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3
We will start by defining it on R and follow up with
a general definition on n-D.
Definition 1

123
 1 i, j, k  1,2,3
totally skewsymmetric
Definition 2

123
n
1
 
ijk
abc
 (ijk )  ( jki)  (kij )  (i  j)
(ijk )   ai bj ck
 
ijk
abc
  ai bj ck   aj bk ci
 ak bi cj  (i  j )
 
ijk
abk
  ai bj   bi aj

u
r
u
r
i
3
i
ijk j k
A  R  ( A  B)   A B

 
 
v v
v v
v vi v v
A B  C  D  A B  C  D

 A B  C D
ijk
j
k
iab
a
i
b
  ja kb   jb ka  A B C D
j


ijk
6
 
k

a
b
v v v v
v v v v
 AC B  D  A D B C
 
ijk

  (B  v)    j (B  v)
i
=  j (
ijk
kab
ijk
a
k
b
Bv )
=  ia jb   ib ja   j (B v )
a
= j (B v )- j (B v )
i
j
j i
=v B -v (  B)
i
j  j
i
b
Time rate of energy change of particle with v
by a force F is:
for a charge particle with added field E':
Summing all the electron in a circuit, we find that the
sources do work to maintain the current at the rate
- sign is the Lenz's law. This is in
addition to the Ohmic losses in the
circuit, which should be excluded
from the magnetic energy content
Thus, if the flux change is δF, the work done by
sources is:
The problem of the work done in establishing a general
steady-state distribution of currents and fields is shown
in Fig.5.20
The current distribution can be broken up into small
current loops.A loop of current of cross-section area Δσ
following a closed path C and spanned by a surface S
with normal n as shown in Fig. 5.20.
The work done against the induced EMF in terms of
the change in magnetic induction through the loop is:
Express B in terms of the vector potential A, we have
Stokes’s theorem implies that
Since J∆σdl =Jd3x, the sum over all loops gives:
Ampère’s law
implies that:
The identity
with P→A ; Q→H gives:
Assuming that the field distribution is localized, the
second term (surface integral) vanishes. Hence we
have
This is the magnetic equivalent of the electrostatic
equation (5.147)
Assuming that the medium is para- or diamagnetic ,
such that a linear relation exists between H and B, then
Hence the total magnetic energy will be
This is the magnetic analog of electrostatic equation
If we assume that a linear relation exists btwn J and A,
(5.144) implies that the total magnetic energy is:
This is magnetic analog of
If an object of permeability μ1 is placed in a magnetic
field whose current source are fixed, the change in
energy can be treated in close analogy with the
electrostatic discussions of section 4.7 by replacing
D→H; E→B.
H٠B-H0٠B0=B٠H0 - H٠B0 + surface terms
Hence we have:
This can also be written as
Both μ0 and μ1 can be functions of position, but they
are assumed independent of the field strength.
If the object is in otherwise free space , the change in
energy can be written as:
This is equivalent to the electrostatic equation
pf/
(5.81)
(5.84)
The force acting on a body can be derived from a
generalized displacement and calculate
with respect to displacement.
From (5.149) the total energy of N distinct circuits can
be expressed as :
by converting (5.149) to
with the help of (5.32).
Breaking up the integrals into sums of separate
integrals over each circuit, we have:
Hence the coefficients L,M of inductance are given by
Note that (5.32) reads:
The integral over d3x' (5.14) can be written as
integral of A. If the cross-section of the ith circuit is
negligible, then mutual inductance becomes:
Ai induced by Jj
Since curl of A = B, the mutual inductance is:
Flux at i induced by Jj
The self inductance is :
If the current density is uniform throughout the interior,
from Ampère’s law:
the magnetic induction, close to the circuit, is:
2 πρdρ
The inductance per unit length inside and outside the
wire out to ρmax is:
Because the expression BΦ fails at ρ>
At distances large compared to A1/2 , the 1/ρ magnetic
induction can be replaced by a dipole field pattern
Thus the magnetic induction can be estimated to be:
If we set
Hence
Upon combining the different contributions, the
inductance of the loop can be estimated to be:
Consider quasi-static magnetic field in conducting
media, the relevant equations are:
Laplace equation gives Φ=0
for uniform, frequency-independent permeable media.
We can estimate the time τ for decay of an initial
configuration with typical spatial variation defined by
length L , then
hence
(5.161) can be used to estimate the distance L over
which fields exist in a conductor, subjected externally
to fields with harmonic variation at frequency
For copper sphere of radius 1 cm,

τ~5-10 m sec
molten iron core of the earth

τ~105 years
Evidence:

earth magnetic field reverse ~106 years ago,

0.5*104 years, B goes to 0
Consider a semi-infinite conductor of uniform
permeability and conductivity occupies the space z>0 :
Because the diffusion equation (5.160) is second order
in spatial derivatives and first order in time, the steadystate solution for Hx(z,t) can be written as the real part
of
By eq(5.160) , h(z) satisfies :
A trial solution
gives
Dim[ k ]~ 1/length 1/δ.
This length is the skin depth δ:
Ex: Seawater
Copper at room temperature
Hence the solution of Hx(z,t) is real part of
With the boundary condition at z→∞, for z>0,
Since H varies in time, there is an electric field:
,
Taking the real part, together with (5.165),
To compare the magnitude of electric field and
magnetic induction, the dimensionless ratio is
by quasi-static assumption. The small tangential
electric field is associated with a localized current
density
The integral in z is an effective surface current:
The time-averaged power input, for the resulting
resistive heating (P=IV), per unit volume is
With (5.167),(5.168), we have
A simple example:
Two infinite uniform current sheets, parallel to each
other and located a distance 2a apart, at z=±a. The
current density J is in the y direction:
z=a
z=-a
z=a
z=-a
At time t=0, the current is suddenly turned off. The
vector potential and magnetic field decay according to
(5.160) , with variation only in z and t. Let, from
Laplace transform,
From (5.160),
, we have
With a change of variable from p to k:
The initial condition can be used to determine h(k):
==> partial_z H_x = J_y
The initial condition can be used to determine h(k):
Exploiting the symmetry in z, we can express cosine in
terms of exponentials:
Inversion of the Fourier integral yields h(k),
Hence the solution for the magnetic field at all t>0 is:
κ=ka; ν=1/μσaa
The integral can be expressed in terms of the error
function:
Hence
Error function can be expanded in Taylor series, the
result is: