Download Constant magnetic solenoid field

Peeter Joot
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Constant magnetic solenoid field
In [2] the following vector potential
A=
Bρ2a
φ̂,
2ρ
(1.1)
is introduced in a discussion on the Aharonov-Bohm effect, for configurations where the interior
field of a solenoid is either a constant B or zero.
I wasn’t able to make sense of this since the field I was calculating was zero for all ρ 6= 0
B=
∇×A
Bρ2a
φ̂
φ̂
= ρ̂∂ρ + ẑ∂z + ∂φ ×
ρ
2ρ
φ̂
Bρ2a
= ρ̂∂ρ + ∂φ ×
φ̂
ρ
2ρ
Bρ2a
1
Bρ2a φ̂
=
ρ̂ × φ̂∂ρ
+
× ∂φ φ̂
2
ρ
2ρ ρ
Bρ2
= 2a −ẑ + φ̂ × ∂φ φ̂ .
2ρ
(1.2)
Note that the ρ partial requires that ρ 6= 0. To expand the cross product in the second term let
j = e1 e2 , and expand using a Geometric Algebra representation of the unit vector
φ̂ × ∂φ φ̂ = e2 e jφ × e2 e1 e2 e jφ
D
E
= −e1 e2 e3 e2 e jφ (−e1 )e jφ
2
= e1 e2 e3 e2 e1
= e3
= ẑ.
(1.3)
So, provided ρ 6= 0, B = 0.
The errata [1] provides the clarification, showing that a ρ > ρ a constraint is required for this potential to produce the desired results. Continuity at ρ = ρ a means that in the interior (or at least on the
boundary) we must have one of
1
A=
Bρ a
φ̂,
2
(1.4)
A=
Bρ
φ̂.
2
(1.5)
or
The first doesn’t work, but the second does
B=∇
×A
φ̂
Bρ
= ρ̂∂ρ + ẑ∂z + ∂φ ×
φ̂
ρ
2
B
Bρ φ̂
= ρ̂ × φ̂ +
× ∂φ φ̂
2
2 ρ
= Bẑ.
(1.6)
So the vector potential that we want for a constant Bẑ field in the interior ρ < ρ a of a cylindrical
space, we need
( 2
Bρ a
if ρ ≥ ρ a
2ρ φ̂
A=
(1.7)
Bρ
φ̂
if
ρ
≤
ρ
.
a
2
This potential is graphed in fig. 1.1.
Figure 1.1: Vector potential for constant field in cylindrical region.
2
Bibliography
[1] Jun John Sakurai and Jim J Napolitano. Errata: Typographical Errors, Mistakes, and Comments,
Modern Quantum Mechanics, 2nd Edition, 2013. URL http://www.rpi.edu/dept/phys/Courses/
PHYS6520/Spring2015/ErrataMQM.pdf. 1
[2] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014. 1
3