Transcript Slide 1

Magnetic potential and field of a SOLENOID
• Infinite length
•N spires/cm
•Current I
•Radius R
J  ( J sin , J cos ,0)  A  ( A , A ,0)
x
y
A
The problem -for Ax –is mathematically
similar to finding the electrostatic
potential Φ with a surface charge ρ0sinθ.
J
This type of surface charge is what is
given by two parallel cylinders of uniform
and equal, but of opposite sign, charge
densities, with their axes displaced by an
infinitesimal distance d: the calculation is
shown below. Let V be the potential of a
cylinder of radius R and charge density
uniform. Two such cylinders of the same
radius and charge density opposite in sign
have a superficial charge density :
  0 dr  0 d sin 
Their potential has a
V
d behaviour
y
V
r 2
y

d 
 2r
y
y
r
(ln(1 / r )
y


y
r2
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 2 y for r  R
for r  R
1
Owing to the azimutal symmetry of the system, we can calculate the
Φ for x=0, where y=r . We get a r dependence of Φ ~r inside the
wire and ~(1/r) outside.
Then, A being always along the direction of the θ axis, it turns
around the solenoid axis like the current does. It varies with r like r
for r<R and like 1/r for r>R. Once we put all the coefficients right,
we obtain:
2NIr 
A
c 
2NIR2 
A
cr 
for r  R
for r  R
Having A written as a function of r we can calculate its curl, i.e. B.
A
A 1 Ar
1 Az A
A A

)r  ( r  z )θ  (    
)k
r 
z
z
r
r
r r 
B    A  (0,0, Bz )
B   A  (
Bz  {
Bz  0
4NI
c
for r  R
for r  R
Having Bz finite inside the solenoid and at its radius suddenly
zero is not realistic. But… also an infinite solenoid is not quite
realistic. And anyway the coil has a finite thickness.
Note at this point an important physical fact: outside
the solenoid the vector potential is not zero nor constant, BUT:
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BUT:
const 
A
r 
While not being trivial still has curl equal zero. On the other hand it
is quickly verified that this A has zero divergence. We are then led
by this paradoxical result to ask ourselves where the Physics is,
in the field B or in the potential A. The answer has been given by
the Bohm – Aharonov experiment, 1956. Which has already been
described.
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EM fields varying in time
 Ε  4 
 Ε  
 Β  0
1 Β
c t
4
1
 Β 
J {
c
c
Ε
}
t
This is the term added by Maxwell to
preserve charge conservation. It allows
the equations to become “Wave
equations”.
With this new term the electrical and magnetic fields are now coupled
through their dependence on time. Without it, a variation of B with
time would have caused a variation of E. End of the story. Now, a
variation of E changes B as well: the process is self-sustaining.
Now a question comes out immediately: beforehand (in electrostatics)
we knew that we could express E as the gradient of a potential and B
as the curl of a “vector potential”. How does that change now?
B  0
B   A
This still holds, then
this is still valid
We then insert it in the second Maxwell equation and obtain
E  
1  (  A)
1 A
   (
)
c
t
c t
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1 A
  E    (
)
c t
This equation is satisfied, by the formula
on the right: the 3rd and 2nd EofM are
satisfied since the condition on the curl of
E is satisfied, and this determines E
E
1 A
 
c t
up to the gradient of any function! We call that function Φ; it will be
determined by the 1st EofM which, upon substitution of this value for E
becomes
1 
(  A)  4
c t
while, substituing E and B in the4th EofM
 2 
Vector
formula
4
1 2A 1 
(A)  J  2 2 
( )
c
c t
c t
4
1 2A 1 
2
(A) A  J  2 2 
( )
c
c t
c t
and, reordering,
 A
2
1 2A
c 2 t 2
( A 
1 
4
) J
c t
c
We still have to choose the gauge. For the time-dependent case
we choose the Lorentz gauge, defined by:
A  
1 
c t
The Lorentz gauge.
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Inserting now the Lorentz gauge condition in the last 2 equations in
Φ and in A they become much simpler:
1
  2
c
1
2A  2
c
2
 2
  4
2
t
2A
4


J
2
t
c
Which is (are, actually) the classical wave equations. With wave
velocity = c !!!
These equations are of course NOT independent: beside the
Lorentz gauge which has been used to obtain them, and
constrains Φ and A, J and
conservation. In formulas:
A  
1 
c t

J  0
t
ρ
are also related by the charge
T heLorentzgauge
Charge conservation
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Faraday’s law
The second EofM, also called Faraday’s induction law, usually
given in its differential form:
E  
1 B
c t
can also be written in integral form (using the Stokes theorem in
the passage from line to surface integral) if we define the
“electromotive force” E
E   E  dl   (  E)  n dS  
S

1 B
 n dS

c S t
1 
( B  n dS)   1   S (B)
c t S
c t
Now, let the closed line over which we calculate the line integral
be a real electric wire. The electromotive force E is generated
by the variation (in time) of the flux of B through any surface
enclosed by the closed line. Such variation can be generated in
three different ways:
•Varying the current through the magnets.
•Moving the circuit.
•Moving the magnets.
Matter of fact, what is actually measurable is the force on a
charge in the wire, so the equation becomes:
E

1
v
F

d
l

(
E

 B)  dl


q
c
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In the laboratory reference system, if the circuit is moving with
velocity v, a free charge inside it will feel the force
F  q(E  v  B)
Let us now study these 3 different cases and see what we get from them.
1.
What changes in time is the current running in the magnets. Then:
E   E  dl   (  E)  n dS 
S
  (
S
1 B
)  n dS   1   (B n)   1  S
c t
c t S
c t
2. The circuit moves with velocity
not change.
B
v0
E
t
0
Case verified
v while the magnetic field does
Then a free charge (electron) will
feel the Lorentz force:
v
  (  B)  dl    ( v  dl )  B
c
( V х dl ) is a vector which is
V
Then:
E
normal to the surface element dS (yellow in
the drawing) covered by the circuit in the
time dt.
= -(1/c) · [time variation of the
flux of B due to the change of surface
covered by the circuit (in blue and in
yellow in the drawing)].
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3. The magnets are moved: the currents which generate B do not
change, the circuit sits still, but the magnets are moved. This case is a
sort of bridge between the two previous ones: from the mathematical
point of view in the formulas it is B that has changed; but it is also
identical to the second case, because if what has moved is the circuit or
the magnets the result is the same.
BUT…
But, the choice of the reference system, causes the choice in the
mathematics of either of the term
E
1 A
v
or of the Lorentz force  B
c t
c
in the calculation of E. The result is, however, the same!
Another remark on the case a): currents varying with time. Let us
imagine a transformer.
C
A
An electromotive force is
generated on the circuit C when
the current is changed on A. But,
where C is located there is hardly
a magnetic field. There is,
however a significant vector
1
potential.
A~-
r
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The conservation of energy
We have already seen the mathematical expression for the
charge conservation

t
J  0
Since we know that energy is also conserved, we would like to
find another similar equation for the energy conservation – of
the electromagnetic field, of course. We are therefore looking
for two physical quantities, let us call them U and S, which can
be related to energy density (of the EM field) and to energy
flow, which satisfy the following equation.
U
   S  {work done on charges}
t
The term on the right-hand side of this equation has the
form:
{work done on charges} E J
What we shall do now is to try and find a formula – in which only the
fields appear - which looks like the first term in the equation above which
states the conservation, i.e. an equation which has the form:
“(the derivative of a scalar plus the divergence of a vector) = -E·J”.
We shall do that starting from E·J, by substituting J with its
expression taken from the 4th EofM:
J
1
E
(c  B  )
4
t
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Here goes the calculation: we shall use the vector formula
  (R  K)  K  (  R)  R  (  K) Vector Formula
The calculation goes as follows:
1
[cE  (  B)  E  E ] 
4
t
1
1 E 2

[c  (B  E)  cB  (  E) 
]
4
2 t
   ( c E  B)  1  ( E 2  B 2 )
4
8 t
EJ 
This is Poynting’s theorem. Note that it equals E·J to the sum of
a time derivative of a scalar + the divergence of a vector,
precisely what we were looking for. It is therefore immediate to
interpret this formula as follows:
1
( E 2  B 2 ) Energydensit y
8
c
S
(E  B) P oynt ingvect or: energy flow
4
U
And the equation for the conservation of energy is:
U
 S  EJ  0
t
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So, we have now, in regime of full time dependence of the fields
and of the charges and currents, found a formula for the energy
distribution of the electromagnetic field. But… hadn’t we already
met a formula for this energy density? Well, we had found two.
We write for simplicity only the electric density. The general
formula adds the electric and the magnetic energies
1
U
8
U
2
E
 dV
and
space
1
 (r )(r )dV

2V
Only one of them is valid in general: the other one (that with
the charge density ρ ) is only valid in the electrostatic case.
Examples of Poynting’s vector
.
E
x
.
.
The energy enters the wire
from outside!!!
k
B
S
j
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The charging of a capacitor
What we want to explain in this slide is how are the fields and the
potentials in the case of a parallel plate capacitor being charged with a
constant current. NB: between the capacitor plates, J=0 !!
E and dE/dt are directed along the vertical axis. To find
out how is B directed in the gap, let’s write
1 E
 B  dl   (  B)  n dS  c  t  ndS 
S
J
S
1
dE
 2rB  r 2
c
dt

B
r dE
2c dt
A is the integral of J, therefore
both Ax and Ay are zero. Az only
depends on r, and so the curl of A
is directed along the θ axis.
Then S, the Poynting’s vector, is
R
h
directed along the axis orthogonal
S
E
B
to E and B, therefore is directed
towards the axis of the capacitor,
and has value
S 
c
r
dE
BE 
E
4
8 dt
The electrostatic energy has uniform density between the plates,
U=(1/8π)E2, the total energy is R2πhU and its rate of change is
(Rh/4)E(dE/dt) . The S vector is directed radially from outside towards
the capacitor gap: ALL THE ENERGY STORED IN THE CAPACITOR
HAS ENTERED LATERALLY FROM THE OUTSIDE AIR!
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