Electromagnetic Theory

PHYS 402

Electrodynamics

• Ohm’s law

• Electromotive Force

• Electromagnetic Induction

• Maxwell’s Equations

1

7.1.1 Ohms Law

the Ohm’s law

=

J fσ

Current density conductivity force per unit charge

1=σ

ρ resistivity

For the EM force

Usually v is small so = J Eσ

=

JE

σ For good conductors σ >>1 so E≃0

Empirical, approximate and not applicable to all media.

10 0∇ ⋅ = ∇ ⋅ = ⇒ = ∇ ⋅ =

E J Eρ ε

σ

for steady state currents and uniform conductivity

Just like in the static case charge density is zero in a conductor

2 0 Laplace equationV∇ =So can apply 2

V

I=JAIn case of a conductor electric field is

proportional to the potential difference I=JA= σEA= σA(V/ℓ)

V=IR Ohm’s Law

potential current resistance [ in ohms (Ω) ]

ℓ

3

Example:

I=?

R=?

uniform

uniformσ

VA resistor has across sectional area A and conductivity σ.if the potential difference between ends is V what is the current.

Electric field is uniform within the wire

= = = = ⇒ =V A L

I JA EA A V RL L A

σσ σ

σ

parallelin

seriesin 1 2 1 2,L L R R R= +

211 2

1 1 1,A A

R R R= +

with the same σ and A you are adding L1 to L2

with the same σ and L you are adding A1 to A2

1

Rpar

=1

R1

+1

R2

=1

ρLA1 + A2( )

I =V

R1

+V

R2same V

4

Example:

To prove the field E is uniform

i.e.,

V=0 V=V0

A=const

σ =const

ˆ0 0 on the cylindrical surfaceJ J n∇ ⋅ = ⋅ =

∵

ˆ ˆ0 0E n n V∴ ⋅ = ⇒ ⋅∇ =

0V

n

∂=

∂

2 0 has only z dependanceV∇ =

0( )V z

V zL

∴ =

0 ˆV

E V zL

= −∇ = −

5

• In an electric field E, charges accelerate.

• So why doesn’t the current is increasing with time ?

– charges (electrons) undergo collisions and lose speed so their

speed won’t increase indefinitely, have an average constant velocity.

If the average distance a charge travel in between collisions is λ, acceleration a and the average time it takes is t,

=

⇒ average velocity =

=

∝

not proportional to E!

Charges have random thermal velocities which are much larger than the change

in speed due to field. Time between collisions are determined by the thermal

speed

Microscopic aspects of the Ohms law

1

2 2= ⇒ = =

thermalave

thermal

at v at

v v

λ λ

6

2

2 2

= = =

thermal thermalave

nfq F nf qJ n f q v E

v m v

λ λ

molecule density charge (e)

free electrons per molecule

Mass (of e)

σ

2

2 thermal

nf qJ E

mv

λ=

Thermal velocity increases with temperature, so according to this simple model

conductivity decreases with temperature which agree with observations.

Collisions convert work done by the field to heat.

Work done per unit charge = V ;

charge following per unit time =I

⇒ = ! = ! Joule’s law

7

8

• There are two forces driving the current around the circuit.

– driving force from the source (ie. From battery, generator ….)

– electrostatic forces (that communicate the influence of the force and smooth

out the flow)

"# = "$ %

electromotive force (emf)

7.1.2 Electromotive Force

If the current is not the same, fields produced by accumulating

charges acts in such a way to even out the flow.

The current is the same all the way around the loop.

( ) 0= = ⋅ = ⋅ ⋅ =∫ ∫ ∫ ℓ ℓ ∵ ℓ sf d f d E dε

9

Within an ideal source net force on charges is zero

(i.e. & = '" '>>1 ⇒f=0)

The potential difference between terminals a,b

equal to the electromotive force of the source.

So the function of the source (ie. a battery) is to

maintain a voltage difference equal to the emf.

The resulting electrostatic field drives the current

around the rest of the circuit.

The battery operates like a pump that moves charges

from a lower to a higher electrical potential.

sE f= −

b b

ab s sa aV E d f d f d ε= − ⋅ = ⋅ = ⋅ =∫ ∫ ∫

ℓ ℓ ℓ

a

b

10

7.1.3motional emf

,,

mag vmag v

Ff vB

q= =

⊗B

,mag vF qvB=

,= ⋅ =∫

ℓ mag vf d vBhε

Suppose a conductive loop is moved at a velocity v in a magnetic field B as shown

in the figure .

Magnetic force on charges in the conductor ab:

magnetic force per unit charge

This produces an emf

Which moves charges up along the conductor at a velocity u

11

12

So the motion of charges is the sum of v and u.

u motion of charges produce a Lorenz force toward left = -.

to move the conducting loop to right this force has to be overcome by a similar force "/011 = -.

The amount of work done on the loop

⊗B

pull

hf d (uB)( )sin

cos

uBh tan vBh

= ⋅ = θθ

= θ = = ε

∫

ℓ

(note: work is done by the pull force, not by B)

( ) ( ) ( )= = − = − = − = − Φ dx d d d

vBh Bh Bhx BAdt dt dt dt

ε ( magnetic flux)Φ13

∴electromotive force in the loop 3 = 456

57

General proof:

( ) ( )

=

Φ = Φ + − Φ = Φ

⋅∫

ribbon

ribbon

d t dt t

B da

( ) ( )

( )

( )

= × = + × = ×

Φ= ⋅ ×

= − × ⋅ = − ⋅ = −

∫

∫ ∫

ℓ ℓ ℓ

ℓ

ℓ ℓ

mag

da v d dt v u d dt w d dt

dB w d

dt

w B d f d ε

magf

d

dt

Φ⇒ ε = −

14

Example:

=?

0

a

magf dsε = ⋅∫

0

awsB ds= ∫

2

2

wBa=

2

R 2

wBaI

R

ε= =

ˆ( )= × = × = magf v B s w B sB sω ω

A metal disk of radius a is in a uniform magnetic field B and rotates with

angular velocity ω as shown in the figure. Find the current in the resister.

15

• In the first experiment a loop of wire in a magnetic field was moved,

and a current in the loop was observed.

This is due to motional emf 3 = 456

57

• In the second experiment the magnet was moved holding the loop still,

a current in the loop was observed.

Here charges are not moving, so there isn’t a magnetic force to create

an emf. So Faraday suggested that The changing magnetic field must

be creating an electric field that produced the observed current (emf).16

7.2 Electromagnetic Induction

Faraday experiments and Faraday’s law:

A changing magnetic field induces an electric field.

Experimentally Faraday found that the emf is equal to the rate of change of

flux: 3 = ∮ ⋅ ;< = 456

57

17

Faraday Law

( )

BE d da

t

B BE d E da da E

t t

∂⋅ = − ⋅

∂

∂ ∂⋅ = ∇ × ⋅ = − ⋅ ⇒∇ × = −

∂ ∂

∫ ∫

∫ ∫ ∫

ℓ

ℓ

In experiment 3 When the strength of the magnetic field was changed (using an

electromagnet) a current in the loop was observed.

Again the changing magnetic field induces an electric field producing an emf

according to the Faraday law. So whenever the magnetic flus through a loop

changes, an emf 3 456

57will be induced in the loop.

18

Example 7.5: The emf produced when a long cylindrical magnet is passed

through a circular wire ring of slightly larger diameter.

Lens Law: The induced current flow in such a direction that the flux it

produces tend to cancel the change.

Nature abhors a change in flux

19

In a time varying magnetic field

If there are no charges (pure Faraday field)

These are mathematically equivalent to magnetostatic equations

and similar to Ampere law

So similar techniques can be applied as the symmetry permits.

20

7.2.2The Induced Electric Field

BE

t

∂∇ × = −

∂

0 ( 0)E ρ∇ ⋅ = =

0B Jµ∇× =

0B∇ ⋅ =

∫Φ

−=⋅dt

ddE ℓ

0 encB d Iµ⋅ =∫ℓ

Example:

If the uniform magnetic field B is changing find the induced electric field.

If B is increasing E runs clockwise as viewed from above

21

2 2.2 [ ( )]2

d d dB s dBE d E s s B t s E

dt dt dt dtπ π π

Φ⋅ = − ⇒ = − = − ⇒ = −∫

ℓ

• A charge ring of linear density λ and radius b is suspended horizontally so that it

is free to rotate. In the center there is a magnetic field B up to radius a . What

happens when the field is turned off.

22

dt

dBa

dt

ddE

2π−=Φ

−=⋅∫ ℓ

2 2

ˆ )

ˆ ˆ ˆ[ ]

dN b F zb Ed

dB dBN dN zb E d zb a zb a

dt dt

λ

λ λ π λπ

= × =

= = ⋅ = − = −∫ ∫

ℓ

ℓ

Torque on dl

Change in the angular momentum on the wheel

0

02 2

0B

N dt b a dB a bBλπ λπ= − =∫ ∫

Example:

• Example: A infinity straight wire carries a slowly varying current I(t). Find

the induced electric field.

23

0 ˆ2

IB

s

µϕ

π≅

0 00

0

0 0

'2 '

1 ( ) ( ) ' (ln ln )

2 ' 2

s

s

Id dE d B da ds

dt dt s

dI dIE s E s ds s s

dt s dt

µ

πµ µ

π π

⋅ = − ⋅ = − ⋅ ⋅

− = − == − −

∫ ∫ ∫

∫

ℓ ℓ

ℓ ℓℓ ℓ

0 ˆ( ) [ ln ]2

dIE s s K z

dt

µ

π∴ = + ℓ

K is independent of s but could depend on t.

This diverge with s,: at large distances this technique is not valid sine EM

disturbances propagate at a finite speed, not instantly.

⦿B

similar to a solenoid induced E field

should be parallel to axis (wire)

24

25

7.2.3 Inductance

2 1 2 1 2

0 1 12

0 1 221

( )

4

4

A da A d

I dd

R

d dM

R

µ

π

µ

π

Φ = ∇× ⋅ = ⋅

= ⋅

⋅∴ =

∫ ∫

∫ ∫

∫ ∫

ℓ

ℓℓ

ℓ ℓ

M21 mutual inductance

0 11 1 12

ˆ

4

dB I I

µ

π

×= ∝∫

ℓ

r

r

0 1 11

4

I dA

R

µ

π= ∫

ℓ

Consider two conductive loops, some

of the fields lines produced by the

current I1 in loop 1 pass through loop 2

Flux through loop2 2 1 2 1

21 1 =

B da I

M I

Φ = ⋅ ∝∫

A1 is given by

The mutual inductance is a purely

geometrical quantity

M21 = M12 = M Φ1 = M12 I2 Φ1 = Φ2 if I1 = I2

Calculation flux due to the field of inner solenoid is complicated, but

Φ1 = Φ2 if I1 = I2 , and the field of the long solenoid is uniform.

Example

20 1 2M a n nµ π= ℓ

A short solenoid (length l, radius a n1 turns per unit length) lies in the axis of a long

solenoid ( radius b n2 turns per unit length). If current I flows through the small solenoid what is the flux through the long solenoid.

1 1 1,

2 21 2 0 1 2 2

20 1 2

2

per turnn

n a B a n n I

a n n I

π µ π

µ π

Φ = ⋅Φ

= ⋅ ⋅ =

=

= Φ

ℓ

ℓ ℓ

ℓ

0 2B n Iµ=

I

In fact when the current through loop 1 is changed flux in loop 1 also changes, so according

to Faraday's law that should induces an emf in loop 1 itself. Since again flux is

proportional to current

27

dt

dIM

dt

d 122 −=

Φ−=ε

If the current in the loop 1 is changed since that will change the flux through

the loop 2 that induces current in loop2

dI

LI Ldt

εΦ = ⇒ = −

L self-inductance depends on the geometry of the loop

unit: Hendry (H) unit: volt/ampere/second volt.second/ampere

This emf in a direction to

counter the change in current

(back emf)

Example: Find the self-inductance of a toroidal coil with rectangular cross

section (inner radius a, outer radius b and total N turns)

28

∫ ⋅=Φ adBN

s

NIB

π

µ

20=

0

2 20 0

1

2

ln ( ) ln ( )2 2

b

a

NIN h ds

s

N Ih N hb bL

a a

µ

πµ µ

π π

=

= ⇒ =

∫

If the circuit is disconnected (cut the wire) the current drops

instantaneously, producing a large emf (reason for spark when tyrn off the

circuit) 29

Example:

IRdt

dIL =−0ε

0( )

Rt

LI t keR

ε −= +

(particular solution)

)1()1()( 00 τεεt

tL

R

eR

eR

tI−−

−=−=

find the current in the circuit ( ) ?I t =

0if (0) 0 ,I kR

ε= = −

time constantL

Rτ =

(general solution)

R0ε

30

7.2.4 Energy in a Magnetic Field

∵

In electrostatics, the energy in an electric field is given by:

2 21 1 1( ) 2 2 2

mm

dW dI dI L I LI W LI I

dt dt dtε= − = = ∴ = = Φ

( )s s loopB da A da A dΦ = ⋅ = ∇ × ⋅ = ⋅∫ ∫ ∫

ℓ

1 1( )

2 2m loop loop

W I A d A I d= ⋅ = ⋅∫ ∫ ℓ ℓ

( ) 20 01τ

2 2τ

2eW Vd E Vd E d

ερ τ

ε= = ∇ ⋅ =∫ ∫ ∫

magnetic fields do no work on charges, but to drive (start) a current back

emf has to be overcome. This requires some work to be dome and that

energy is stored in the magnetic field produced.

1( )

2m

VW A J dτ= ⋅∫

Similarly for a volume current density J

32

since )()()( BAABBA

×∇⋅−×∇⋅=×⋅∇

∫ ⋅×s

adBA

)(∞→s

0

∫∫ == τε

τρ dEdVWelec20

2)(

2

1

∫∫ =⋅= τµ

τ dBdJAWmag2

02

1)(

2

1

1 1

2 2

1 1 = .( )

2 2

mV V

V V

W A Jd A Bd

B d A B d

τ τµ

τ τµ µ

= ⋅ = ⋅∇×

− ∇ ×

∫ ∫

∫ ∫

0

2

0 0

1=

2m

VW B dτ

µ ∫2

0

Example:

Find the magnetic energy stored in a coaxial cable of length l carrying a

current I.

33

bsas

IB φ

π

µ ˆ2

0=

＜ ＜ 0=B

20

0

1( ) (2 )

2 2B B

IW dW sds

s

µπ

µ π= =∫ ∫ℓ ℓ

s a<s b>

20 ln( )4

I b

a

µ

π=

ℓ

2 01ln ( )

2 2B

bW L I L

a

µ

π= ⇒ =

ℓ

20

4

b

a

I ds

s

µ

π= ∫ℓ

34

7.3 Maxwell's Equations

0

Eρ

ε∇ ⋅ =

0B∇ ⋅ =

BE

t

∂∇× = −

∂

0B Jµ∇ × =

(Gauss Law) (Faraday’s Law)

(Ampere’s Law)

So far:

as expected0 0B B

Et t

∂ ∂∇ ⋅∇ ⋅∇ × = = ∇ ⋅ − = − = ∂ ∂

0 00B Jt

ρµ µ

∂∇ ⋅∇ × = = ∇ ⋅ = −

∂

Which is zero for steady

currents but not always zero.

Consider a charging capacitor, here the current

passing through the loop is ill defined.

Ienc is the current passing through a surface that has

loop as the boundary.

For the surface in between plates, which has Ienc=0

For surface on the other side Ienc=I

35

0

0 0 0

EJ

t t

E EJ

t t

ρε

ε ε

∂ ∂∇ ⋅∇ ⋅ = − = −

∂ ∂

∂ ∇ ⋅ ∂ ∇ ⋅= −∇ ⋅ ⇒ ∇ ⋅ + =

∂ ∂

So if the Ampere’s law is modified such that t

EJB

∂

∂+=×∇

000 εµµ

000 0

EB J

t

εµ

∂ ∇ ⋅∇ ⋅∇ × = = ∇ ⋅ + =

∂

Which would be true all situations

The extra term is called the displacement current

the modified Ampere’s law implies that a

changing electric field induces a magnetic field.

0dE

Jt

ε∂

=∂

36

for the capacitor problem

The E field between capacitorsA

QE

00

11

σσ

ε== I

Adt

dQ

At

E

00

11

εε==

∂

∂

0 0 0encE

B dl I dat

µ µ ε∂

⋅ = + ⋅∂∫ ∫

Therefore in between plates Ienc=0 and 0

00

.E

da I A It A

εε

ε

∂⋅ = =

∂∫

Same as any surface intersecting the wire.

37

Maxwell’s Equations

0=⋅∇ B

Et

JB

∂

∂+=×∇

000εµµ

t

BE

∂

∂−=×∇

0ε

ρ=⋅∇ E

Gauss’s law

Faraday’s law

Ampere’s law with Maxwell’s correction

Force law

continuity equationt

J∂

∂−=⋅∇

ρ

( the continuity equation can be obtained from Maxwell’s equations )

( )F q E v B= + ×

Example: Two concentric spherical shells carry charges Q(t) and –Q(t), space in between

them is filled with a material with conductivity σ. Show that the magnetic field inside is zero.

How is it expanded in terms of displacement currents.

38

20 0

1ˆ .

4

Q QJ E r I Q J da

r

σσ σ

πε ε= = ⇒ = − = =∫ ɺ

Due to radial symmetry, magnetic field only has a radial component B.

20 . (4 ) 0 0B B da B r Bπ∇ ⋅ = ⇒ = = ⇒ =∫

ohmic currant alone cannot explain B=0

0 2 20

ˆ ˆ 04 4

d dE d Q Q

J r r J Jt dt r r

σ σε

π πε

∂ = = = − ⇒ + = ∂

Q(t)

-Q(t)

In matter there are:

39

7.3.5 Maxwell’s Equations in Matter

bound charges bound currentsPb ⋅−∇=ρ MJb

×∇=

Change in polarization results in a flow of a current which must be

included in the total current

-+ -+ -+ -+-+ -+ -+ -+ -+ -+ -+ -+

- + - + - + - +- + - + - + - +- + - + - + - +

P increased 0

p p b bp p

J JJ J

t t t t

σ σ∂ ∂∇ ⋅ ∂ ∂∇ ⋅ = ∇ ⋅ = = − ⇒∇ ⋅ + =

∂ ∂ ∂ ∂

Continuity equation satisfied.

Pfbf

⋅∇−=+= ρρρρcharge density

current density Pt

MJJJJJ fPbf

∂

∂+×∇+=++=

ˆb

P

P n PdI da da da

t t t

PJ

t

σ⊥ ⊥ ⊥

∂ ∂ ⋅ ∂= = =

∂ ∂ ∂

∂⇒ =

∂

40

00

1Gauss's law ( ) ( )f fE P D D E Pρ ρ ε

ε∇ ⋅ = − ∇ ⋅ ⇒∇ ⋅ = = +

Et

Pt

MJB f

∂

∂+

∂

∂+×∇+=×∇ 000 )( εµµAmpere’s law ( with Maxwell’s term )

)()( 0000 PEt

JMB f

+

∂

∂+=−×∇ εµµµ

Dt

JH f

∂

∂+=×∇

0

1 H B M

µ

= −

fD ρ=⋅∇

Dt

JH f

∂

∂+=×∇0=⋅∇ B

t

BE

∂

∂−=×∇

In terms of free charges and currents, Maxwell’s equations become

ExP e

0ε= ED

ε=

HxM m

= BH

µ

1=

)1(0 ex+= εε

)1(0 mx+= µµ

, and , are mixed.D H E B

In these equations

41

7.3.6 Boundary conditions

Maxwell’s equations in integral form can be used to deduce boundary

conditions

Over any closed surface S

for any surface S bounded

by the closed loop L

L s

dE d B da

dt⋅ = − ⋅∫ ∫ ℓ

,f encsD da Q⋅ =∫

0sB da⋅ =∫

fencL s

dH d I D da

dt⋅ = + ⋅∫ ∫ ℓ

applying above to thin

Gaussian pillbox

42

1 1,D B

2 2,D B

aaDaD fσ=⋅−⋅

21

fDD σ=− ⊥⊥21

021 =− ⊥⊥BB

0→S∫ ⋅−=⋅−⋅ 021 adB

dt

dEE

ℓ

ℓ

021 =− EE

= =

nKHH f ˆ21 ×=−= =

nKBB f ˆ11

2

2

1

1

×=−

µµ

= =

applying to a thin Ampereian loop

1 2 ˆ ˆ( ) ( )free f fH l H l I K n l l K n⋅ − ⋅ = = ⋅ × = ⋅ ×

43

1 2 fD D σ⊥ ⊥− =1 2 0E E− =

1 2 0B B⊥ ⊥− =

1 1 2 2 fE Eε ε σ⊥ ⊥− =

1 21 2

1 1ˆfB B K n

µ µ− = ×

(i)

(ii)

(iii)

(iv)

44