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Jackson 5.3 Homework Problem Solution
Dr. Christopher S. Baird
University of Massachusetts Lowell
PROBLEM:
A right-circular solenoid of finite length L and radius a has N turns per unit length and carries a current
I. Show that the magnetic induction on the cylinder axis in the limit NL → ∞ is
B
z
0
N I
cos 1
cos 2
where the angles are defined in the figure.
SOLUTION:
Let us first find the on-axis magnetic induction produced by one circular loop of current. Place the
observation point such that the line from the observation point to the loop makes an angle θi with the
solenoid's axis and is a distance r i
away. In this set-up, then the observation point can be considered at
the origin of a spherical coordinate system.
If the current is flowing into the paper at the bottom and out of the paper at the top, the current density
in spherical coordinates becomes:
J = I
r − r i
r
i
Plug this into the Biot-Savart Law:
B x =
0
∫
J x '× x − x '
∣ x − x '∣
3
d
3
x '
B x =
0
∫
2
∫
∫
∞ (^) [
I
r '− r i
r '
i
]
× x − x '
r
2
r '
2
− 2 r r ' cos cos 'sin sin ' cos − '
3 / 2
r '
2
sin ' d r ' d ' d '
θ 2
θ 1
z
θ i
z
r i
Spherical unit vectors are not constant or fixed, but change as we integrate. The safest thing to do is
express the vector directions in Cartesian coordinates, using relations such as
ϕ '=−sin ϕ '
i +cosϕ '
j
and
x − x '= r sin cos − r ' sin ' cos '
i r sin sin − r ' sin ' sin '
j r cos − r ' cos '
k
B x =
0
∫
0
2
∫
0
∫
0
∞
r '
2
sin ' d r ' d ' d '
[
I
r '− r i
r '
i
−sin '
i cos '
j
]
×
r sin cos − r ' sin ' cos '
i r sin sin − r ' sin ' sin '
j r cos − r ' cos '
k
r
2
r '
2
− 2 r r 'cos cos 'sin sin ' cos − '
3 / 2
Now evaluate the Dirac delta's
B x =
I
0
∫
0
2
r i
sin i
d '[ −sin '
i cos '
j ]×
r sin cos − r i
sin i
cos '
i r sin sin − r i
sin i
sin '
j r cos − r i
cos i
k
r
2
r i
2
− 2 r r i
cos cos i
sin sin i
cos−'
3 / 2
Evaluate the cross product
B x =
I
0
∫
0
2
r i
sin i
d '
r cos − r i
cos i
cos '
i sin '
j r i
sin i
− r sin cos−'
k
r
2
r i
2
− 2 r r i
cos cos i
sin sin i
cos − '
3 / 2
This expression is general for any point in space in spherical coordinates. All that is left to do is
perform the integral over the azimuthal angle and we would have the final solution to the total
magnetic field (magnetic induction) B as any point in space created by a loop of current. The integral is
messy however, and we don't need to do it for this particular problem. For an observation point at the
origin r = 0 , this reduces to:
B x =
I
0
sin i
r i
[
−cos i
i (^) ∫
0
2
cos ' d '
j (^) ∫
0
2
sin ' d 'sin i
k (^) ∫
0
2
d '
]
We can now perform the integrals and find
B x =
0
I
sin
2
i
r i
k
If we recognize that the ring has some fixed radius no matter where we place the ring, we can simplify
this using r^ i
sin i
= a
B x =
k
0
I a
2
r i
3
or
B x =
k
0
N I
[
cos 2
cos 1 ]
We can check our answer by taking limiting cases. If the solenoid's length becomes infinite, both angles
become zero and we get our usual answer for a long solenoid B ^ x =^
k N I
