Magnetic Field of a Steady Current

Magnetic field of a steady
current
Section 30
Previously supposed zero net current.
Then
For a conductor, there can be non-zero net current. Now we suppose there is such.
Then
“Conduction” current density
Magnetization gives no
contribution to net
current, even though
there is surface current
Rate of energy dissipation per unit volume = 
E 
. 
j
1. Now we allow there to be a conduction current density
Distribution of 
j
 is determined from section 21 by
If we ignore the effect of the 
j
-generated 
H
-field on the conductivity of the body, then 
These equations do not involve the magnetic field created by 
j 
itself.
Then, 
H
 is determined from the 
j 
distribution according to equations of section 21.
Recipe:  Find 
j
.  Then find 
H
.  Ignore the effect that this 
H
 has on 
j
. 
2.  We next need the differential equations for 
B
 and 
H
 in the presence of conduction
current 
j
.  Boundary conditions are unchanged: 
H
t
, and B
n
 are continuous.
H
 determined from given 
j
.
Same boundary
condition as when 
j
 = 0
New differential equations in the presence of
conduction current
Boundary condition 
H
1t
 = 
H
2t
 still holds, as when 
j
 = 0
Proof:
Homogeneous interface between conductors.
For homogeneous interface,            Is finite
 
J
 is finite
everywhere,
including
boundaries
finite
H
y
 is continuous.  (
H
t
 is continuous.)
For the given 
j
, we want to find 
H
 and 
B
 
These are the equations we
need to solve.
Introduce the vector potential 
A
 and find the equation for it.
Due to gauge invariance, we can place a condition on 
A, 
namely
Assume isotropic linear medium (
B
 = 
 
H
), or else there are too many unknowns
A differential equation for 
A
 with 
j
 as the source.
Recipe:  Solve for 
A
, then find 
B
, then 
H
.
3.  The vector potential 
A
 and differential equations for it.
Special case:  homogeneous medium where 
 is spatially uniform
Differential equation for 
A
 with given 
j
 source term
Recipe: Find 
A
, then find 
B
, then 
H
.
4.  Special case: Homogeneous media.
Special case:  Piecewise homogeneous medium
Different parts have different 
holds in each part.
Boundary condition between the parts
is continuous.
is continuous.
B
oundary conditions for 
A
 at interface between different homogeneous media.
For example, systems with translational invariance along one axis, like a straight wire.
z
y
By assumption
One possibility for current:
Then we expect: 
holds automatically
Condition
Magnetic field lies in
the xy plane, e.g.
cylindrical wire.
(but not assumed
homogeneous in the xy plane)
5. Special case: Two-dimensional problems, and boundary conditions for 
A
 between
different 2D homogeneous media.
What equation does 
A
 satisfy in this 2D problem?
Homework
But
Scalar differential equation for 
scalar
 A(x,y) when 
 = 
(x,y) and 
j
 = 
j
(x,y)
e
z
What if 2D problem is piecewise homogeneous?
holds in each region with own 
Boundary conditions are continuity of A,   
 
    
 
, and (curl
A
)
t
/

Homework
6. Special case of 2D problem with current along axis:  cylindrical symmetry.
If current is symmetrical about the z-axis, j
z
 = j(r)
Then magnetic field lines are circles (r = constant)
(the field must have the same symmetry as its sources.
r = const
Any surface bound by contour r = const
Stokes 
Total current enclosed within radius r
Solve for 
A
, then
 
7.  Another kind of 2D problem is the axially symmetric current distribution
We expect
Magnetic induction (field)
H
-field
8.  Non magnetic media:  
 = 1
Now there are no conditions at interfaces or surfaces since all media are
magnetically identical
(v.2 section 43)
Since 
 = 1, 
H
 = 
B
Acts on coordinates 
r
 of field point P
HW
Thin wires, neglect internal regions of wire:  “linear currents”
Total current in the conductor
Biot-Savart law.
9.  Linear circuits, and the Biot-Savart law
If thin conducting material is surrounded by magnetic material of permeability 
e.g. wire wound iron core
10.  Linear circuit embedded in a magnetic medium
If 
j
 = 0 except for linear currents
Except for line singularities
Then
Magnetic scalar potential
Contrast with electrostatics
 is single
valued.
In all space
11.  Magnetic potential theory
In magnetic case
at a line current                 
If the line current penetrates the
surface of integration.
Magnetic scalar potential 
 is many valued.

 = -
B
 + 
A
 = (4
/c) J
enc                              
If we are at large distances from a finite
current distribution…
HW
Magnetic moment
What is magnetic moment of a flat circular wire loop?
12.  Magnetic moment
Linear current
Total current in the wire               
Closed loop of current           
= Area of triangular surface bounded by 
r
 and 
dl
Enclosed area if
current lies in a plane
is independent of the particular surface bounded by
the given current loop.
13.  Magnetic moment of linear circuit
Electromagnetic field energy is dissipated as Joule Heat
Poynting vector for
field in vacuum
HW:
14.  Energy flux in a conductor
Rate at which energy density leaves
the EM field by energy continuity
Rate at which Joule heat is
generated per unit volume
(21.6)
This energy flux must get
burned up as Joule heat
because it doesn’t flow out
the other side
Direction is from vacuum
into the conductor
Inside of a current-carrying conductor
E
t
 and H
t
 are
continuous at
the surface
Slide Note
Embed
Share

Exploring the behavior of magnetic fields in the presence of conduction currents, discussing differential equations for B and H, considering boundary conditions and vector potential A. The content emphasizes solving equations to find H and B, with special cases for homogeneous and piecewise homogeneous media.

  • Magnetic Field
  • Steady Current
  • Conduction Current
  • Homogeneous Medium
  • Vector Potential

Uploaded on Oct 03, 2024 | 0 Views


Download Presentation

Please find below an Image/Link to download the presentation.

The content on the website is provided AS IS for your information and personal use only. It may not be sold, licensed, or shared on other websites without obtaining consent from the author. Download presentation by click this link. If you encounter any issues during the download, it is possible that the publisher has removed the file from their server.

E N D

Presentation Transcript


  1. Magnetic field of a steady current Section 30

  2. 1. Now we allow there to be a conduction current density Previously supposed zero net current. Then For a conductor, there can be non-zero net current. Now we suppose there is such. Then Magnetization gives no contribution to net current, even though there is surface current Conduction current density Rate of energy dissipation per unit volume = E . j

  3. 2. We next need the differential equations for B and H in the presence of conduction current j. Boundary conditions are unchanged: Ht, and Bn are continuous. Distribution of j is determined from section 21 by These equations do not involve the magnetic field created by j itself. If we ignore the effect of the j-generated H-field on the conductivity of the body, then Then, H is determined from the j distribution according to equations of section 21. Recipe: Find j. Then find H. Ignore the effect that this H has on j.

  4. New differential equations in the presence of conduction current H determined from given j. Same boundary condition as when j = 0

  5. Boundary condition H1t = H2t still holds, as when j = 0 Proof: Homogeneous interface between conductors. For homogeneous interface, Is finite J is finite everywhere, including boundaries finite Hy is continuous. (Ht is continuous.)

  6. 3. The vector potential A and differential equations for it. For the given j, we want to find H and B These are the equations we need to solve. Introduce the vector potential A and find the equation for it. Due to gauge invariance, we can place a condition on A, namely Assume isotropic linear medium (B = H), or else there are too many unknowns A differential equation for A with j as the source. Recipe: Solve for A, then find B, then H.

  7. 4. Special case: Homogeneous media. Special case: homogeneous medium where is spatially uniform Differential equation for A with given j source term Recipe: Find A, then find B, then H.

  8. Special case: Piecewise homogeneous medium Different parts have different holds in each part. Boundary conditions for A at interface between different homogeneous media. Boundary condition between the parts is continuous. is continuous.

  9. 5. Special case: Two-dimensional problems, and boundary conditions for A between different 2D homogeneous media. For example, systems with translational invariance along one axis, like a straight wire. (but not assumed homogeneous in the xy plane) By assumption y One possibility for current: z Then we expect: Condition holds automatically Magnetic field lies in the xy plane, e.g. cylindrical wire.

  10. What equation does A satisfy in this 2D problem? Homework But Scalar differential equation for scalar A(x,y) when = (x,y) and j = j(x,y)ez

  11. What if 2D problem is piecewise homogeneous? holds in each region with own , and (curlA)t/ Boundary conditions are continuity of A, Homework

  12. 6. Special case of 2D problem with current along axis: cylindrical symmetry. If current is symmetrical about the z-axis, jz = j(r) Then magnetic field lines are circles (r = constant) (the field must have the same symmetry as its sources. Stokes Any surface bound by contour r = const r = const Total current enclosed within radius r

  13. 7. Another kind of 2D problem is the axially symmetric current distribution We expect Magnetic induction (field) H-field Solve for A, then

  14. 8. Non magnetic media: = 1 Now there are no conditions at interfaces or surfaces since all media are magnetically identical (v.2 section 43) Since = 1, H = B Acts on coordinates r of field point P HW

  15. 9. Linear circuits, and the Biot-Savart law Thin wires, neglect internal regions of wire: linear currents Total current in the conductor Biot-Savart law. R3

  16. 10. Linear circuit embedded in a magnetic medium If thin conducting material is surrounded by magnetic material of permeability e.g. wire wound iron core

  17. 11. Magnetic potential theory If j = 0 except for linear currents Except for line singularities Then Magnetic scalar potential Contrast with electrostatics In all space is single valued.

  18. In magnetic case at a line current If the line current penetrates the surface of integration. = - B + A = (4 /c) Jenc Magnetic scalar potential is many valued.

  19. 12. Magnetic moment If we are at large distances from a finite current distribution Magnetic moment HW What is magnetic moment of a flat circular wire loop?

  20. 13. Magnetic moment of linear circuit Total current in the wire Linear current Closed loop of current = Area of triangular surface bounded by r and dl is independent of the particular surface bounded by the given current loop. Enclosed area if current lies in a plane

  21. 14. Energy flux in a conductor Electromagnetic field energy is dissipated as Joule Heat HW: Poynting vector for field in vacuum

  22. Inside of a current-carrying conductor Et and Ht are continuous at the surface This energy flux must get burned up as Joule heat because it doesn t flow out the other side Direction is from vacuum into the conductor Rate at which Joule heat is generated per unit volume (21.6) Rate at which energy density leaves the EM field by energy continuity

More Related Content

giItT1WQy@!-/#giItT1WQy@!-/#giItT1WQy@!-/#giItT1WQy@!-/#giItT1WQy@!-/#giItT1WQy@!-/#