂Ez ( -- + -- + -- ) dv = Net flux leaving whole volume ∂x ∂y ∂z

The flux leaving per unit volume is that divided by dv, thus:

∂Ex ∂Ey ∂Ez ( -- + -- + -- ) = div E ∂x ∂y ∂z

...which can be written ∇.E

Now the net flux leaving is equivalent to the charge enclosed divided by ε0:

div E dv = ρ(v) dv / ε0

...giving:

div E = ρ/ε0
Electromotice Force (EMF)

Consider a charged capacitor:

EQ B|- <- +|A |- <- +| |- <- +| +---------|- <- +|---------+ : |- <- +| : : |- <- +| : : |- <- +| : : : :..........................: mathematical circuit

Inside the capacitor there is an electric field EQ due to the charges.

B VB-VA = - ∫ E.dl = EQ x A E.dl = 0 closed loop

(This is all electrostatics -- no moving charges.)

Consider now an electrochemical cell.

| | | | B|- <- +|A |- →→ +| Key: |- <- +| +---------|- →→ +|---------+ <- EQ due to charges : |- <- +| : : |- →→ +| : →→ Field set up by : |- <- +| : chemical interactions : ↑ | : : Electrolyte | : : | : : : :..........................: mathematical circuit

Owing to chemical interactions, - charges are driven to A, and + charges are driven to B.

This continues until the electric field EQ due to charge separation balances the field set up by the chemical interaction.

Therei s then no net electric field inside the cell.

The A terminal is at a higher potential than terminal B (as evidenced by the fact that current flows).

The difference VA - VB is called the EMF ε of the cell (on an open circuit).

B ∫ E.dl (via circuit) = VA-VB = ε A

But:

B ∫ E.dl (via cell) = 0 A

...because the two fields cancel each other inside the cell. Hence:

E.dl = ε + 0 = ε whole circuit

The general definition of the EMF in a circuit is therefore:

ε = ∫ E.dl whole circuit

Note that the circuit could be fictional, this is just any mathematical closed path.

Magnetism

There are only magnetic dipoles — e.g. small permanent magnets, small coils of current carrying wire. — no magnetic monopoles. (At least, not in the macroscopic world.)

For small coils carrying a circuit I the magnetic dipole moment is given by:

m = I × area . /|\ m __|___ / | |\ I \_____/ | ↑ | area dS m = IdS (for small areas)
The Magnetic B-field

Direction is that in which a freely suspended dipole points (a compass). Magnitude is obtained by measuring the torque needed to hold the compass needle at π/2 rads to the field.

_____________________\ __ / B /| / ' ________ /___________\ α(/ / / / _____________________\ / Γ = m × B

If α = π/2, Γ = mB, ∴ B = Γ/m (where m is the magnetic dipole magnitude).

Magnetic Field set up by a Dipole . /|\ Bθ |_ \ r | | \-----------------+------> Br \'θ \ _\| m (magnetic dipole)

In E-fields,

2p cos θ ER = -------- 4πε0r3 p cos θ Eθ = ------- 4πε0r3

...and by analogy, the magnetic dipole case is similar:

μ0 2m cos θ BR = -- . -------- 4π r3 μ0 m cos θ Bθ = -- . ------- 4π r3

μ0 = permeability of free space.

Magnetic Flux | _ normal-->| α /| B |,^./ | / ___|_/ / r|/\ \_____/<-- arbitrary area dS / / /

Magnetic flux passing through dS is BndS = B cos α dS = B.dS

Gauss Theorem in Magnetostatics

Magnetic flux leaving a closed surface is zero.

∫∫ B.dS = 0 closed surface ∴ div B = 0
Magnetic Fields set up by Current Carrying Circuits
Biôt-Savart Law (_ B |\ I / ,' \ / ,' | r/ ,' / / ,' \__ θ /,' \^/' dl ,\ ,'_/ ,' / ,' \__ \ (_

What is the field dB at the point marked B due to dl.

We find that:

μ0 I dl sin θ dB = -- . ---------- 4π r2

...and it is into the diagram at the point B, so:

μ0 I dl × r dB = -- . -------- 4π r3

We can integrate this law to give:

Field due to a long straight wire | /|\ I | | | ...''|''... .'' | ''. : |-. h : : +-+-------x : | __: '.. | ../ B '''.....''' | | | μ0 I B = --- 2πh Field at the centre of a circular coil. ..'.. .' '. : : /:\ I : B : : ---<----:------+-+----+---<--- : |_| : : | : : |a : '. | .' ''|'' ↑ COIL μ0 I B = --- 2 a Field inside a cylindrical coil. ,o8888888888888888888888888888888888888o. ,dP' `9b. `9b. `9b. `9b. `9b. `9b. ,8' `8. `8. `8. `8. `8. `8. 8' `8. `8. `8. `8. `8. `8. oP 'o 'o 'o 'o 'o 'o B 8 8 8. 8. 8. 8. 8 --<--O]------ - -O] /O]\ /O]\ /O]\ /O]\ O] 8. 8 / 8 |' / 8 |' / 8 |' / 8 |' 8 `o ,P ,P ,P ,P ,P ,P 8. ,8 ,8 ,8 ,8 ,8 ,8 `8. ,8' ,8' ,8' ,8' ,8' ,8' `9b. .dP' .dP' .dP' .dP' .dP' .dP' `98888888888888888888888888888888888888P' ↑ n turns per unit length

B = μ0nI inside (uniform and independent of radius).

The Ampère Circuital Law

The line integral of B around a closed path is equal to μ0 times the current enclosed.

B.dl = μ0 × (current enclosed)

This is still just for a vacuum.

Current passing through area ds is:

dI = Jn dS = J cos α dS = J.dS | _ normal→| α /| J |,-./ ,---|--/-. / _|_/_ \ / / |/ / \ ( /____/←dS ) \ / `--->------' path B . dl = μ0 ∫∫ J . dS line surface integral integral around path Field due to a long straight wire. | /|\ I | | | ...''|''... .'' | ''. : |-. h : : +-+-------x : | __: '.. | ../ B '''.....''' | | |

The lines of B are circles, by symmetry. Around a circle of radius h, B is a constant. Therefore:

B . dl = B ∫ dl closed closed = B 2πh

But ∫B.dl - μ0 × (current enclosed) = μ0I ∴ B 2πh = μ0I and B = μ0I/(2πh) as before.

Field inside a long solenoid. n turns per unit length ,o8888888888888888888888888888888888888o. ,dP' `9b.⨀⨀⨀⨀⨀⨀⨀⨀⨀⨀⨀⨀⨀⨀⨀⨀⨀⨀⨀⨀⨀⨀⨀⨀⨀⨀⨀⨀⨀⨀⨀`9b. ,8' `8. `8. 8' `8. -->-->-->-->-->-->-->-->-->-- `8. oP 'o 'o 8 8 -->-->-->-->-->-->-->-->-->-- `8 O O O 8. 8 -->-->-->-->-->-->-->-->-->-- 8 `o ,P 1 2 ,P 8. ,8 -->- +-->-->-->--+ ->-->-->-- ,8 `8. ,8' | | ,8' `9b. .dP'⨂⨂⨂⨂⨂⨂⨂⨂|⨂⨂⨂⨂⨂⨂⨂⨂⨂⨂⨂|⨂⨂⨂⨂⨂⨂⨂⨂⨂⨂,dP' `98888888888888|88888888888|88888888888P' | | 4 +--<--------+ 3 l path

We perform the line integral in steps.

2 1→2 ∱ B.dl = Bl 1 3 2→3 ∱ B.dl = 0 (B at right angles to dl) 2 4 3→4 ∱ B.dl = 0 (B = 0 outside) 3 1 4→1 ∱ B.dl = 0 (B at right angles to dl) 4

Using the Ampère theorem,

B.dl = Bl

We perform the line integral in steps.

12 12 B.l = B l 23 23 B.l = 0 B at π2 to l 34 34 B.l = 0 B=0 outside
continued

Thus, if we make Z s = Z 0 2 Z 1 , all power from the source enters the line.

Revision
Relationships between B, H, E, D

We may approximate: B = μ r μ 0 H D = ε r ε 0 E (linear approximation).

Mnemonic: DeeDee.

Electric Dipole Moment Per Unit Volume p = ε 0 χ E

Mnemonic: p-e-chi-e. Pixie.

Magnetic Dipole Moment Per Unit Volume M = χ H

Mnemonic: mmm-chi-'h. Mickey.

Waves
Poynting Vector N = E × H

Mnemonic: Neh.

Wave In A Vacuum Refractive Index n = ε r μ r = c v p Phase Velocity v p = c ε r μ r = ω k = f λ Power Power Per Unit Area = E 2 Z
Wave In A Lossy Medium Electromagnetic Wave E = E 0 e i ω ( t - n x c ) E = E 0 e i ω t - α x - i β x E = E 0 e i ω t - γ x E = E 0 e i ( ω t - β x ) wave part e - α x decaying amplitude Complex Refractive Index = n′ - i n″ i ω c = α + i β α = ω n″ c β = ω n′ c α β = n″ n′
Faraday Lenz Law

The EMF is given by:

ε = - Φ t

where

Φ = B . s ε = - Φ t