1Definition123 1 , , 1, 2,3i j k

2Definition123 1n

3We will start by defining it on R and follow up with

a general definition on n-D.

ricskewsymmettotally

( ) ( ) ( ) ( )ijk abc ijk jki kij i j

( ) ai bj ckijk

ijk abkai bj bi aj

j k i

k i j ( )

ijk abcai bj ck a b c

a b c i j

i iA B C D A B C D v v v vv v v v

ijk j k iab a bA B C D j k a b

ja kb jb ka A B C D

A C B D A D B C v v v vv v v v

6ijk ijk

3 ( )i i ijk j kA R A B A B ur ur

i kj

kj

i j i j j

i j j ij j

i i

j j

(B v) (B v)

= ( B v )

= (B v )

= (B v )- (B v )

=v B -v ( B)

ijk

ijk ab a b

a ba b b a

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 thecircuit, 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

Since J∆σdl =Jd3x, the sum over all loops gives:

Stokes’s theorem implies that

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-H0٠B

0=B٠H

0 - H٠B

0 + 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

(5.81)

(5.84)

pf/

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 J

j

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:

The inductance per unit length inside and outside the wire out to ρmax is:

2 πρdρ

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:

for uniform, frequency-independent permeable media.

Laplace equation gives Φ=0

We can estimate the time τ for decay of an initial configuration with typical spatial variation defined by length L , then

(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

hence

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 steady-state 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

With the boundary condition at z→∞, for z>0,

Since H varies in time, there is an electric field:

Hence the solution of Hx(z,t) is real part of

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

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,

z=a

z=-a

, we haveFrom (5.160),

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:

The integral can be expressed in terms of the error function:

κ=ka; ν=1/μσaa

Hence

Error function can be expanded in Taylor series, the result is: