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C:\User\Teaching\505506\F10\HW08sol.doc
PHY 505
Classical Electrodynamics I
Fall 2010
Homework 8 with solutions
Problem 8.1
(to be graded of 10 points). Calculate the magnetic field distribution along the axis of a
straight solenoid (Fig. 5.6a
of the lecture notes) with a finite length
l,
and a round crosssection of
radius
R
. Assume that the solenoid has many wire turns (
N
>> 1) which are uniformly distributed along
its length.
Solution
: Because of the large number of wire turns, and hence the almost circular shape of each turn,
the solenoid field may be presented as a sum of field of
N
circular current loops. The field induced by a
single loop has been calculated in class  see Eq. (5.23) of the lecture notes. Generalizing it to the case of
arbitrary position (
z’
) of the loop at the
z
axis, we get
2
/
3
2
2
2
0
1
)
'
(
2
z
z
R
R
I
B
.
Now we can find induced by the solenoid as a whole as
2
/
2
/
2
/
3
2
2
2
0
2
/
2
/
2
/
3
2
2
2
0
)
/
(
2
)
(
2
N
N
j
N
N
j
j
N
N
jl
z
R
R
I
z
z
R
R
I
B
,
where index
j
numbers wire turns (starting from the middle of the solenoid). For
N
>> 1 this sum is
equivalent to integral
1
.
)
2
/
(
2
/
)
2
/
(
2
/
2
)
'
(
'
2
'
)
'
(
2
)
/
(
2
2
2
2
2
0
2
/
'
2
/
'
2
2
0
2
/
2
/
2
/
3
2
2
2
0
2
/
2
/
2
/
3
2
2
2
0
l
z
R
l
z
l
z
R
l
z
l
N
I
z
z
R
z
z
l
N
I
dz
z
z
R
R
l
N
I
dj
N
jl
z
R
R
I
B
l
z
l
z
l
l
N
N
N
Well inside a long solenoid (
l
>>
z
,
R
), this expression reduces to Eq. (5.40). On the other hand,
at
z
>>
R
,
l
, this result gives
1
2
3
0
3
2
0
,
2
4
2
Nm
INR
m
z
m
z
INR
B
N
N
N
,
i.e. the dipole field, with the dipole moment
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 Fall '08
 Stephens,P
 Work, Magnetic Field, Energy density, 0 L, 0 2 l, 0 2m, 2 2 j

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