Symbolic Analysis of Magnetic Field with Symbolic Analysis of Magnetic Field with the Biot-Savart-Laplace Method

Daniel Ioan and Irina Munteanu

9.5  Problem 5 - The Helmholtz coil

Let us consider two thin circular conductors, of radius a, carrying the current i and placed coaxially, at the distance d. Determine d such that the magnetic field in the center of this system (called Helmholtz coil) be as uniform as possible.

Solution

The field produced by the two circular conductors, on their common axis, is obtained by superposition:

> restart:Bsp:=mu*i*a^2* (1/(2*sqrt(a^2+(z-d/2)^2)^3) + 1/(2*sqrt(a^2+(z+d/2)^2)^3)); # Field on the axis


Bsp : = m i a2 ( 4
(4 a2 + 4 z2 -z d + d2)3/2
+ 4
(4 a2 + 4 z2 + 4 z d + d2)3/2
)

> f1:=plot(subs(i=1,a=1,d=0.8,mu=1,Bsp),z=-1..1 ,color=blue):

> f2:=plot(subs(i=1,a=1,d=1,mu=1,Bsp),z=-1..1):

> f3:=plot(subs(i=1,a=1,d=1.2,mu=1,Bsp),z=-1..1 ,color=black):

> with(plots):display(f1,f2,f3);

pictures/bsl_en21.gif
To determine the optimal distance, we will develop Hsp(z) in Taylor series around the origin:

> s := simplify(taylor(Bsp,z=0,6));


s : = 8  m i a2
(4 a2 + d2)3/2
+ 192  m i a2 ( - a2 + d2)
(4 a2 + d2)7/2
 z2 + 1920  m i a2 (2 a4 -a2 d2 + d4)
(4 a2 + d2)11/2
 z4 + O(z6)

> p0:=op(1,s); # Field in the origin


p0 : = 8  m i a2
(4 a2 + d2)3/2

> p2 := op(3,s); # Second degree term


p2 : = 192  m i a2 ( - a2 + d2)
(4 a2 + d2)7/2

> solve(p2=0,d);


a,  - a

In conclusion, the distance between the two conductors must be equal to their radius, in order to have a field as close to the uniform one as possible (i.e., in order to have zero 2nd order Taylor term, which has the biggest contribution to the non-uniformity).

The relative deviation from uniformity is:

> d:=a: err := (Bsp - p0)/p0;


err : =
5
8
 
(m i a2 ( 4
(5 a2 + 4 z2 -z a)3/2
+ 4
(5 a2 + 4 z2 + 4 z a)3/2
) - 8
25
  m i a2 5
(a2)3/2
5 (a2)3/2

m i a2

> plot( subs(i=1,a=1,mu=1, abs(err)), z=-1..1);

pictures/bsl_en22.gif
> plot( subs(i=1,a=1,mu=1, z=10^t, log10(abs(err))), t=-2..0,axes=boxed); # decimal logarithm of the deviation, versus the decimal logarithm of the distance to the center

pictures/bsl_en23.gif
> fsolve( subs(i=1,a=1,mu=1, abs(err)) = 0.01,z=0..1);


.3137460252

Consequently, the error (deviation from uniformity) is lower than 1% for distances |z| < 0.3 a.

9.5.1  The field outside the axis

To analyze the uniformity of the field outside the axis, we will use approximate expressions based on truncated power series in r and z, valid in the vicinity of the axis.

> sp:=convert(s, polynom); # Approximation of the field on the axis


sp : = 8
25
  m i a2 5
(a2)3/2
- 1152
3125
  m i a6 z4
(a2)11/2

> A:=mu * (r*sp/2 + sum((r/2)^(2*n+1)*(-1)^n* (diff(sp,zn,zn)) /(n! * (n+1)!),n=1..7)): # Vector potential in the vicinity of the axis

> Br:=-diff(A, z): Brc := simplify(subs(i=1,mu=1,a=1,Br)): # Components of the field

> Bz:=diff(r*A,r)/r: Bzc := simplify(subs(i=1,mu=1,a=1,Bz)):

> err := sqrt((Bz - p0)^2 + (Br)^2)/p0; # Module of the relative deviation with respect to the central field


err : = 5
8
(((
m ( 1
2
 r ( 8
25
  m i a2 5
(a2)3/2
- 1152
3125
  m i a6 z4
(a2)11/2
) + 864
3125
  r3 m i a6 z2
(a2)11/2
- 72
3125
  r5 m i a6 5
(a2)11/2
)
+ r m ( 4
25
  m i a2 5
(a2)3/2
- 576
3125
  m i a6 z4
(a2)11/2
+ 2592
3125
  r2 m i a6 z2
(a2)11/2
- 72
625
  r4 m i a6 5
(a2)11/2
)) / r
- 8
25
  m i a2 5
(a2)3/2
)2 + m2 ( - 2304
3125
  r m i a6 z3
(a2)11/2
+ 1728
3125
  r3 m i a6 z
(a2)11/2
)2)1/25 (a2)3/2 / (m i a2
)

> contourplot(subs(i=1,a=1,mu=1, log10(err)),r=0..1, z=-0.5..0.5,filled=true,coloring=[cyan,blue]); # Magnetic field uniformity domains

pictures/bsl_en24.gif
> plot3d(subs(i=1,a=1,mu=1, abs(err)),r=0..0.3, z=-0.3..0.3, axes=boxed,orientation=[149,54]); # Variation of the relative deviation in the rz plane

pictures/bsl_en25.gif
>


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On 16 Feb 2000, 02:04.