Download Lecture Notes 01: Conservation Laws: Continuity Equation (Charge Conservation), Poynting's Theorem (Energy Conservation), Poynting's Vector

UIUC Physics 436 EM Fields & Sources II
Fall Semester, 2015
Lect. Notes 1
Prof. Steven Errede
LECTURE NOTES 1
CONSERVATION LAWS


Conservation of energy E, linear momentum p , angular momentum L and electric charge q
are of fundamental importance in electrodynamics (n.b. this is also true for all fundamental
forces of nature – the weak, strong, EM and gravitational force, both microscopically (locally),
and hence macroscopically (globally - i.e. the entire universe)!
Electric Charge Conservation
Previously (i.e. last semester in Physics 435), we discussed electric charge conservation:
Surface area element,
free
Enclosing
surface, S
Volume, v
Electric current flowing outward from volume v



I free  t    J free  r , t da Amperes 
S

Q free  t     free  r , t  d  Coulombs 
through closed bounding surface S at time t:
Electric charge contained in volume v at time t:
v
An outward flow of current through surface S corresponds to a decrease in charge in volume v:
I free  t   
dQ free  t 
dt
 Amperes  Coulombs sec  i.e.
But:
dQ free  t 
dt

d

 free  r , t  d  

v
v
dt
t
dt
 0 , I free  t   
dQ free  t 



I free  t    J free  r , t da  
Global conservation of electric charge:

 free  r , t 
dQ free  t 
S
0
dt
dQ free  t 
dt
d
Use the divergence theorem on the LHS of the global conservation of charge equation:

 
 free  r , t 

v  J free  r , t  d  v t d  Integral form of the continuity equation.
This relation must hold for any arbitrary volume v associated with the enclosing surface S;
hence the integrands in the above equation must be equal – we thus obtain the continuity

equation (in differential form), which expresses local conservation of electric charge at  r , t  :

 
 free  r , t 

 J free  r , t   
 Differential form of the continuity equation.
t
n.b. The continuity equation doesn’t explain why electric charge is conserved
– it merely describes mathematically that electric charge is conserved!!
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2015. All Rights Reserved.
1
UIUC Physics 436 EM Fields & Sources II
Fall Semester, 2015
Lect. Notes 1
Prof. Steven Errede
 
Poynting’s Theorem and Poynting’s Vector S  r , t 
We know that the work required to assemble a static charge distribution is:
o


E 2  r , t  d  o

v
2
2
WE  t  




1




  E  r , t  E  r , t   d  2   D  r , t  E  r , t   d
v
v
SI units:
Joules
Linear Dielectric Media
Likewise, the work required to get electric currents flowing, e.g. against a back EMF is:
WM  t  
1
2 o
1
2 
v B  r , t  d  2o




1




  B  r , t  B  r , t   d  2   H  r , t  B  r , t   d
v
v
SI units:
Joules
Linear Magnetic Media
Thus the total energy, UEM stored in EM field(s) is (by energy conservation) = total work done:
U EM  t   Wtot  t   WEM  t   WE  t   WM  t  
1 
1

 

 o E 2  r , t   B 2  r , t   d   uEM  r , t  d


v
2 v
o

1 
1 2  


U EM  t    u EM  r , t  d     o E 2  r , t  
B  r , t   d
v
2 v
o

SI units:
Joules
1
1 2  


B  r , t   (SI units: Joules/m3)
where uEM = total energy density: u EM  r , t     o E 2  r , t  
2
o

 

Suppose we have some charge density   r , t  and current density J  r , t  configuration(s)
 
 
that at time t produce EM fields E  r , t  and B  r , t  . In the next instant dt, i.e. at time t + dt, the
charge moves around. What is the amount of infinitesimal work dW done by EM forces acting
on these charges / currents, in the time interval dt ?
The Lorentz Force Law is:
 
 
 
 
F r ,t   q E r ,t   v r ,t  B r ,t 


The infinitesimal amount of work dW done on an electric charge q moving an infinitesimal
 
distance d   vdt in an infinitesimal time interval dt is:

n.b. to v !!!

 
   
 
 
  
dW  F d   q E  v  B d   qE vdt  q v  B vdt  qE vdt





(n.b. magnetic forces do no work!!)
0
But:
2


q free  r , t    free  r , t  d
and:

 


 free  r , t  v  r , t   J free  r , t 
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2015. All Rights Reserved.
SI units:
Joules
UIUC Physics 436 EM Fields & Sources II
Fall Semester, 2015
Lect. Notes 1
Prof. Steven Errede
The (instantaneous) rate at which (total) work is done on all of the electric charges within the
volume v is:
 
 
   
dW  t 
    
  F  r , t  d   r , t  dt   F  r , t v  r , t    q free  r , t  E  r , t v  r , t 
v
v
v
dt


  


   free  r , t  d E  r , t v  r , t  using : q free  r , t    free  r , t  d
v
 
  
  

  E  r , t   free  r , t  v  r , t  d but : J free  r , t    free  r , t  v  r , t 

v



 

dW  t 

  E  r , t  J free  r , t  d  P  t  = instantaneous power (SI units: Watts)
v
dt
 


The quantity E  r , t  J free  r , t  is the (instantaneous) work done per unit time, per unit volume –



i.e. the instantaneous power delivered per unit volume (aka the power density).
 

dW  t 

Joules
  E  r , t  J free  r , t  d (SI units: Watts =
)
v
dt
sec
 
We can express the quantity E  J free in terms of the EM fields (alone) using the Ampere
Maxwell law (in differential form) to eliminate J free .

P t  
Thus:



Ampere’s Law with Maxwell’s Displacement Current correction term (in differential form):
 
  

 

E  r , t 


  B  r , t   o J free  r , t   J D  r , t   o J free  r , t   o o
t
 

E  r , t 
1   

J free  r , t  
  B r ,t   o
Thus:
t
o
 
 

   1   
E  r , t  

E  r , t  J free  r , t   E  r , t 
  B r ,t   o

t 
 o
Then:
 
  
  E  r , t 
1  
E  r , t    B  r , t    o E  r , t 

o
t







Now: 

Thus: E 

 

 


 E  B   B   E   E    B 
 
     
  B   B  E    E  B 


Griffiths Product Rule #6 (see inside front cover)
 
  
B  r , t 
But Faraday’s Law (in differential form) is:   E  r , t   
t

  
 B   

E    B   B    E  B
t


 B 1   
 E 1   
1  2
1  2

B B 
B  and similarly: E 

EE 
However: B 

E 
t 2 t
2 t
t 2 t
2 t








© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2015. All Rights Reserved.
3
UIUC Physics 436 EM Fields & Sources II
Fall Semester, 2015
Lect. Notes 1
Prof. Steven Errede
Therefore:
  
 

 
1  1  2 


1  2 

E  r , t  J free  r , t    
B  r , t     E  r , t   B  r , t     o 
E  r , t  


o  2 t

 2 t

 
1 
1 2   1   
2 

B  r , t     E  r , t   B  r , t 
oE r ,t  
o
2 t 
 o
Then:
 

dW  t 

P t  
  E  r , t  J free  r , t  d
v
dt
 
1 d 
1 2  
1   
2 







E
r
t
B
r
t
d
E
r
t
B
,
,
,







 r , t  d


o
v
o
o 
2 dt v 










Apply the divergence theorem to this term, get:
Poynting’s Theorem = “Work-Energy” Theorem of Electrodynamics:
P t  
dW  t 
d  1 
1 2   
1

B  r , t    d 
    oE2 r ,t  
dt
dt v  2 
o
o
 
 
 

  E  r , t   B  r , t  da
S
1 
1 2  
2 

E
r
t
B  r , t   d = instantaneous energy stored in the EM fields

,



o
o
2 v 

 
 
E  r , t  and B  r , t  within the volume v (SI units: Joules)
 
 
1

Physically, the term   E  r , t   B  r , t  da = instantaneous rate at which EM energy is
Physically,


o
S


carried / flows out of the volume v (carried microscopically by virtual (and/or real!) photons


across the bounding/enclosing surface S by the EM fields E and B  i.e. this term represents/is
the instantaneous EM power flowing across/through the bounding/enclosing surface S
(SI units: Watts = Joules sec ).
Poynting’s Theorem says that:
The instantaneous work done on the electric charges in the volume v by the EM force is equal to


the decrease in the instantaneous energy stored in EM fields ( E and B ), minus the energy that is
instantaneously flowing out of/through the bounding surface S .
 
 
1  
E  r , t   B  r , t  = energy / unit time / unit area,
We define Poynting’s vector: S  r , t  
o




transported by the EM fields ( E and B ) across/through the bounding surface S

n.b. Poynting’s vector S has SI units of Watts/m2 – i.e. an energy flux density.
4
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2015. All Rights Reserved.
UIUC Physics 436 EM Fields & Sources II
Thus, we see that:
P t  
Fall Semester, 2015
Lect. Notes 1
Prof. Steven Errede
 
dW  t 
dU EM  t 


  S  r , t da
S
dt
dt
 

where S  r , t da = instantaneous power (energy per unit time) crossing/passing through an

ˆ , as shown in the figure below:
infinitesimal surface area element da  nda
n̂ = outward pointing unit normal
vector (everywhere  to surface S )
n̂
EM energy flowing out of volume v
through enclosing surface S
da
ẑ
ŷ

x̂
Volume v
Enclosing surface S
 
Poynting’s vector: S  r , t  
1
o
 
 
E  r , t   B  r , t  = Energy Flux Density (SI units: Watts/m2)
The work W done on the electrical charges contained within the volume v will increase their
mechanical energy – kinetic and/or potential energy. Define the (instantaneous) mechanical

energy density umech  r , t  such that:

 


dumech  r , t   
dU mech


 E  r , t  J free  r , t  Hence:
  E  r , t  J free  r , t  d
v
dt
dt

Then: P  t  

 

dW  t  dU mech d



  umech  r , t  d   E  r , t  J free  r , t  d
v
v
dt
dt
dt


However, the (instantaneous) EM field energy density is:
1
1 2  


u EM  r , t     o E 2  r , t  
B  r , t   (Joules/m3)
2
o

Then the (instantaneous) EM field energy contained within the volume v is:

U EM  t    u EM  r , t  d (Joules)
v
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2015. All Rights Reserved.
5
UIUC Physics 436 EM Fields & Sources II
Thus, we see that:
Fall Semester, 2015
Lect. Notes 1
Prof. Steven Errede
 
  
d



umech  r , t   uEM  r , t   d    S  r , t da     S  r , t  d


S
v
dt v


Using the
Divergence
theorem
The integrands of LHS vs. {far} RHS of the above equation must be equal for each/every space
time point  r , t  within the source volume v associated with bounding surface S. Thus, we obtain:
  



umech  r , t   uEM  r , t     S  r , t 
The Differential Form of Poynting’s Theorem:
t
Poynting’s theorem = Energy Conservation “book-keeping” equation, c.f. with the
Continuity equation = Charge Conservation “book-keeping” equation:
  


  r , t    J  r , t 
The Differential Form of the Continuity Equation:
t


u  r , t   

Since mech
 E  r , t  J free  r , t  , we can write the differential form of Poynting’s theorem as:
t

  
 

uEM  r , t 

  S  r , t 
E  r , t  J free  r , t  
t

 

uEM  r , t    

  S  r , t   0
E  r , t  J free  r , t  
Or:
t
 
Poynting’s Theorem / Poynting’s vector S  r , t  represents the (instantaneous) flow of EM


energy in exactly the same/analogous way that the free current density J free  r , t  represents the
(instantaneous) flow of electric charge.
In the presence of linear dielectric / linear magnetic media, if one is ONLY interested in FREE
charges and FREE currents, then:
 
 
 
1  
free 
uEM
 r , t   E  r , t  D  r , t   B  r , t  H  r , t 
2

 
 
D r ,t    E r,t 
   o 1   e 
 
 
 
 
 
S  r , t   1 E  r , t   B  r , t   E  r , t   H  r , t 
 
 
B r ,t   H r ,t 
  o 1   m 

6
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2015. All Rights Reserved.
UIUC Physics 436 EM Fields & Sources II
Fall Semester, 2015
Lect. Notes 1
Prof. Steven Errede
Griffiths Example8.1:
Poynting’s vector S , power dissipation and Joule heating of a long, current-carrying wire.
When a steady, free electrical current I (≠ function of time, t) flows down a long wire of
length L  a (a = radius of wire) and resistance R   L  a 2 C  , the electrical energy is
dissipated as heat (i.e. thermal energy) in the wire.
Electrical power dissipation: P  V  I  I 2 R
I
V

S
Long wire of
resistance R

S   ˆ
a
I
I
Battery
(or power supply)
V1
1
o
 
EB

E   zˆ

B  ˆ
La
V


J free   C E   I  a 2  zˆ  Amps m 2 

 J free V

zˆ Volts m 
Longitudinal Electric Field: E 
C
L
Free Current Density:
ẑ
V2

E
V  V1  V2   0  Volts 
Potential Difference:
V
L
0
 a


n.b. The {steady} free current density J free (   C E  I  a 2 ) and the longitudinal electric field

E   V L  zˆ are uniform across (and along) the long wire, everywhere within the volume of
the wire    a  .  Thus, this particular problem has no time-dependence…

 I
B inside    a   o 2 ˆ   x 2  y 2 in cylindrical coordinates
From Ampere’s Law:
2 a
 outside
  
o I
B


a

ˆ (Tesla)


B
r
d


I




o
encl
 C
2


n.b. for simplicity’s sake, we have approximated the finite length wire by an ∞-length wire.
This will have unphysical, but understandable consequences later on….
o I
 

1    
E r  B r 
Poynting’s Vector: S  r  
B
2 a
o
 ˆ

 inside
V  I 
V  I 
S
   a 
 zˆ  ˆ  
  ˆ 
2
2 a L
2 a 2 L

Poynting’s vector S oriented radially inward for   a .


S outside    a   0 {because E    a   0 !!!}
0
 a
 in

B    a  varies B out    a 
linearly with 
varies as 1


© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2015. All Rights Reserved.
7

UIUC Physics 436 EM Fields & Sources II
Fall Semester, 2015
Lect. Notes 1
Prof. Steven Errede
Note the following result for Poynting’s vector evaluated at the surface of the long wire, i.e. @   a :

V  I
S inside    a  
  ˆ  (SI units: Watts/m2)
2 aL


S outside    a   0   a discontinuity in S at   a !!!

Since E outside    a   0 :

S 
V  I
2 aL
 a
0

V  I 
V  I 
ˆ
Slong    a  
  ˆ   
2
2 a L
2 a 2 L
wire

Now let us use the integral version of Poynting’s theorem to determine the EM energy flowing
through an imaginary Gaussian cylindrical surface S of radius   a and length H  L :
 

dW  t  dU mech d



  umech  r , t  d   E  r , t  J free  r , t  d
v
v
dt
dt
dt
 
  
dU EM  t 
d



  S  r , t da    u EM  r , t  d    S  r , t  d
S
v
dt
dt v

P t  



Since this is a static/steady-state problem, we assume that dU EM  t  dt  0 .

 ˆ , nˆ2 , da2
a
da2


S   ˆ

 zˆ, nˆ1 , da1
da1
H
V
V1

 zˆ, nˆ3 , da3
Gaussian Surface S
da3
x̂

ẑ
ŷ
V2
Then for an imaginary Gaussian surface taken inside the long wire (   a ):
Pwire

0

0

 



   S wire da   LHS S da1  cyl S da2  RHS S da3
S
endcap
surface
endcap
 



 





S    ˆ  is  to da1    zˆ  ;
da1  da1   zˆ 
da2  da2 ˆ
da3  da3   zˆ 


S    ˆ  is anti-  to da2    ˆ  ;


S    ˆ  is  to da3    zˆ 
Only surviving term is:

  2  V  I 
z  H
 2 


 V  I 
ˆ
ˆ
2





S   da2    H 2 


d

dz



H
V
I






2
2
2 
 0 
z 
2
surface
 2 a H 
 2 a H 
a 
Pwire      cyl
8
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2015. All Rights Reserved.
UIUC Physics 436 EM Fields & Sources II
Fall Semester, 2015
2

Thus: Pwire     V  I   (Watts)
a
And:
Pwire    a   V  I
Lect. Notes 1
Prof. Steven Errede

V  I  
a
Pwire   
2
(Watts)
 a
0

This EM energy is dissipated as heat (thermal energy) in the wire – also known as Joule
heating of the wire. Since Pwire      2 , note also that the Joule heating of the wire occurs
primarily at/on the outermost portions of the wire.
From Ohm’s Law: V  I  Rwire where Rwire = resistance of wire = Cwire L Awire  L  Cwire Awire
Joule Heating
of currentcarrying wire

Pwire      I 2 Rwire  
a
2
Pwire    a    I Rwire
Power losses in wire show up / result
in Joule heating of wire. Electrical
energy is converted into heat
(thermal) energy – At the microscopic
level, this is due to kinetic energy
losses associated with the ensemble of
individual drift/conduction/free
electron scatterings inside the wire!
2
Again use the integral version of Poynting’s theorem to determine the EM field energy
flowing through an imaginary Gaussian cylindrical surface S of radius   a and length H  L .
We expect that we should get the same answer as that obtained above, for the   a Gaussian


cylindrical surface. However, for   a , S outside    a   0 , because E outside    a   0 !!!


Thus, for a Gaussian cylindrical surface S taken with   a we obtain: Pwire    S wire da  0 !!!
S
What??? How can we get two different Pwire answers for   a vs.   a ??? This can’t be!!!
 We need to re-assess our assumptions here…
It turns out that we have neglected an important, and somewhat subtle point...

The longitudinal electric field E   V L  zˆ formally/mathematically has a discontinuity at   a :

J free

E 
C
0
0
 a

I  a2
C

V
L


i.e. The tangential ( ẑ ) component of E is discontinuous at   a .
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2015. All Rights Reserved.
9
UIUC Physics 436 EM Fields & Sources II
Fall Semester, 2015
Lect. Notes 1
Prof. Steven Errede
Formally/mathematically, we need to write the longitudinal electric field for this situation as:


J free

J free
E  
1      a   
1      a   zˆ
 
 
C
C
0 for   a
where the Heaviside step function is defined as:     a   
as shown below:
1 for   a
1
   a
0
 a
0
Furthermore, note that:   x      t  dt and that:
x

where   x  is the Dirac delta function.

d
  x    x ,
dx
Now, in the process of deriving Poynting’s theorem (above), we used Griffith’s Product Rule # 6
  
     
to obtain E    B  B   E   E  B , and then used Faraday’s law (in differential form)


 

 B 1   
 E 1   
1  2
1  2
  E   B t and then used B  
B B 
B  and E 

EE 

E 
2 t
2 t
t 2 t
t 2 t
with uEM  12  o E 2  1o B 2 to finally obtain:












 
dW  t  dU mech d

  umech d   E  J free d
v
dt
dt
dt v
 
  
dU EM  t 
d

  S da    uEM d    S  r , t  d
S
v
dt
dt v
 
So here, in this specific problem, what is   E ???
P t  
In cylindrical coordinates, the only non-vanishing term is:






 
J
J
J
a






 
B
free
free
free

Ezˆ 
ˆ 
    a  ˆ  
 E  
1      a    ˆ  





   C 

t
C
C



 
 
B  
In other words:   E  
   
t  



10
0
for   a

J free 
 ˆ for   a
C 

0
for   a
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2015. All Rights Reserved.
UIUC Physics 436 EM Fields & Sources II
Fall Semester, 2015
Lect. Notes 1
Prof. Steven Errede
Thus, {only} for   a integration volumes, we {very definitely} need to {explicitly} include
the  -function such that its contribution to the integral at   a is properly taken into account!
 
dW  t 
d
   uEM d   S da
S
dt
dt v
 
d
   12  o E 2  1o B 2 d   S da
S
dt v
 
d 2
d 2
  12  o 
E d  2 1o 
B d   S da
v dt
v dt
S


 
 dE
 dB
  o  E  d  1o  B d   S da
v
v
S
dt
dt

 
 dE
  
  o  E  d  1o  B  Ed   S da
v
S
v
dt


J free
 
 dE

ˆ
  o  E  d 






B
a
d
S



 S da
v
o C v
dt
P t  




For this specific problem: dE dt  0 and for   a , S    a  
P t  
o C

v


E    a B    a  0 .


0
Thus for   a :

J free
1
o


J free 
J free

B     a  ˆ d  2 aL
B    a   2 a L
o C
o I
o  C 2 a


J free
C
I L

 J free V
V
zˆ , and thus, finally we obtain, for   a : P  t  
But: E 
I  L  V  I ,

L
C
L
which agrees precisely with that obtained earlier for   a : P  t   V  I !!!
For an E&M problem that nominally has a steady-state current I present, it is indeed curious that


  J free
B
is non-zero, and in fact singular {at   a }! The singularity is a
    a  ˆ  
 E 
C
t

consequence of the discontinuity in E on the   a surface of the long, current-carrying wire.
The relativistic nature of the 4-dimensional space-time world that we live in is encrypted into
Faraday’s law; here is one example where we come face-to-face with it!
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2015. All Rights Reserved.
11
UIUC Physics 436 EM Fields & Sources II
Fall Semester, 2015
Lect. Notes 1
Prof. Steven Errede
Let’s pursue the physics of this problem a bit further – and calculate the magnetic vector
 
potential A  r  inside    a  and outside    a  the long wire…
 
 
   
o J  r  
d 
In general, we know/anticipate that {here}: A  r   J  r    zˆ since: A  r  
4 v r
  
where r  r  r  r  .
 
We don’t need to carry out the above integral to obtain A  r   a simpler method is to use
 
 
  

B  r     A  r  in cylindrical coordinates. Since A  r   Az  r  zˆ (only, here), the only non
 
Az  r 
zero contribution to this curl is: B  r   
ˆ .



Az    a 
A    a 
o I 
1
1
ˆ 
o J ˆ  
ˆ 
  o J  zˆ
For   a : B    a  
2
2 a
2


2


A    a 
A    a 
I
1
1
1
1
ˆ 
For   a : B    a   o ˆ  o Ja 2   ˆ   z
  o Ja 2   zˆ
2
2

2



Using   a as our reference point for carrying out the integration {and noting that as in the

case for the scalar potential V  r  , we similarly have the freedom to e.g. add any constant vector
 
to A  r  }:

1
1
1
1
A    a    o J   d  zˆ
  o J   2  c12  zˆ   o J   2  c12  zˆ
2
2
2
4

1
1
1
A    a    o Ja 2    d  zˆ   o Ja 2 ln   c2  zˆ
2
2

where c1 and c2 are constants of the integration(s).

Physically, we demand that A    be continuous at   a , thus we must have:

1
1
A    a    o J  a 2  c12  zˆ   o Ja 2 ln  a c2  zˆ
4
2

Obviously, the only way that this relation can be satisfied is if c1  c2   a , because then A    a   0
{n.b. ln 1  ln e0  0 }.
   
Additionally, we demand that A  r   J  r    zˆ , hence the physically acceptable solution is
 
c1  c2  a , and thus the solutions for the magnetic vector potential A  r  for this problem are:

1
1
A    a    o J   2  a 2  zˆ   o J  a 2   2  zˆ
4
4

1
1
A    a    o Ja 2 ln    a  zˆ   o Ja 2 ln   a  zˆ
2
2
12
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2015. All Rights Reserved.
UIUC Physics 436 EM Fields & Sources II

A  
Fall Semester, 2015
Lect. Notes 1
Prof. Steven Errede

A  0   14 o Ja 2
 a
0


Ain    a 

Aout    a 
varies as 1(/a)2
varies as ln(/a)

1
Note that: A    a   o J ln   a  zˆ has a {logarithmic} divergence as    , whereas:
2


1
1
B         A       o Ja 2   ˆ  0
2

This is merely a consequence associated with the {calculationally-simplifying} choice that we
made at the beginning of this problem, that of an infinitely long wire – which is unphysical.
It takes infinite EM energy to power an infinitely long wire… For a finite length wire carrying a
steady current I, the magnetic vector potential is mathematically well-behaved {but has a
correspondingly more complicated mathematical expression}.

It is easy to show that both of the solutions for the magnetic vector potential A    a  satisfy

  
the Coulomb gauge condition:  A  r   0 , by noting that since A    a   Az    a  zˆ are
 
functions only of  , then in cylindrical coordinates:  A    a   Az    a  z  0 .
Let us now investigate the ramifications of the non-zero curl result associated with Faraday’s

law at   a for the A -field at that radial location:


  J free
B
    a  ˆ  
 E 
C
t
  
A
Since B    A   z ˆ {here, in this problem}, then:






J free
B    A
  Az 
  Az  J free

 
    a  ˆ or:
    a
 ˆ  


t
t
t   
C
t     C




J
J
J
A
free
A
free
free

    a  zˆ .
Then: z 
    a d  
    a  or:

C
t
 C 
t
 C


    a 
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2015. All Rights Reserved.
13
UIUC Physics 436 EM Fields & Sources II
Fall Semester, 2015
Lect. Notes 1
Prof. Steven Errede
Now, recall that the {correct!} electric field for this problem is:

J free

1      a   zˆ
E  
 
C
However, in general, the electric field is defined in terms of the scalar and vector potentials as:
 
 
 
A  r , t 
E  r , t   V  r , t  
t



J free

A J free
zˆ

    a  zˆ , we see that: V 
Since {here, in this problem}:
C
C
t
and hence {in cylindrical coordinates} that: V  z   

J free
C
z , then:



J free 
J free

  J free 
V   
z  zˆ 
zˆ .
 z  zˆ 
 C z
C
z   C 


Note that the {static} scalar field V  z   

J free
C

z pervades all space, as does A    a    zˆ .
Explicitly, due to the behavior of the Heaviside step function     a  we see that the electric

for   a
  0

A J free
A  
 J

    a  zˆ is:
field contribution
.
t  free zˆ for   a
C
t
 C
Explicitly writing out the electric field in this manner, we see that:


 J free
J free
 

zˆ  0
zˆ for   a


A
a









C
C
E    a   V    a  
 

t
J free
 J free
zˆ 
zˆ  0
for   a

C
 C


Thus, for   a we see that the  A    a  t contribution to the E -field outside the wire
 
{which arises from the non-zero   E of Faraday’s law at   a } exactly cancels the


V    a  contribution to the E -field outside the wire, everywhere in space outside the wire,

despite the fact that A    a  varies logarithmically outside the wire!!!!
14
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2015. All Rights Reserved.
UIUC Physics 436 EM Fields & Sources II
Fall Semester, 2015
Lect. Notes 1
Prof. Steven Errede
The long, current-carrying wire can thus also be equivalently viewed as an electric flux tube:
 


 E   E da  J free  C  1      a   zˆ da  I  C
S
S



The electric field E is confined within the tube ( = the long, current carrying wire) by the

 A    a  t contribution arising from the Faraday’s law effect on the   a boundary of the
flux tube, due to the {matter geometry-induced} discontinuity in the electric field at   a !
 


The   E  J free  C     a  ˆ   B t effect at   a also predicts a non-zero “induced” EMF


in a loop/coil of wire:     m t . The magnetic flux through a loop of wire is:

 
 m   Ad    B da  B  Aloop where Aloop is the cross-sectional area of a loop of wire {whose plane
C
S
is perpendicular to the magnetic field at that point}. Note further that the width, w of the coil only needs

to be large enough for the coil to accept the B t contribution from the -function at   a . Then, here
in this problem, since the magnetic field at the surface of the wire is oriented in the ̂ -direction, and:


 loop

loop
J
J
 m
B  A
B
free  A
free
ˆ



    a   , then we see that:   
    a
t
t
t
C
C
For a real wire, e.g. made of copper, how large will this EMF be – is it something e.g. that we
could actually measure/observe in the laboratory with garden-variety/every-day lab equipment???
A number 8 AWG (American Wire Gauge) copper wire has a diameter D = 0.1285” = 0.00162 m
(~ 1/8” = 0.125”) and can easily carry I = 10 Amps of current through it.
The current density in an 8 AWG copper wire carrying a steady current of I = 10 Amps is:
J 8 AWG 
I
4 I
4 10


 4.8 106
2
2
2
 a  D   0.001632 
The electrical conductivity of {pure} copper is:
 Amps
m2 
 CCu  5.96 107  Siemens m  .
If our “long” 1/8” diameter copper wire is L  1 m long, and if we can e.g. make a loop of ultrafine gauge wire that penetrates the surface of the wire and runs parallel to the surface, then if we
approximate the radial delta function     a  at   a as ~ a narrow Gaussian of width
w ~ 10 Å  1 nm  109 m (i.e. ~ the order of the inter-atomic distance/spacing of atoms in the copper
lattice { 3.61 Å }), noting also that the delta function     a  has physical SI units of inverse length
(i.e. m-1) and, neglecting the sign of the EMF, an estimate of the magnitude of the “induced” EMF is:
 Cu 
J 8 AWG  Aloop
 CCu
    a 
J 8 AWG  L  w
 CCu
J
 w   8 AWG
Cu
 C
 4.8 106  Amps m 2  

 1 m  80 mV !!!
  L  
6  107  Siemens m  



This size of an EMF is easily measureable with a modern DVM…
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2015. All Rights Reserved.
15
UIUC Physics 436 EM Fields & Sources II
Fall Semester, 2015
Lect. Notes 1
Prof. Steven Errede
Using Ohm’s Law: V  I  R , note that the voltage drop Vdrop across a L  1 m length of
8 AWG copper wire with I  10 Amps of current flowing thru it is:
1m
drop
V
 I  R1m  I 
CCu L
wire

A


 J 8 AWG  Awire 
L
Cu
C
wire

A

J 8 AWG
 CCu
 L   Cu !!!

In other words, the “induced” EMF,   J free  Aloop  C     a  in the one-turn loop coil of

length L {oriented as described above} is precisely equal to the voltage drop Vdrop  J free  C  L




along a length L of a portion of the long wire with steady current I flowing through it, even though
the 1-turn loop coil is completely electrically isolated from the current-carrying wire!!!
 
This can be easily understood... Using Stoke’s theorem, the surface integral of   E can be

converted to a line integral of E along a closed contour C bounding the surface of integration S ;




likewise, a surface integral of B t    A t can be converted to a line integral of A t along
a closed contour C bounding the surface of integration S:



  
 
  A  
 m
B 
A 
da       da   
d 
     E da   E d   
 
S
C
S t
S
C t
t
t 



n.b.:  V d   0


C
Then for any closed contour C associated with the surface S that encloses the Faraday law
 
  E -function singularity at   a , e.g. as shown in the figure below:
the “induced” EMF  can thus also be calculated from the line integral

C
 
E d  taken around
the closed contour C. From the above discussion(s), the electric field inside (outside) the long




current-carrying wire is Ein  J  C Eout  0 , respectively {n.b.  tangential E is


discontinuous across the boundary of a {volume} current-carrying conductor!}. Then:
16
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2015. All Rights Reserved.
UIUC Physics 436 EM Fields & Sources II


2



3
Fall Semester, 2015


4
Lect. Notes 1


1
Prof. Steven Errede


3 4
   E d    Ein12 d 12   E 23 d 23   Eoutside
d 34   E 41 d 41  E    V12
C
1
 J   C  E  V12
2



3
0



0
4

0
 


The presence of a non-zero Faraday’s law   E   B t  J free  C     a  ˆ term at the surface

of the long current-carrying wire implies that the “induced” EMF   J free  Aloop  C     a  can also




be viewed as arising from the mutual inductance M  Henrys  associated with the long wire and the coil
{oriented as described above}, and a non-zero I t :


loop
 m
B  Aloop
I J free  A
 
    a

 M

C
t
t
t

We can obtain a relation between B t and I t using the integral form of Ampere’s law:
 
sides of this equation with respect to time:
C Bd   o I encl . Taking the partial derivative of both

 
I

B 


d   o encl
B
d





C
C
t
t
t


The contour of integration C needs to be taken just outside the surface of the long wire, along


the ̂ -direction, since B  ˆ at   a , i.e. d   ˆ in order to include the non-zero Faraday’s law
effect at the surface of the long wire.


J
J




B  o  I
I
2 a B
2 a
free
free

 
    a
    a  or:


Then:



C
t  2 a  t
t  o  t
 o   C


loop
 m
B  Aloop
I J free  A
    a

 M

Then:   
C
t
t
t
 Aloop 
Solving for the mutual inductance, we obtain a rather simple result: M  o     Henrys 
 2 a 
Note that the mutual inductance, M involves the magnetic permeability of free space
o  4 107  Henrys m  {n.b. which has SI units of inductance/length} and geometrical
aspects {only!} of the wire (its radius, a) and the cross-sectional area of the loop, Aloop .

What is astonishing {and unique} r.e. the “induced” Faraday’s law EMF   J free  Aloop  C     a 


associated with a long, steady current-carrying wire is that “normal” induced EMF’s only occur in electrical
circuits that operate at non-zero frequencies, i.e. f  0 Hz . However, here, in this problem, we have an
example of a DC induced EMF – i.e. an induced EMF that occurs at f  0 Hz , arising from the non-zero
 



Faraday’s law effect   E   B t  J free  C     a  ˆ due to the longitudinal E -field discontinuity


at the surface    a  of a long, steady current-carrying wire!!!
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2015. All Rights Reserved.
17
UIUC Physics 436 EM Fields & Sources II
Fall Semester, 2015
Lect. Notes 1
Prof. Steven Errede
Instead of using a long wire to carry a steady current I to observe this effect, one might instead
consider using e.g. a long, hollow steady current-carrying pipe of inner (outer) radius a, (b) respectively.
Following the above methodology, one can easily show that for such a long, hollow current-carrying
 
pipe, two opposing non-zero Faraday law   E radial -function contributions occur – one located at
the   a inner surface, and the other located at the   b outer surface of the long hollow currentcarrying pipe:
 


  E   B t   J free  C     a       b   ˆ

The E -field is:




E  V  A t  J free  C 1      a       b   zˆ




1 for   a
where:     a   
is the complement of the Heaviside step function, such that:
0 for   a
d   x  dx    x  and:   x       t  dt where:   x  is the Dirac delta-function.
x

Hence, a 1-turn coil {oriented as described above} enclosing the   a inner surface .and.
the   b outer surface of a current-carrying hollow pipe will have a “null” induced EMF, i.e.
  0 due to the wire loop simultaneously enclosing the two opposing non-zero Faraday law
 
  E radial -function contributions, one located at   a , the other at   b :


loop
 m
B  Aloop J free  A
 


    a       b    0
C
t
t
In general, any penetration/hole made into the metal conductor of a long, steady current 
carrying wire will result in a non-zero Faraday law   E -function on the boundary/surface of


that penetration/hole! Since the current density J free  0 in the penetration/hole, E  0 there and

thus a discontinuity in E exists on the boundary of the penetration/hole, hence a non-zero
 
Faraday law   E -function exists on the boundary of the penetration/hole!
This fact {unfortunately} has important ramifications for the experimental detection /
observation of the predicted non-zero DC induced EMF in a coil {oriented as described above},
Embedding a portion of a physical wire loop inside the long, steady current-carrying wire
requires making a penetration/hole {no matter how small} into the wire, which will result in a
 
non-zero Faraday law   E -function on the boundary/surface of that penetration/hole in the
 
wire! Thus, the wire loop will in fact enclose not only the Faraday law   E radial singularity at
  a on the surface of the wire, but will also enclose another, opposing singularity located on
the boundary/surface of the penetration/hole made into the long wire {which was made to embed
a portion of the wire loop in a long, steady current-carrying wire in the first place}, thus
experimentally, a “null” induced EMF, i.e.   0 is expected/anticipated, because of this…
Hence, in the real world of experimental physics, it appears that embedding a portion of a real wire
loop in a long, steady current-carrying wire in an attempt to observe this effect is doomed to failure…
18
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2015. All Rights Reserved.