Jackson 5 17 Homework Solution

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Jackson 5.17 Homework Problem SolutionDr. Christopher S. Baird

University of Massachusetts Lowell

PROBLEM:

A current distribution J(x) exists in a medium of unit relative permeability adjacent to a semi-infiniteslab of material having relative permeabilityr and filling the half-space,z< 0.

(a) Show that forz> 0 the magnetic induction can be calculated by replacing the medium of

permeabilityr by an image current distribution J*, with components,

r1 r1Jx x , y ,z , r1

r1Jy x , y ,z , r1

r1Jzx , y ,z(b) Show that for z < 0 the magnetic induction appears to be due to a current distribution [2r/(r+ 1)]Jin a medium of unit relative permeability.

SOLUTION:

(a) Using the method of images, we replace the effects of the actual currents on the material interfacewith an image current deep within the magnetic material. We will label the original current distribution

J(x) and the image current distribution as J*(x), where J is known but its image at this point needs to

be determined. Because the mirror surface is a flat plane, we can safely assume that a piece of current

in J at (x,y,z) will be mirrored by a piece of current in J* at (x,y, -z). Therefore:

Ji*

x , y , z=Ai Jix , y ,z for each component, so that i =x,y,z

The magnetic B field forz> 0 created by these currents are:

Bz0 x=04 J x 'J*x 'xx '

xx '3

d3x '

For thez< 0 region, place an additional image current distribution J** = aJ in its positivezlocation,

which creates the field:

Bz0 x=

0

r

a

4 J x'xx

'xx '

3 d3x '

Now apply boundary conditions:

[B2B1n=0]on S

[Bz0z=Bz0z ]z=0


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[ 04 J x 'J *x 'xx 'xx '3 d3x ']z=0z=[

0

ra

4J x 'xx '

xx '3

d3x ']

z=0

z

Because each piece mirrors each other piece, and because we are on the z = 0 plane, which is

equidistant from each current piece and its image, the integrands must be equal.

[J x 'J*x 'xx '

xx '3 ]z=0z=[ ra

J x 'xx '

xx '3 ]z=0z

[J x 'J *x 'xx ']z=0z=[ra J x 'xx ']z=0z

If we break each current into azcomponent and transverse component, J=JzzJ t , we have:

[JzzJ tJz* zJ t*xx ']z=0z=[ ra JzzJ txx ']z=0z

Use the identity abc =cab

[ zJz zJ tJz* zJ t*xx '= ra zJz zJ txx ']z=0

[ zJ tzJ t*xx '= ra zJ txx ']z=0

[Jx jJy iJx* jJy* i xx '= ra Jx jJy i xx ']z=0

JyJy* xx 'JxJx

*yy '= ra J y xx ' ra Jx yy '

This must be true for allx andy, so they must be independent:

JyJy*= ra Jy , JxJx

*= ra Jx

Jy*= ra1Jy , Jx

*= ra1Jx

Now apply the other boundary condition:

[nH2H1=K]on S

There is no free current on the interface, so K= 0.

[ 10

zBz0=1

r 0zBz0]z=0

[ 10 z 04 J x 'J *x 'xx 'xx '3 d3x '= 1 r0 z

0 r a

4 J x 'xx '

xx '3

d3x ']

z=0


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[ zJ x 'J*x 'xx '=za J x 'xx ' ]z=0

Use the identity abc =ac babcbeing careful to remember thatz' becomes -z' whenmultiplied with J*

z'JJ *JzJz*xx '=z'a Ja Jzxx '

Expand everything into components and recognize that components as well as functional pieces are allindependent.

Jz*=a1Jz , Jx *=1a Jx , Jy*=1aJy , Jz*=a1Jz

Now solve fora using any two appropriate equations from the boxed ones above:

a=2

r+1

So that finally:

Jx *= r1 r1Jx x , y ,z , Jy * r1

r1Jyx , y ,z , Jz*=r1

r1Jzx , y ,zWhat does this mean physically? If the material is a very strong paramagnetic material, than therelative permeability is effectively infinite. In this special case, the image current equals the original

current in magnitude and loops in the same direction in the x-y directions, but in the opposite direction

in thezdirection:

Paramagnetic materials tend to suck in magnetic field lines, no matter the shape or orientation of thecurrent distribution. This means that paramagnetic materials attract permanent magnets and current

distributions, no matter their orientation. A small loop of current like shown in the diagram above can

be thought of as a small magnetic dipole, with North facing up. The image current can also be thoughtof as a small dipole magnet with North facing up. The South end of the top magnet is therefore facing

J

0

Original Problem

J

Image Method

J*

0

Final Solution for z>0

J


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the North end of the bottom magnet, so they will be attracted. Let us turn the loop of current to see this

effect and to see thezcomponent's behavior:

For a weak paramagnetic material, the same patterns result, but the overall effect is less.

For a diamagnetic material,r < 1, a quick glance at our equations shows us that the image current will

have thex andy components going in the opposite direction as the original current, but thez

component will be going in the same direction. Note that diamagnetic materials are typically veryweak.

As we see, diamagnetic materials tend to repel magnetic field lines, current distributions, and

permanent magnets, no matter their orientation.

(b) As found above, the field in the lower region is:

Bz0 x= 0 ra

4 J x 'xx '

xx '3

d3x '

J

0

Original Problem

J

Image Method

J*

0


J

J

0

< 0

Original Problem

J

Image Method

J*

0


< 0

J


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Bz

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Jackson 5 17 Homework Solution