PH317 MAGNETISM
WINTER Quarter 2009-2010
Solution to Assignment 9
Assigned: Monday 2/08/10
Due: Monday 2/15/10
Text: 7-58
SOLUTION
Let the width w be large enough so that the magnetic field between
the plates is uniform. Also, the results of Example 5.8 (page 227) can be
applied directly. Let the x-axis be parallel to h, the y-axis be'
parallel to w and the z-axis beparallel to l.
(a) Let l be the length of a section of a plate.
C
=
eoA
h
=
eowl
h
C
=
C
l
=
eow
h
(b) The magnetic flux between the plates
FM=LI=Bnet·a
where the surface area involved is parallel to the x-z plane. Take the
direction of the normal vector for each plate pointing into the region between
the plates.
Bnet
=
mo
2
Ktop×
^
n
top
+
mo
2
Kbot×
^
n
bot
=
mo
2
(-K
^
z
)× (-
^
x
)+
mo
2
(K
^
z
)×(
^
x
)
=
(m0K/2)
^
y
+(m0K/2)
^
y
=
(m0K)
^
y
a
=
lh
^
y
FM
=
(m0K)
^
y
·(lh)
^
y
=
mohl
æ è
I
w
ö ø
=
mohlI
w
=
LI
L
=
L
l
=
moh
w
(c) The product gives
LC
=
æ è
moh
w
ö ø
æ è
eow
h
ö ø
=
moeo
=
1
c2
(d) In the above calculation, make the replacements
eo® e
mo® m
LC
=
me
=
1
v2
Text: 9-9
SOLUTION
The fields for both parts are
E
=
Eocos(k·r - wt)
B
=
1
c
^
k
×E
Let [^(n)] be the polarization unit vector.
(a) Identify the vectors:
^
n
=
^
z
^
k
=-
^
x
k=-k
^
x
^
n
·
^
k
The electric field can be written
E
=
Eo
^
n
cos(k·r - wt)
The dot product is
k·r
=
-k
^
x
·(x
^
x
+y
^
y
+z
^
z
)
=
-kx
The final expression is
E
=
Eo
^
z
cos(-kx-wt)
=
Eo
^
z
cos(kx+wt)
The magnetic field is
B
=
1
c
(-
^
x
)×
é ë
Eo
^
z
cos(kx+wt)
ù û
=
æ è
Eo
c
ö ø
^
y
cos(kx+ wt)
(b) The equations required are
^
k
=
^
x
+
^
y
+
^
z
Ö3
^
n
=
nx
^
x
+nz
^
z
^
n
·
^
k
=
0
The values of the variables nx and nz must be determined. Using the
dot product of the polarization and propagation unit vecors,
æ è
nx
^
x
+nz
^
z
ö ø
·
æ ç
è
^
x
+
^
y
+
^
z
Ö3
ö ÷
ø
=
0
nx+nz
=
0
nz
=
-nx
^
n
=
nx
^
x
-nx
^
z
=
nx
æ è
^
x
-
^
z
ö ø
Take the dot product of the polarization vector with itself and solve for
nx.
^
n
·
^
n
=
nx2
æ è
^
x
-
^
z
ö ø
·
æ è
^
x
-
^
z
ö ø
=
2nx2
=
1
nx
=
1
Ö2
^
n
=
^
x
-
^
z
Ö2
The dot product of the propagation vector with the positipon vector is
k·r
=
k
^
k
·r
=
k
Ö3
æ è
^
x
+
^
y
+
^
z
ö ø
·
æ è
x
^
x
+ y
^
y
+z
^
z
ö ø
=
k(x+y+z)
Ö3
The electric field becomes
E
=
Eocos(k·r-wt)
=
Eo
^
n
cos(k·r-wt)
=
Eo
æ ç
è
^
x
-
^
z
Ö2
ö ÷
ø
cos
é ë
k(x+y+z)
Ö3
-wt
ù û
=
Eo
æ ç
è
^
x
-
^
z
Ö2
ö ÷
ø
cos
é ë
w(x+y+z)
cÖ3
-wt
ù û
The magnetic field is
B
=
1
c
^
k
×E
=
Eo
c
æ è
^
k
×
^
n
ö ø
cos
é ë
w(x+y+z)
cÖ3
-wt
ù û
=
Eo
c
æ ç
è
-
^
x
+2
^
y
-
^
z
Ö6
ö ÷
ø
cos
é ë
w(x+y+z)
cÖ3
-wt
ù û
Text: 9-13
SOLUTION
The reflection and transmission coefficients are determined as a ratio
of intensities where
I =
1
2
ev E2
This result was arrived at by calculating the Poynting vector and then,
assuming a sinusoidal variation of the fields, taking a time average. In the
following calculation, the fields will be considered real rather than complex.
R
=
EoR2
EoI2
=
æ è
1-b
1+b
ö ø
2
T
=
e2v2
e1v1
EoT2
EoI2
=
e2v2
e1v1
æ è
2
1+b
ö ø
2
For the transmission coefficient, the modifying factor can be written
e2v2
e1v1
=
æ è
m1
m2
ö ø
æ è
e2m2
e1m1
ö ø
æ è
v2
v1
ö ø
=
æ è
m1
m2
ö ø
æ è
v1
v2
ö ø
2
æ è
v2
v1
ö ø
=
m1v1
m2v2
=
b
The sum of the reflection and transmission coefficients become
R + T
=
æ è
1-b
1+b
ö ø
2
+b
æ è
2
1+b
ö ø
2
=
1-2b+b2+4b
(1+b)2
=
1+2b+b2
(1+b)2
=
æ è
1+b
1+b
ö ø
2
=
1
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