A Flux Problem — The Hard Way
B
dA
phi
phi’
r
2r
r
2r
The above figure shows the layout. We are to calculate the flux above the line connecting the two infinite
wires, through the curved cylindrical surface of length L. I showed in class how this can be done using the
fact that the magnetic flux through a closed surface is always zero. But let’s check this the hard way. In
so doing, you will both appreciate the shortcut we used in class, and you will hopefully learn some vector
analysis.
The flux integral is,
Z
~·B
~ ,
ΦB =
dA
(1)
∩
where I indicate only the curved portion by ∩. Let the plane of the figure be the x − y plane. Clearly, the
magnetic field does not depend on the z-coordinate for infinite wires. As each wire contributes the same
~ in terms of φ is:
flux, we may simply consider the leftmost wire. For our situation, dA
~ = Lrdφ(cos φî + sin φĵ) .
dA
(2)
For an infinite wire, the magnetic field magnitude is:
B1 =
µ0 i
,
2πd
(3)
where d is the distance from the wire to a point on the surface. In this case, using the law of cosines gives
us,
p
(4)
d(φ) = r2 + (2r)2 − 2(2r) cos(π − φ),
p
= r 5 + 4 cos φ .
(5)
~ and dA
~ are not parallel, and change as we move along the
As you can see from the figure, the directions of B
~
surface. We therefore need to express the direction of B also in terms of φ. First, however, we will express
it in terms of φ0 . Notice how the radial vector from the wire to the surface can be written,
ρ
~ = d(cos φ0 î + sin φ0 ĵ) .
(6)
~ which is in the circumferential direction, is necessarily perpindicular to
Now notice that the direction of B,
~ must be zero. This leads to the identification:
ρ
~. Therefore, the dot product ρ
~·B
~ = B(− sin φ0 î + cos φ0 ĵ) .
B
(7)
Finally, before we integrate, we must express φ0 in terms of φ. This can be done using the law of sines,
sin φ
sin φ0 = √
,
5 + 4 cos φ
s
sin2 φ
2 + cos φ
.
cos φ0 = 1 −
=√
5 + 4 cos φ
5 + 4 cos φ
(8)
(9)
Substituting (9) into (7) and taking the dot product with (2) gives us the flux integral in the form we want,
Z
2µ0 iLr π − sin φ cos φ + (2 + cos φ) sin φ
ΦB =
dφ
,
(10)
2πr
5 + 4 cos φ
Z0 π
2µ0 iL
sin φ
=−
.
(11)
dφ
π
5
+
4 cos φ
0
To evaluate the integral, make the substitution y = cos φ. This leaves us with,
Z
2µ0 iL 1 dy
ΦB = −
,
π
−1 5 + 4y
µ ¶
2µ0 iL 1
=−
ln 9 ,
π
4
µ0 iL
ln 3 .
=−
π
This is precisely the negative of the flux through the bottom we calculated in class:
Z
2µ0 iL r dx
ΦBOT
=
,
B
2π
−r 2r − x
µ0 iL
=
ln 3 .
π
I will live it to you which way you would prefer to agonize.
(12)
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