SEE

.

-3-

it'

, We need to set

up ou -

I I

I' r

I Ampe , an loop and integrate.

i → II I the current flux through

- that loop .

.

,

I-

..

.-

'

A circular loop is need to respect the symmetry of

the were .

§B.dim. f) 5.DA' ( T = ¥ . )

Moi Itr'

for r'

> r

IT r2

Ztr

'D? {q÷

# r,

-

for rkr

If I is in the I direction we car easily write

this as !

Btw ) = { 7¥.I rsr

I÷÷i war

The flux of a

region

veryclose to the wire loop

will be'

.

R

!÷fdtdw-z.IT#RlnF=m.IRlnw-roEF'

"

arms"

Tenpin

For w-

- % → Ola=p.IR/n(o#)=rIRlnll07--m.IR.2lnl0=m.IR

(

4.61. -

- )

for W' - R → loath

.IR/n/E/--M.IR.3ln10

=p .IR ( 6.91 . .

)

What is interesting to notice here is that in both

cases,

the flux sales according to R,

while the

width of theregion

only contributes a numericalfactorby virtue of being a part of the log .

With this in mind

,we expect the flux will be

proportional tog.

I R with a constant of

proportionality that is > 461. .

.

and 26.91.

. .

If we want to venture a guessthat well work with

other values for Mr,

we still need to keepthis ratio in our log .

Thus,

a rough approximationMight be :

§=mIRln ) .

[ = Io. # =

m.HR/nt-9R/rl=m.Rlnfo.rI/TIL--meRln/o-9rR-

)

Zn Check that V. I= O

b and Do D= O.

÷÷¥i÷÷÷÷÷÷¥÷÷¥÷÷

-

a

> x

Because the fields are constant everywhere theyare well - defined

,there derivatives are zero as well .

This 17 . E=o and R . 13=0 with in the slab,

and

we just need to consider the boundaries .

If we imaginea small cubic volume

straddling

z=b,

we on integrate thedivergence as a flux

and see that only two sides will contribute .

non - Zero

i' ←

non - Zero

F- b fix ft .Ed3x=SEdF

what's more since the flux is in one

side and out the other,

these two contributions

cancel.

The exact same logic will apply for to B

at the boundary ,

and at the plane z=a,

with onlysuperficial differences .

Check the curl conditions .

-

Again ,since the derivatives of E and B

are zero,

it is trivial to show that the curl of each is

zero within the slab .The trick then is to

check the boundary with an appropriatelyoriented curve .

e

-

a. y X→¥* } w ( The direction of

⇐ athe curve is purely

( for convenience )

f@xE-i3III.dA-fE.dltf2fIdAI1uE.l

-10+0+0 III - ¥3. e. D= - Bea.

El - Bei -_ o → IgaA similar integral using

17×13 - =0 wellgive

:

Be - IftEla . )= Be- Iz Ela.

-

- O

• 2

Plugging in E- Be → Bfl- Iz )=O→ in =c

@ E- 0 then : ¥= C j E=c

Now looking at

E- b with the same integrals .

÷b T⇐ b

* x→}w§Erdl= -

2¥11t v

El = - Zet -

Blu)=i ( + Blk ) → Ez=b

.

( Boy ,this looks family

. )

Onceagain looking

at the other equation .

§ B - de = IT IET It

B- e = to # ( Elw ) = to El 5 = EI Be

ESo,

also at z=b,

we have : b. = CB-

= C

So on both boundaries,

we have the same conditions.

Namely , that E/B=c and that the changing front

must propagate at the speed of light .