Introduction: So far we have These equations are OK for static fields, i.e. those fields independent of time. When fields vary as a function of time the

ELEC 3105 Basic EM and Power

Engineering

Faraday’s LawLenz’s Law

Displacement Current

Faraday’s Law

Introduction: So far we have

E

0 E

JH�

0 B

�

These equations are OK for static fields, i.e. those fields independent of time. When fields vary as a function of time the curl equations acquire an additional term.

0 E

gets a t

B

gets a t

D

JH�


Engineering

Faraday’s Law

Consider the following experiment. Pull a wire loop through a region of non-uniform magnetic field .

v

Wire loop, path encloses a surface SB

Magnetic field vector points into the page

1B

2B

tvtv

area gained in time t

area lost in time t

tv

tv

21BB

Faraday’s Law

Consider the following experiment. Pull a wire loop through a region of non-uniform magnetic field .

v

B

Magnetic field vector points into the page

1B

2B

tvtv

21BB

v

Charge in wire

11BvqF

Charge in wire

22BvqF

1F

2F

Faraday’s Law

Consider the work done on +1C test charge moved around the loop; this is the “emf” electromotive force.

v1

B

2B

tvtv

21BB

v

Charge in wire

11BvqF

Charge in wire

22BvqF

1F

2F

21* FFqemf

t

tvBtvBemf

21

Faraday’s Law𝑊𝑞=𝑉Early definition of potential

ExamineExpression

vBvBemf 21

Consider the work done on +1C test charge moved around the loop; this is the “emf” electromotive force.

v1

B

2B

tvtv

v1

F

2F

t

tvBtvBemf

21Now: tvB

1Flux change at right side of loop

tvB2

Flux change at left side of loop

t

Fluxemf

S

adB flux

Note that B and da are in opposite directionsdt

demf

Faraday’s Law

1B

2B

t

Fluxemf

dt

demf

This is a general result. Even if we hold the loop stationary and change B, the emf is still given by the negative rate of change of the flux.

Move magnet

N S

Faraday’s Law

1B

2B

Further generalization is possible. Consider moving loop in time varying magnetic field.

Move magnetN

S

Move loop

S

adt

BdBvemf

S

adt

B

dBv MOTIONAL emf

TRANSFORMER emf

The induced emf always opposes the change in

flux

Faraday’s Law


Engineering

Lenz’s Law

magnetB

loopB

N S

MOVE LOOP

dBvemf

motional emf

The induced emf always opposes the

change in flux

v

• Move loop towards magnet• B increases in loop• Flux increases in loop• Current induced through emf• Current produces magnetic field in loop: 2nd postulate• This magnetic field in opposite in direction to magnetic field of magnet

I

Faraday’s Law / Lenz’s Law

magnetB

loopB

N S

MOVE LOOP

dBvemf motional emf

The induced emf always opposes

the change in flux

v

• Move loop away from magnet• B decreases in loop• Flux decreases in loop• Current induced• Current produces magnetic field in loop• This magnetic field in same direction to magnetic field of magnet

I


magnetB

loopB

N S

MOVE MAGNET


change in flux

v

• Move magnet towards loop• B increases in loop• Flux increases in loop• Current induced• Current produces magnetic field in loop• This magnetic field in opposite direction to magnetic field of magnet

I

S

adt

Bemf

transformer emf


magnetB

loopB

N S

MOVE MAGNET


change in flux

v

• Move magnet away from loop• B decreases in loop• Flux decreases in loop• Current induced• Current produces magnetic field in loop• This magnetic field is in same direction to magnetic field of magnet

I

S

adt

Bemf

transformer emf


Suppose loop is stationary so we have only transformer emf.

MOVE MAGNET

N S

S

adt

Bemf

I

dEemf Equivalent battery to drive current around loop

S

adt

BdE

S

adt

BadE

t

BE

Faraday’s Law

t

BE

Faraday’s Law in derivative form: Valid for all points in space

The induced emf always opposes the change in

flux

S

adBt

emf

Note on: Faraday’s Law / Lenz’s Law

dt

demf

WE KNOW THAT �� (𝑡 )∝𝐼 (𝑡 )

constant

�� (𝑡 )=�� ∗ 𝐼 (𝑡 )

t

tIadKemf

S

Inductance: relates the induced emf to the time rate of change of the current 𝑣 (𝑡 )=−𝐿

𝜕 𝐼 (𝑡 )𝜕𝑡

S

adBt

emf


tBAemf sin

Rotating current loop in constant magnetic field

��X

O

𝜃

𝜔

�� ∙ ��𝑎=𝐵 𝑑𝑎𝑐𝑜𝑠 (𝜔 𝑡 )

Sinusoidal voltage change

Amplitude of voltage depends on B, A and POWER GENERATOR


vBLemf

Moving rod in constant magnetic field

��

dBvemf

X

X

X

X

X

X

X

X

X

X

X

X

vL

Which end of the rod is positive and which end is negative?

+

-

+

-

?


Engineering



Capacitor

WIRE

enclosedIdH

So far we have the following expression for Ampere’s law.

Certainly there is no problem in evaluating the integral. The path shown encloses an area A through which the wire cuts.

We are in fact in the process of charging the capacitor through the current I of the wire.

IdH

For surface A chosen

Capacitor

WIRE

enclosedIdH

So far we have the following expression for Ampere’s law.

Certainly there is a problem in evaluating the integral. The path shown encloses an area A’ through which the wire does not cut. Yet the integral value evaluated here for the surface that does not contain the wire A’ and the surface of the previous slide A which does contain the wire must be the same since the surfaces A’ and A are arbitrary.

We are in fact in the process of charging the capacitor through the current I of the wire.

0

dHFor path A’ chosen

Both integrals should give the same result??????


Capacitor

WIRE

IdH

For different surfaces

Capacitor

WIRE

0

dH

We have a problemDisplacement Current

A A

adt

DadJdH

Solution to the problem

Between the capacitor plates we have a changing electric flux density. This changing electric flux density is equivalent to a current density.

t

DJ

D


A A

adt

DadJdH

Solution to the problem

A A

adt

DadJadH

t

DJH


Ampere’s law in integral form

Ampere’s law in derivative form

Apply Stoke’s theorem

Ampere’s Law in derivative form: Valid for all points in space

t

DJH


Figure 8.1 Application of Ampere’s law to a circuit with an �air-filled capacitor and time-varying current.

NOW it might make sense, if

i = dD/dt Sp

Taken from ELEC 3909

t

DJH

�

Ratio Jc to JDSome materials are neither good conductors nor perfect dielectrics, so that both conduction current and displacement current exist. A model for a poor conductor or lossy dielectric is shown below. Assume a time dependence of ejwt for the electric field.

E

s e

Then

Remember this result (for ELEC 3909)

Ratio Jc to JDA circular cross-section conductor of radius 1.5 mm caries a current ic = 5.5 sin(4E10t)(m A). What is the amplitude of the displacement current density if s = 35 MS/m and r = 1.

E

s e

Then

Remember this result (for ELEC 3909)

79.9

369

104

75.3E

EE

E

wJ

J

D

c

s

2

2

386.779.9

35.1/65.5

m

AE

E

EEJ D

m

Faraday Homopolar Generator

B

a

V

The circular metal disk rotates at (rad/s) in a uniform flux density B. Sliding contacts connect a voltmeter to the disk. What voltage is indicated on the meter.Solution: One radial element of the disk is examined. A general point on the radial element has velocity:

ˆrv

r

Faraday Homopolar GeneratorIn steady state the magnetic force on the free charges in the disk will equal the

electric force induced by charge migration. We can then equate electric and magnetic forces to solve for the electric field in the rotating disk.

ˆrv me FF

qvBqE

)ˆ( rrBE

aa

wrBdrdEV00

Faraday Homopolar Generator

B

a

V

The circular metal disk rotates at (rad/s) in a uniform flux density B. Sliding contacts connect a voltmeter to the disk. What voltage is indicated on the meter.

aa

wrBdrdEV00

2

2BaV

Super-Matrix Formulation

H

E

H

E

ro

ro

mm

01

10

tje

E

cE

r

r

m

01

10

HjZo

'FFBeeeR

rJC

Eo

qpn

EzjkzjGjqpoqpn

qpn

E zn

zjGjqpoqpn

qpno

o

r

r neeR

rJC

m

m

m

1

1

Solve for allowed Frequencies

Super-Matrix Formulation

E

cE

r

r

m

01

10

Wav

elen

gth

Tabulation Index

Band Gap

Resonator States

Super-Matrix FormulationW

avel

engt

h

Tabulation Index

Band Gap

Resonator States

Axial magnetic field

Documents

Introduction: So far we have These equations are OK for static fields, i.e. those fields independent of time. When fields vary as a function of time the