The Cryogenic Neutron EDM Experiment:
Calculation of the Magnetic Field in the vicinity of the Ramsey Cell
(with trim coils)
Introduction
The “B0” field in the cryogenic neutron EDM experiment is maintained by a superconducting
solenoid wound inside the cylindrical helium tank. It is clearly useful for us to be able to predict
both the magnitude and direction of this field throughout the volume of the Ramsey cell, so that we
can make any modifications necessary to ensure that it complies with the various experimental
constraints.
The chief constraint is that the geometric phase induced false EDM signal should be no greater than
the overall error that the experiment is aiming to achieve. This signal is caused by the presence in
the neutron’s rest frame of a magnetic field component due to the relativistic “E x v” effect, which
combines with any radial component of field to produce a rotating magnetic field and hence a false
EDM.1
Inside a solenoid of finite length, there will indeed be a small radial component of the magnetic
field everywhere except on the axis. As the size of the Ramsey Cell will be of the same order of
magnitude as the solenoid, we cannot approximate the field to its on-axis value, and therefore need
to investigate this radial component at all points in the volume occupied by the cell. In terms of
actually measuring it, however, there is a problem: since the radial component in our solenoid is
several orders of magnitude smaller than the axial component, in any orientation of the probe that is
not entirely along the radius vector, the former will be swamped by the latter; and even if the probe
could be aligned precisely along the radius vector, its finite width would probably still cause
problems.
Luckily, the radial component is related to the field gradient in the axial direction. Writing the
equation div B = 0 in cylindrical polar co-ordinates, we have
1 
rBr   1 B  Bz  0
r r
r 
z
Assuming azimuthal symmetry we can omit the middle term, and expanding the first term this
therefore becomes
Br Br Bz


0
r
r
z
If we can further assume a linear dependence of Br on r so that
this becomes
2 Br Bz

0
r
z
whence
Br Br

,
r
r
Br 

r Bz
2 z
Using this relation, Pendlebury et al.1 derive a maximum permissible value of 1nT/m for the axial
B z
field gradient
.
z
Calculating the Off-Axis Field
The standard formula for the field at a point on the axis of a finite solenoid
Bz 
0 NI
2l
cos 1  cos  2 
where ,  are the angles subtended at the point in question by the radii at the ends,
is clearly of no use when the point is not on the axis. Instead we will have to derive a formula from
first principles.
Consider a small current element dl at a point P(xP,yP,zP) on a circular loop of wire:
y
P
x

I
According to the Biot-Savart Law, the magnetic field due to this element at a point Q (xQ,yQ,zQ) is
given by the formula
dB 
0 I  dl  PQ 
4  PQ 3 
(1)
If the radius of the circular loop is a, the length of the element dl is ad and so in cartesian coordinates the vector dl is given by
dl   a sin d i  a cos d  j
To find the components of the vector PQ , we consider the projection of the circular loop in the
plane of the point Q:
P'
Q
R
The vector PP' is clearly zQ  z P k , so that the vector PQ can be written
PQ  PP'  P' R  RQ  zQ  zP .k  a sin  . j  a cos   xQ .i
Hence

 
dl  PQ   a sin d i  a cos d  j  zQ  z P k  a sin   j  a cos   xQ i

 zQ  z P a sin d  j  a 2 sin 2 d k  zQ  z P a cos d i  a cos d a cos   xQ k
 zQ  z P a cos d i  zQ  z P a sin d  j  a 2  axQ cos  d k
Furthermore,
(2)
PQ 2  PP'2  P' R 2  RQ 2  zQ  z P 2  a sin  2  a cos   xQ 2
 zQ  z P   a 2  2axQ cos   xQ2
2
(3)
Substituting from (2) and (3) into (1), therefore, we obtain


2
0 I  zQ  z P a cos d .i  zQ  z P a sin d . j  a  axQ cos  d .k 
dB 

 (4)
3
4 
2
2
2 2

zQ  z P   a  2axQ cos   xQ




Thus the field at Q due to this single loop of wire has x, y and z components given by
0 Ia zQ  z P  2
Bx 
0
4
0 Ia zQ  z P  2
By 
0
4
0 Ia 2
Bz 
4 0
cos d
z
 z P   a  x  2axQ cos 
2
Q
z
 z P   a  x  2axQ cos 
2
Q
Q
2
2
Q

(5)

(6)
3
2
3
2
cos  d
 z P   a  x  2axQ cos 
2
Q
2
Q
sin d
a  x
z
2
2
2
Q
(7)

3
2
(6) can easily be integrated to give the answer zero, which is to be expected since we have assumed
azimuthal symmetry and hence B  0 , so that at Q, B y  0 . (5) and (7) cannot be integrated
analytically, so we will have to use a numerical method. Before we do this, however, we note that
substituting xQ  0 in (5) gives zero, as expected (since we don’t expect a radial component on the
axis) and also that the same substitution in (7) gives
Bz 
0 Ia 2
4 0
ad
z
 zP   a 2
2
Q

3
2


0 Ia 2
4 zQ  z P   a 2
2
2
 d 

3
2
0

0 Ia 2
2 zQ  z P   a 2
2

3
2
which is the standard formula for the axial field due to a single turn of radius a carrying a current I
at a distance (zQ – zP).
Having identified Bx at Q as the radial component, and noted that it will be the same at all points a
distance xQ from the axis, again due to azimuthal symmetry, we will henceforth write xQ as r, so that
(5) and (7) become
0 Ia zQ  z P  2
Br 
0
4
cos d
z
 z P   a  r  2ar cos 
2
Q
2
2

3
2
(8)
and
0 Ia 2
Bz 
4 0
a  r cos  d
z
 z P   a  r  2ar cos 
2
Q
2
2
(9)

3
2
We will now assume that our single turn of wire forms part of a solenoid of length l, and with a
constant number of turns per unit length, N/l. (We will avoid the common practice of using the
single symbol N to represent “turns per unit length” as this can become confusing, especially when
we later introduce the trim coils). We will take a thin slice of the solenoid, in a plane perpendicular
NdzP
to its axis, of thickness dzP. Thus the number of turns this contains is
, and hence the field at
l
Q due to this slice has components
dBr 
0 NIa zQ  z P dz P
4l
2
cos d

z
0
 z P   a  r  2ar cos 
2
Q
2
2
(10)

3
2
and
 NIadz P
dBz  0
4l
a  r cos  d
2

0
z
 z P   a  r  2ar cos 
2
Q
2
2
(11)

3
2
Thus the total field due to the solenoid, whose ends are at zP = ± l/2, has components
0 NIa l / 2 2
Br 
4l l/ 2 0
z
z
Q
 z P cos ddz P
Q
2

0 NIa 2 
4l 0 

z
1
 l / 2  a 2  r 2  2ar cos 
2
Q
3
2
2
 NIa 2 
zQ  z P 2  a 2  r 2  2ar cos 
 0


4l 0 


 z P   a  r  2ar cos 
2


l
1
2
2
 l cos d

2




cos d (12)
2
2
2
zQ  l / 2  a  r  2ar cos  

1

and
0 NIa
Bz 
4l
2
l /2
0
l / 2
 a  r cos  d 
z
dz P
 z P   a 2  r 2  2ar cos 
2
Q

3
2