10 Lecture Perfect Metals in Magnetism and Inductance
In this lecture you will learn: Some more about the vector potential Magnetic field boundary conditions The behavior of perfect metals towards time-varying magnetic fields Image currents and magnetic diffusion Inductance
ECE 303 Fall 2007 Farhan Rana Cornell University
The Vector Potential - Review
In electroquasistatics we had:
r E = 0
Therefore we could represent the E-field by the scalar potential: In magnetoquasistatics we have:
r E = -
r r . B = . o H = 0 r r r B = o H = A
( )
(
)
Therefore we can represent the B-field by the vector potential:
A vector field can be specified (up to a constant) by specifying its curl and its divergence Our definition of the vector potential A is not yet unique we have only specified its curl For simplicity we fix the divergence of the vector potential A to be zero:
r
r
r .A = 0
ECE 303 Fall 2007 Farhan Rana Cornell University
1
Magnetic Flux and Vector Potential Line Integral - Review
The magnetic flux through a surface is the surface integral of the B-field through the surface r r B-field
= B . da r r = o H . da
Since:
r r r B = o H = A
We get:
= B . da
r
r
r r = A . da r r = A . ds
(
)
r ds
Stoke's Theorem
A closed contour
The magnetic flux through a surface is equal to the line-integral of the vector potential along a closed contour bounding that surface
ECE 303 Fall 2007 Farhan Rana Cornell University
Vector Potential of a Line-Current
Consider an infinitely long line-current with current I in the +z-direction The H-field has only a -component Using Ampere's Law:
y
^ Iz
r r 2 A = - o J
H (2 r ) = I I H = 2 r
r
x
Work in cylindrical co-ordinates
If the current has only a z-component then the vector potential also only has a z-component which, by symmetry, is only a function of the distance from the line-current But
H =
r A
o
=-
1 Az (r ) o r
Az (r ) I =- o r 2 r
Integrating from ro to r :
Az (r ) - Az (ro ) =
o I ro ln 2 r
ECE 303 Fall 2007 Farhan Rana Cornell University
2
Vector Potential of a Line-Current Dipole r
Consider two infinitely long equal and opposite line-currents, as shown
r
y r- d x
^ -Iz
^ +Iz
The vector potential can be written as a sum using superposition:
r+
r I r I r Az (r ) = o ln o - o ln o 2 r+ 2 r- =
o I r- ln 2 r+
The final answer does not depend on the parameter ro
Question: where is the zero of the vector potential? Points for which r+ equals r- have zero potential. These points constitute the entire y-z plane
ECE 303 Fall 2007 Farhan Rana Cornell University
H-Field of a Line-Current Dipole
y
H
x d
Something to Ponder Upon Poisson equation: Vector Poisson equation (only the zcomponent relevant for this problem)
=- o
2
2 Az = - o J z
Vector potential of a line-current dipole:
Potential of a line-charge dipole:
(r ) =
r
2 o
r ln - r +
r I r Az (r ) = o ln - 2 r+
Notice the Similarities
ECE 303 Fall 2007 Farhan Rana Cornell University
3
Magnetic Field Boundary Conditions - I
The normal component of the B-field at an interface is always continuous Maxwell equation:
oH1 oH2
r r . B = . o H = 0
The net magnetic flux coming into a closed surface must equal the magnetic flux coming out of that closed surface (since there are no magnetic charges to generate or terminate magnetic field lines)
Therefore:
oH2 = oH1
ECE 303 Fall 2007 Farhan Rana Cornell University
Magnetic Field Boundary Conditions - II
The discontinuity of the parallel component of the H-field at an interface is related to the surface current density (units: Amps/m) at the interface
K
H1 H2 - H1 = K
H2
This follows from Ampere's law:
r r H = J
or
r r r r H . ds = J . da
K
The line integral of the magnetic field over a closed contour must equal the total current flowing through the contour
H2 L - H1 L = K L H2 - H1 = K
H1
H2
L
ECE 303 Fall 2007 Farhan Rana Cornell University
4
Perfect Metals and Magnetic Fields - I
A perfect metal can have no time varying H-fields inside it
Note: Recall that in magnetoquasistatics one can have time varying H-fields (as long as the time variation is slow enough to satisfy the quasistatic conditions)
The argument goes in two steps as follows: A time varying H-field implies an E-field (from the third equation of magnetoquasistatics)
r r . o H (r , t ) = 0
r r r H (r , t ) = J (r , t )
r r r r o H (r , t ) E (r , t ) = - t
Since a perfect metal cannot have any E-fields inside it (time varying or otherwise), a perfect metal cannot have any time varying H-fields inside it
ECE 303 Fall 2007 Farhan Rana Cornell University
Perfect Metals and Magnetic Fields - II
At the surface of a perfect metal there can be no component of a time varying H-field that is normal to the surface
r H (t )
perfect metal
The argument goes as follows: The normal component of the H-field is continuous across an interface So if there is a normal component of a time varying H-field at the surface of a perfect metal there has to be a time varying H-field inside the perfect metal Since there cannot be any time varying H-fields inside a perfect metal, there cannot be any normal component of a time varying H-field at the surface of a perfect metal
ECE 303 Fall 2007 Farhan Rana Cornell University
5
Perfect Metals and Magnetic Fields - III
Time varying currents can only flow at the surface of a perfect metal but not inside it
r K (t )
perfect metal
The argument goes as follows: Time varying currents produce time varying H-fields So if there are time varying currents inside a perfect metal, there will be time varying H-fields inside a perfect metal Since there cannot be any time varying H-fields inside a perfect metal, there cannot be any time varying currents inside a perfect metal
ECE 303 Fall 2007 Farhan Rana Cornell University
Current Flow and Surface Current Density
So how does a (time varying) current flow in perfect metal wires? there Remember cannot be any time varying currents inside a perfect metal......
y
Consider an infinitely long metal wire of radius a carrying a (time varying) current I in the +z-direction The current flows entirely on the surface of the perfect metal wire in the form of a uniform surface current density K( t )
K
a
x
K (t ) =
I (t ) 2 a
Can use Ampere's law to calculate the H-field outside a perfect metal wire carrying a (time varying) current I
y r
H
K
(2 r ) H (t ) = (2 a ) K (t ) a I (t ) H (t ) = K (t ) =
r
a
x
2 r
ECE 303 Fall 2007 Farhan Rana Cornell University
6
Parallel Plate Conductors
Consider two infinitely long (in the z-direction) metal plates, of width W and separated by a distance d, as shown The top plate carries a (time varying) current I in the +z-direction and the bottom plate carries a (time varying) current I in the zdirection
y W
^ Kz
d
Hx
x ^ -Kz
Surface current density on the top plate = K =
I W
If W >> d, then the H-field inside the plates in very uniform and can be calculated by using the boundary condition:
H x outside metal - H x
0
inside metal
=K
H x outside metal = K =
I W
One can also use Amperes law directly - see if you can identify an appropriate contour for using Ampere's law to get the same answer
ECE 303 Fall 2007 Farhan Rana Cornell University
Image Currents - I
Consider a perfect metal with a wire carrying a time varying current I(t ) in the +zdirection at a distance d above the perfect metal, as shown below H( t )
d perfect metal
Surely this picture cannot be right............there is time varying H-field inside the perfect metal
ECE 303 Fall 2007 Farhan Rana Cornell University
7
Image Currents - II
So what does really happen ...... H
d perfect metal
Currents flow on the surface of the perfect metal that completely cancel the time varying H-field inside the perfect metal In other words, surface currents screen out the time varying H-field from the perfect metal
Question: Is there a better way to understand what the resulting H-field looks like outside the perfect metal?
ECE 303 Fall 2007 Farhan Rana Cornell University
Image Currents - III
The magnetic field outside the perfect metal can be obtained by imagining a fictitious current element that is a mirror image of the actual current element but carrying a current in the opposite direction
H
d perfect metal d
image current
ECE 303 Fall 2007 Farhan Rana Cornell University
8
Image Currents - IV
Example: A current loop carrying a time varying current over a perfect metal
I(t ) perfect metal d d
image current
ECE 303 Fall 2007 Farhan Rana Cornell University
Not So Perfect Metals - I
So what does really happen when a current is suddenly switched on at time t=0 Time = t = 0 d
H
H Time = t = If you wait "long enough" magnetic field will penetrate real metals! d
ECE 303 Fall 2007 Farhan Rana Cornell University
9
Not So Perfect Metals Magnetic Diffusion
Question: How long does it take for the magnetic field to penetrate into real metals? Answer: Start from these magnetoquasistatic equations:
r r H = J r r oH E = - t
r r J =E
r r r r H H = J = E = - o t r r r H 2 . H - H = - o t r r H 2 H = o Magnetic diffusion equation t
(
)
In time "t " the magnetic field will diffuse a distance "d " into the metal, where:
t
o d 2
2
ECE 303 Fall 2007 Farhan Rana Cornell University
Inductance
The magnetic flux enclosed by current carrying conductors is directly proportional to the current carried by the conductors H
r r r r Magnetic Flux = = B . da = o H . da
I
The constant of proportionality is called the Inductance (units: Henry)
I
=LI
or
L=
I L= d dI
ECE 303 Fall 2007 Farhan Rana Cornell University
More accurately:
10
Inductance of the Parallel Plate Conductors
y
From previous slide:
W
^ Kz
Hx = K =
I W d Hx x K= I W ^ -Kz
Since the structure is infinite in the z-direction, one can only talk about the inductance per unit length (units: Henry/m) (unit length means unit length in the z-direction)
L = Inductance per unit length = L=
Flux per unit length Total current
o H x d
I
=
o d
W
ECE 303 Fall 2007 Farhan Rana Cornell University
Inductance of the Concentric Cylinders
Consider two perfect metal concentric cylinders infinitely long in the z-direction. The inner one carries a total (time varying) current I in the +z-direction. The outer shell carries a total (time varying) current I in the -z-direction The magnetic field, by symmetry, has only a -component Using Ampere's law: (2 r ) H (r ) = I
b a
r
H (r ) =
I 2 r
Since the structure is infinite in the z-direction, one can only talk about the inductance per unit length (units: Henry/m)
L=
I
=
o H (r ) dr
a
b
I
b = o ln 2 a
ECE 303 Fall 2007 Farhan Rana Cornell University
11
Inductance of Two Metal Wires - I
Consider two infinitely long metal wires carrying equal (time varying) currents but in opposite directions Need to solve: y d >> a d x
^ +Iz
metal wires of radius a
^ -Iz
As long as d >> a, the surface current density on each metal wire is circularly symmetric and uniform, and so the outside field appears as if one had line-currents at the center of each metal wire So the vector potential can be written as: Az (r ) =
r
o I r- ln 2 r+
Now remember the golden formula from a previous lecture for the magnetic flux:
= B . da = A . ds
ECE 303 Fall 2007 Farhan Rana Cornell University
r
r
r
r
Inductance of Two Metal Wires - II
y The line integral of the A-field along the indicated contour gives the net flux passing through the contour d >> a d
= B . da = A . ds
=
r
r
r
^ +Iz
r
x H metal wires of radius a
^ -Iz
o I r- I r ln - o ln - 2 r+ r- = d 2 r+ r- = a
r+ = a r+ = d
I d = o ln a
L=
^ -Iz
d
Looking from the top........ x contour
I
=
o d ln a
z
unit length
ECE 303 Fall 2007 Farhan Rana Cornell University
^ +Iz
12