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 )
-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 .