S2 Appendix: Details on Derivation of QSCR Magnetic
and Electric Fields
In the main text, the integral in Eq (3) has no closed form solution. To approximate the
answer, first define the integrand of Eq (3) as the differential z-component of the
magnetic vector potential:
dAz =
µo Io cos2 [k( h2 − |z 0 |)]
i1
4π h 2
2 2
ρ + (z − z 0 )
(A.1)
Next, it is possible to compute the differential magnetic field from the relationship
dH = µ1o ∇ × dAz az . Evaluating this curl produces
dH =
Io cos2 [k( h2 − |z 0 |)]ρ
i 3 aφ ,
4π h 2
2 2
0
ρ + (z − z )
(A.2)
This approach allows us to recover the magnetic field by integrating Eq (A.2):
H = Hφ a φ =
Io
4π
Z
∞
−∞
cos2 [k( h2 − |z 0 |)]ρ 0
h
i 3 dz aφ
2 2
2
0
ρ + (z − z )
(A.3)
A good approximation to the solution of Eq (A.3) can be made by recognizing that
almost all of the area under the integrand’s curve is accumulated when z 0 = z. Since z
is between ±h, the cos2 term is nearly constant and can be moved outside the integral
in Eq (A.3)
Z ∞
Io
h
ρ
0
Hφ ≈
cos2 [k( − |z|)]
(A.4)
2 3 dz
2
0
4π
2
2
−∞ (ρ + (z − z ) )
Then, integration of what remains in Eq (A.4) yields the H-field due to the QSCR’s
central pole as presented in the main text:
H = Hφ a φ ≈
Io cos2 [k( h2 − |z|)]
aφ
2π
ρ
(A.5)
Next, attention turns to calculation of the E-field. To do this we curl the differential
H-field, dHφ aφ of Eq (A.2) and integrate the result to obtain the E-field. Thus,
dE =
1
∇ × dH
jωo o
(A.6)
where ωo is the frequency and o is the permittivity of free space. Then, using Eq (A.2)
in (A.6) yields
Io
dE = −j
πωo o
("
#
"
# )
2
0
2
0
0
2
0
2
2
0
h
h
h
h
3 ρ cos [k( 2 − |z |)](z)
3 ρ cos [k( 2 − |z |)](z )
1 cos [k( 2 − |z |)]
3 ρ cos [k( 2 − |z |)]
−
aρ +
−
az
(A.7)
4 [ρ2 + (z − z 0 )2 ]5/2
4 [ρ2 + (z − z 0 )2 ]5/2
2 [ρ2 + (z − z 0 )2 ]3/2
4 [ρ2 + (z − z 0 )2 ]5/2
Like with the integral that produced the magnetic field, each of the terms to be
integrated in Eq (A.7) has a maximum when z = z 0 , which is where most of the area
under the integrand’s curve is accumulated when integrating. To yield a non-zero
answer, however, a two term Taylor approximation around z = z 0 can be used to replace
the cos2 term in Eq (A.7), permitting integration over z 0 . The two-term Taylor
approximation for the cos2 function in a neighborhood around z = z 0 is
PLOS
1/2
(
cos
2
[k( h2
0
− |z |)] ≈
cos [k( h2 − z)] + 2k sin [k(2|z| − h)](z 0 − z),
cos [k( h2 + z)] − 2k sin [k(2|z| − h)](z 0 − z),
z0 > 0
z0 < 0
(A.8)
Using this in place of the cos2 terms in Eq (A.7) and integrating over all real z 0 yields
the result for the E-field due to the QSCR’s central pole as presented in the main text:
E=
z3
−jηo Io
ρ2 + 2z 2
sin[k(2 |z| − h)]
a
+
a
ρ
z
2π
(ρ2 + z 2 )3/2 ρ
(ρ2 + z 2 )3/2
(A.9)
where ηo is the impedance of free space.
PLOS
2/2