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Electromagnetism

1 ENGN6521 / ENGN4521: Embedded Wireless

Radio Spectrum use for Communications


Electromagnetism IGauss’s law and Dielectrics


Topics

➤ Surface and line integrals

➤ Maxwells equations (DIV and CURL)

➤ Charge and current

➤ Electric and magnetic field

➤ Gauss’ law for E and B

➤ Dielectrics

➤ Ohm’s law


The surface integral

‹

AE.dA =

‹

A|E||dA| cos θ (1)


The line integral

‰

CE.dl =

‰

C|E||dl| cos θ (2)


Maxwell’s equations in integral form

‹

SE · dS =

Q

ǫo[Gauss] (3)

‹

SB · dS = 0 [Gauss] (4)

‰

E · dl = −∂

∂t

ˆ

SB · dS [Faraday] (5)

‰

B · dl = µo

ˆ

Sj · dS+ µoǫo

∂

∂t

ˆ

SE · dS [Ampere] (6)


Maxwell’s Equations in differential form


Gauss’ Law

For the electric field‹

SE · dS =

Q

ǫo(7)

For the magnetic field‹

SB · dS = 0 (8)

where ǫo is the permittivity of free-space (Farads per m)


Charge: Q (positive and negative)

➤ The ultimate source of all EM fields

➤ Measured in Coulombs

➤ Can be postive or negative - note the direction of the electric field

Q =

˚

Vρ · dV (9)


Charge as a source of the Electric Field: MAXWELL’s first equation


Simple case of a point charge: Coulomb’s Law

‹

SE · dS =

Q

ǫo(10)

Er × 4πr2 =Q

ǫo(11)

Er =Q

4πǫor2(12)


Parallel Charged Plates


Electrostatics: The Electrostatic Potential (Beware - replace this by Faraday’s law

➤ Consider a situation where there is no changing magnetic field in

Faraday’s law.

➤ Definition: The potential difference between two points x1 and x2 is

defined by,

Φ = −

x2ˆ

x1

E.dl = −

ˆ

γ

E.dl

➤ Since the path γ can be any which connects the points x1 and x2 we may

conclude that E = −∇Φ.

➤ Kirchhoffs voltage law: The sum of the voltage drops in a circuit is

zero - BEWARE: to be ’modified’ by ∂B∂t


Parallel Charged Plates

‰

E · dl = 0 [Faraday, no changing B] (13)


Parallel Plate Capacitor

➤ Capacitors are reservoirs of charge - a bit like transient batteries

➤ C = QV (Farads)

➤ Since V = Ed, C = ǫ0Ad


Actual capacitors


The vacuum gap capacitor


Moving charges → Current

➤ The rate of charge crossing a surface is given by (note that charges are

positive even if electrons are flowing!),

∂Q

∂t=

¨

Sj · dS (14)

➤ Current is measured in Amperes (Coulombs per second)

➤ Charge conservation: the charge flows out of a region as a current Q.

Why the ’-’ sign here?

∂Q

∂t= −

‹

Sj · dS (15)

➤ Charge conservation leads to Kirchhoff’s current rule: currents

entering a circuit node sum to zero


Current Density (current per unit area)


Kirchhoff’s current law


Dielectrics and Conductors

➤ Dielectrics are insulating materials VIZ. do not allow D.C. current to flow

through them. Usually we just call them insulators

➤ In conductors, charge carriers (electrons) are free to move. e.g. metals.


Dielectrics 1

➤ Two opposite, cancelling charges separated in space is referred to

collectively as a dipole.

➤ Electrons and nuclei in dielectrics experience opposing forces in the

presence of an imposed electric field.

➤ Electrons move opposite to the field and nuclei move in the direction of the

field. This produces a distribution of dipoles referred to as polarisation.

➤ The charge separation induced by the field acts to reduce the electric field

within the dielectric.


Dielectrics 2


Dielectrics 3: Polarisation

➤ Polarisation P is the dipole moment per unit volume induced by an imposed, externalelectric field

➤ At the edge of a polarised dielectric there is a charge density left over by the displacementof the dipoles given by σp = P.n̂ where n̂ is the unit vector normal to the surface.

➤ σp belongs to the material. It is not a free charge.


Dielectrics 4: Relative dielectric constant

➤ For a slither of dielectric, the surface charge is related to the E-field within the dielectricthat produces it by

σp = P · n̂ = ǫ0(ǫr − 1)E · n̂

➤ ǫr at D.C. is a positive dimensionless number and ǫr > 1 for dielectrics. E.g. ǫr = 2 forteflon, ǫr = 1 for air or vacuum.

➤ When the electric field oscillates, ǫr is a complex function of frequency -lossy

➤ The ratio of the imaginary to real components of ǫr is termed the loss tangent of thedielectric:

tan[δ(ω)] =Im(ǫr(ω))

Re(ǫr(ω))(16)

➤ Even though dielectrics are insulators at DC they can be lossy conductors atradiofrequency.


Exercise 1: Compute the capacitance of a parallel plate air-gap capacitor

(E − 0)×A =Q

ǫo=

σ ×A

ǫo=> E =

σ

ǫo(17)

V = E × d => C =Q

V=

ǫoA

d(18)

+ + + + + + + + +

- - - - - - - - -

E = 0

E > 0

Q = σΑ

electricfieldlinesarea = A

enclosedby dots

capacitorplates

d


Exercise 2: Compute the capacitance for a gap of dielectric constant ǫr

Use the definition of ǫr and comnpute the E − field inside the dielectric

E =σcap − σp

ǫo=

σp

ǫo(ǫr − 1)(19)

(σcap − σp)(ǫr − 1) = σp => σp =σcap(ǫr − 1)

ǫr(20)

V = E × d =σp × d

ǫo(ǫr − 1)=

σcapd

ǫoǫr(21)

Qcap = σcapA => C =Qcap

V=

ǫrǫoA

d(22)


Conductors and Ohm’s law

➤ Ohm’s law: The current density in a conductor is proportional to the electric

field within the conductor.

j = σE (23)

➤ Conductors are completely specified by the conductivity σ.

➤ For copper, σ = 5.80× 107mhos/meter.

➤ Ohm’s law is assumed to be an accurate result for metals at all

radiofrequencies :)

➤ I.E. σ is always a real number and independent of frequency.


Resistance

➤ Resistance is obtained by converting j to I and E to V for a short length of

metal.

I

A= σ

V

l(24)

V =l

σAI (25)

➤ From which we define resistance

R =l

σA(26)

➤ We also define resistivity, η from σ

η =1

σ(27)


Conductors vs Dielectrics

➤ for dielectrics: σp = ǫ0(ǫr − 1)E · n̂

➤ If σp oscillates as a function of time then, jωσp = jωǫ0(ǫr − 1)E · n̂

➤ since the dipoles are reversing sign at rate ω, we may write, jP · n̂ = jωσp,

where jP is the polarisation current.

➤ Dielectrics obey Ohm’s law with conductivity defined by

σ = jωǫ0(ǫr − 1)

➤ The difference between conductors and insulators is simply that σ is

resistive and frequency independent for a conductor, but is mainly

reactive and frequency dependent for insulators.

➤ As we already know, ǫr is a complex function of frequency and the real and

imaginary frequency functions of ǫr can be computed from each other

(Kramers-Kronig).


Charge as a source of the Electric Field: MAXWELL’s first equation

➤ the field can get distorted with complex charge distributions

➤ makes no difference to the integral

The coulomb force...

F = QE (28)


Electromagnetism IIMoving Charges. Faraday’s law and Ampere’s law.

Ferrites


ELECTROMAGNETISM SUMMARY

➤ RF shielding

➤ Magnetostatics

➤ Ampere’s law and Faraday’s law

➤ Inductance

➤ Maxwell’s equations


The Floating Conducting Object

Ohm’s law for metals and Gauss’s law for the electric field give rise to the

concept of electromagnetic shielding


The Earthed Conducting Object


Images

➤ A charge near a conducting surface attracts charge of the opposite sign to

the nearest point on the surface.

➤ These surface charges arrange themselves so that the tangntial

component of electric field is zero on the surface.

➤ One may compute the electric field outside the conductor by assuming

(mathematically) that there is an image charge within the conductor.


Proximity to Floating (coupled) and Earthed Conductors (decoupled)


Magnetostatics: The static magnetic field

➤ Gauss’s law for the magnetic field:‹

A

B · dA = 0 (29)

➤ There is no static sink or source of the magnetic field. Also generally true.

➤ However current is the source of a static magnetic field. Using Ampere’s law (with nochaning E-field)

‰

γ

B.dl = µ0I (30)

The line integral of B around a closed circuit γ bounding a surface A is equal to the

current flowing through A. Ampere’s law


Important Correction to Ampere’s Law

➤ A time varying Electric field is also a source of the time varying magnetic

field,

‰

γB.dl =

¨

Aµ0

(

j+ ǫ0∂E

∂t

)

.dA (31)


Why this Correction to Ampere’s Law? I

➤ Ampere’s law without E contradicts charge conservation

‰

γB.dl =

¨

Aµ0

(

j+ ǫ0∂E

∂t

)

.dA (32)

➤ Consider the new Ampere’s law on a close surface area, A.


Why this Correction to Ampere’s Law? II

➤ If we shrink the closed contour γ on the left hand side to zero then we

obtain

0 =

‹

Aµ0

(

j+ ǫ0∂E

∂t

)

.dA (33)

➤ However the two terms on the right hand side are‹

AE.dA =

q

ǫ0(34)

and‹

Aj.dA = −

∂q

∂t(35)

➤ - Q.E.D. In fact it simply confirms how capacitors work!!


Faraday’s Law

➤ The law of electromagnetic induction or Lenz’s / Faraday’s law

➤ A time varying magnetic field is also the source of electric field,˛

γ

E.dl = −

ˆ

A

∂B

∂t.dA (36)


Inductance I

➤

¸

circuitE · dl = −˜

A∂B∂t · dA = −∂ΦB

∂t

➤ Note that Kirchhoff’s voltage law still works.

V

B field

Iinductor


Inductors II Magnetic field of a solenoid

➤ Consider Ampere’s law for a solenoid with a static current ∂I∂t

= 0.

➤ If the solenoid is long and axially uniform then the magnetic field is directed along the axis.

➤ By choosing a curve along the axis of the solenoid, Ampere’s law gives,

ˆ d

0

dlB = µoNI => B =µoNI

d=> B = µoNlI (37)

where µo = 4π × 10−7Henriess/meter is the permeability of free-space, N is the number of turnsenclosed by the contour and Nl is the number of turns per unit length.


Inductance III Inductance of a coil


Inductance IV: Inductance of a coil

➤ A simple formula defining the inductance of a coil can be found by noting that if there are N turns on a coilthen the total flux linking the coil is Φ = N ×B ×A

➤ From Faraday, the electromotive force (E.M.F.) at frequency ω generated by the changing flux is given byV = jω ×N ×B ×A

➤ Next we use the formula for magnetic field B of an infinite solenoid as an approximation for a coil (a solenoidof finite length)

V = jω ×N × (µoNlI)×A = jω × (µoN2A

d)I (38)

➤ From which we define the inductance, L,

V = jωLI, L =µoN2A

d(39)

➤ More accurate formula (lab2) ...

L(µH) =0.394r2N2

9r +10d(40)


Ferromagnetic Materials

➤ Whereas a dielectric reduce E − field with ǫr ≫ 1, magnetic materials known as ferritesthat can amplify an internal magnetic field.

➤ In Ferrites, magnetic domains align with the imposed field (see the figure below).

➤ If the above solenoid were wound on a ferromagnetic core, then the formulae for the B

field and the inductance L would become B = µrµoNlI and L = µrµoN 2Ad


Inductors and Ferrites


A comparison between ferrites and dielectrics

➤ A dielectric decreases the imposed E-field whereas a ferromagnetic material increasesthe imposed B-field.

➤ ǫr > 1 increases C because it decreases the voltage for a given charge on the capacitorelectrodes.

➤ µr > 1 increases L because it increases the magnetic flux for a given current, thusincreasing the E.M.F.


Self Inductance (I)

➤ In practice the magnetic field linking a circuit can either come from current in the circuititself (as we have just discussed) or from some other external current in another circuit.

➤ So far we have just looked at the case of a magnetic field outside the wire which acts backon the circuit through Faraday’s law. This is self-inductance


Self Inductance (II)

➤ We have already seen the self-inductance of a coil.

➤ The most obvious case (but definitely not the simplest case) is that of a cylindrical straight conductor wherethe circuit of integration is precisely that above

L = 0.002 l

[

loge

(

2l

a

)

−3

4

]

: l (cm) for L (µH) (41)


Mutual Coupling

➤ The magnetic fields of different circuits induce electromotive forces in each other.

➤ Mutual Inductance, M

V1 = MdI2

dt, V2 = M

dI1

dt(42)


Transformers

➤ Two neighbouring coils can couple to each other inductively by Faraday’s law

V21 = jωLiii + jωMio (43)

V43 = jωLoii + jωMii (44)


A summary of units of measure

The following units can be deduced from the formulae thus far assuming that

S.I. units are used for distance (metres), time seconds), charge (Couombs).

➤ Electric field is Newtons per Coulomb but is measured in Volts per metre

➤ Capacitance is Coulombs per volt but is measured in Farads

➤ Current is Coulombs per second but is measured in Amperes

➤ Magnetic field is measured in Tesla

➤ Magnetic flux is (Teslas×m2) but is measured in Webers

➤ The permittivity of free space, ǫo = 8.85× 10−12 Farads per metre

➤ The permeability of free space, µo = 4π × 10−7 Henries per metre


Electromagnetism III

➤ Plane and transverse electromagnetic waves

➤ Power flow

➤ How antennas work


EM Waves

➤ Start with Faraday’s law and Ampere’s law in VACUO.

˛

γ

E.dl = −

ˆ

A

∂B

∂t.dA (45)

˛

γ

B.dl =

ˆ

A

µ0ǫ0∂E

∂t.dA (46)

➤ Recall that the line integral along γ is on the perimeter of the surface A

➤ Thus a one dimensional E and B must have E.B = 0.


Plane Waves 1

➤ Apply F and A to the diagrams below and note that A points in the direction

of the right hand screw rule with respect to the direction of γ


Plane Waves 2


Plane Waves 3

➤ Integrating E around γ in the left figure...

[E(z + dz)−E(z)]L = −∂B

∂tL∆z (47)

∂E

∂zL∆z = −

∂B

∂tL∆z (48)

➤ Integrating B around γ in the right figure...

[B(z + dz)−B(z)]L = −ǫ0µ0∂E

∂tL∆z (49)

∂B

∂zL∆z = −ǫ0µ0

∂E

∂tL∆z (50)


Plane Waves 4

➤ Two equations for E and B

∂E

∂z= −

∂B

∂t(51)

∂B

∂z= −ǫ0µ0

∂E

∂t(52)

➤ Simultaneous solution gives the wave equation for each field:

∂2E

∂z2= ǫ0µ0

∂2E

∂t2(53)

∂2B

∂z2= ǫ0µ0

∂2B

∂t2(54)


Plane Wave Solutions

➤ E = E(X); B = B(X) where X = z ± vt

➤ Propagation speed in VACUO

v = c =

√

1

ǫ0µ0

= 3× 108m/s (55)

➤ For travelling waves , E,B ∝ f(ωt∓ kz) and EB

= ±c

➤ plane waves satisfy E ·B = 0

➤ plane waves propagate in the direction of E×B

➤ plane waves have components Ez = 0 abd Bz = 0 in the direction of propagation of thewave


Transverse electromagnetic waves or T.E.M Waves

➤ Consider plane waves propagating along the z − axis perpendicular to the x− y plane

➤ T.E.M. waves are plane waves whose fields also vary in the x− y plane

➤ Like plane waves, T.E.M. waves also have Ez = 0 abd Bz = 0 in the direction ofpropagation of the wave

➤ E = Eo(x, y) exp j(ωt− kz) and B = Bo(x, y) exp j(ωt− kz)

➤ ∇× Eo(x,y) = 0 and ∇×Bo(x,y) = 0

➤ As a result, ∃ a scalar function Φ(x, y) which gives Eo(x,y) = −∇Φ(x, y) and satisfiesLaplace’s equation: ∇2Φ(x, y) = 0


Plane and T.E.M Waves


Definition of Power

➤ Consider complex vector functions of the form

X,Y = (U(x) + jV(x)) exp jωt (56)

➤ Compute the time average of the product of two complex numbers

< X ·Y > =1

2ℜ[X ·Y∗] (57)

➤ Electromagnetic power dissipation per unit volume (Power density Watts/m2) givenj(A/m2) and E(V/m)

Power density =1

2ℜ[j · E∗] (58)

➤ Only gives dissipated power => Ohmic power


What about transmitted or radiated power?

➤ Already seen this with transmission lines, e.g. Pforward = 12ℜ[VfI

∗f ] - But what about free

space E.M. waves??

➤ Approach: relate this definition of power to the T.E.M. B and E fields in the transmissionline to produce a suitable definition of power flux for T.E.M waves


Poynting’s Theorem (I)


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