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

Daniel Ioan and Irina Munteanu

9.4  Problem 4 - Current-carrying circular wire

Determine the magnetic field strength produced by a filamentary circular conductor of radius a, placed in air and carrying the current i , in a point on the symmetry axis of the conductor (normal on its plane).

Solution

We will consider a Cartesian coordinate system, chosen such that the wire is placed in the yOz plane, with the center in the origin. The field point (in which the field must be computed) is therefore placed on the Ox axis and has the coordinate x on this axis.

The product J dv has in this case the expression i dl = i t dl = i t a d a, in which dl is the length element with the unit vector t tangential to the circle.

Due to symmetry, the components Hy si Hz are zero on the conductor's axis. The denumerator of the integrand for Hz becomes:

i (t x R) a d a = R (i x t) a d a = a2d a,

because the unit vectors i and t are orthogonal, their vector product is the unit vector of the position vector in the yOz plane, and the projection of vector R on this unit vector is the circle's radius, a. Consequently, the integrand is constant, and the value of the integral is:

Hz : = [(I a2 0pda)/(4 p R3)] = [(Ia2)/(2 R3)],

in which R = {a2 + x2} is constant.

> restart: with(linalg): with(plots):

Warning, new definition for norm

Warning, new definition for trace

> Hz := i*a^2/(2*sqrt(a^2+x^2)^3);


Hz : = 1
2
  i a2
(a2 + x2)3/2

Numerical application: I=2A, a = 5cm, x = 0;

> H:=subs(i=2, a=0.05, x = 0, Hz);


H : = 20.00000000

> B:=evalf(4*Pi*10^(-7)*H);


B : = .00002513274123

In conclusion, the magnetic field in the center of the conductor has the strength H=20 A/m and the flux density B=25 mT.

> plot(subs(i=2,a=0.05,Hz), x=-0.1..0.1); # Field variation [A/m] on the conductor's axis

pictures/bsl_en16.gif
>

9.4.1  Comments regarding the field outside the axis

The asymptotic expression of the field at great distance is:

H = 1/(4 p) ( 3 (mR)R/ R5 - m/ R3),

in which the vector m has the magnitude i p a2 and is oriented along the Ox axis.

> restart;

> Hm:= i*a^2/4*[3*x*x/sqrt(x^2+y^2)^5 - 1/sqrt(x^2+y^2)^3, 3*y*y/sqrt(x^2+y^2)^5]; # The field at great distance


Hm : = 1
4
 i a2 [3  x2
(x2 + y2)5/2
- 1
(x2 + y2)3/2
,  3  y2
(x2 + y2)5/2
]

> Hsp:=i*a^2/(2*sqrt(a^2+x^2)^3); # The field of the circular wire - on the axis


Hsp : = 1
2
  i a2
(a2 + x2)3/2

> with(plots):fieldplot(subs(i=1, a=1, Hm), x=1..3, y=1..3);

pictures/bsl_en17.gif
> f:= plot(subs(i=1, a=1,y=0, Hm[1]), x=2..6,color=blue):

> g:=plot(subs(i=1,a=1,Hsp), x=2..6,color=red):

> display(f,g);

pictures/bsl_en18.gif
> errh:= subs(i=1,a=1,y=0,abs((evalm(Hm)[1]-Hsp)/Hsp)); # Error relative to the field in the current point, versus distance


errh : = 2 

 ( 3
4
  x2
(x2)5/2
- 1
4
  1
(x2)3/2
- 1
2
  1
(x2 + 1)3/2
) (x2 + 1)3/2 

> errhc:=subs(i=1,a=1,y=0,abs((evalm(Hm)[1]-Hsp )/subs(i=1,a=1,y=0,x=0,Hsp))); # Error relative to the field in the center, versus distance


errhc : =

  3
2
  x2
(x2)5/2
- 1
2
  1
(x2)3/2
- 1
(x2 + 1)3/2
 

> plot([errh,errhc],x=2..10,color=[blue,red]);

pictures/bsl_en19.gif
> solve(errh=0.01, x);


12.25762249,  -12.25762249

> solve(errhc=0.01, x);


2.634421564,  -2.634421564

In conclusion, the field outside the axis can be calculated with the formula valid at great distance, for distances bigger than 3...12 times the circle's radius.

To calculate the field outside the axis, in the neighbourhood of the center, we need to calculate the integral:

> restart: A:=mu*a*i/(4*Pi)*int(cos(alpha)/ sqrt((a*sin(alpha))^2 + (a*cos(alpha)-y)^2+z^2),alpha=0..2*Pi);


A : = 1
4
 
m a i 
p
0 
cos(a)
{a2 sin(a)2 + (a cos(a) - y)2 + z2}
 da

p

The vector potential can be also expressed in terms of the elliptic integrals E and K.

> k:=sqrt(4*a*y/((a+y)^2+z^2));


k : = 2    
 


a y
(a + y)2 + z2
 

> Ae:=mu*i/k/Pi*sqrt(a/y)*( (1-k^2/2)*EllipticK(k) - EllipticE(z,k));


Ae : = 1
2
m i    
 


a
y
 
((1 - a y
(a + y)2 + z2
) EllipticK(2    
 


a y
(a + y)2 + z2
 
) - EllipticE(z,  2    
 


a y
(a + y)2 + z2
 
))
(   
 


a y
(a + y)2 + z2
 
 p)

> evalf(subs(a=1,y=2,z=1,mu=1,i=1,2*Pi*y*Ae)); # Flux on a circle of radius 2, placed at z=1


.5560336279

> evalf(subs(a=1,y=2,z=1,mu=1,i=1,2*Pi*y*A));


.5560336272

The field in the neighbourhood of the axis can be calculated using an alternative method, based on development in power series in the variable r (distance to the axis).

Assuming that the domain in the vicinity of the axis is not current carrying, the vector potential satisfies the equation rot rot A =0. Due to the axial symmetry, the vector potential has, in cilindrical coordinates, only one non-zero component Af, component which admits a series in odd powers of r:

A(r,z) = k = 0 r(2 k + 1) fk + 1(z)

By writing in cylindrical coordinates the equation satisfied by the vector potential in the vicinity of the axis, it results that a term can be obtained by recurrence, in terms of the second derivative of the preceding term:

fk + 1 = - [([(2)/(x2)] fk - 1)/(k (k + 1))]

Since m Hz(0,  z) = 2 f1(z) , it results that both the vector potential and the flux density's components can be expressed in terms of the field on the axis, Hz(0,z):

A(r,z) = m ( k = 0 [(([(r)/2])(2 k + 1) ( - 1)k Hz(0,  z)(2 k))/(k! (k + 1)!)] )

> Hsp:=i*a^2/(2*sqrt(a^2+z^2)^3); # The field of the circular wire - on the axis


Hsp : = 1
2
  i a2
(a2 + z2)3/2

> A:=mu*(r*Hsp/2 + sum((r/2)^(2*n+1)*(-1)^n* (diff(Hsp,zn,zn)) /(n! * (n+1)!),n=1..5)): # Power series of the vector potential

> tt:=evalf(subs(a=1,y=0.2,z=1,mu=1,i=1,2*Pi*y* Ae)); # Flux on a circle of radius 0.2 placed at z=1


tt : = 1.256637062 Ae

> evalf(subs(r=0.2,a=1,z=1,mu=1,i=1,2*Pi*r*A));


.05555081376

> tt/";


.9999999445

> Br:=-diff(A, z): Brc := simplify(subs(i=1,mu=1,a=1,Br));


Brc : = - 3
524288
r z(10895360 r4 z6 + 1537536 r10 z10 - 693693 r10 - 33030144 z10
- 27525120 z8 - 15728640 z6 - 5898240 z4 - 4259840 r2 z4 + 327680 r2 z18
+ 7884800 r4 z10 + 860160 r6 z14 - 9139200 r6 z10 - 582120 r8 - 470400 r6 - 358400 r4
- 1310720 z2 - 245760 r2 - 131072 z20 - 2007040 r4 z14 - 573440 r4 z16 - 1310720 z18
- 5898240 z16 - 1075200 r6 z12 - 15728640 z14 - 27525120 z12 + 14694400 r4 z8
- 4587520 r2 z6 + 1146880 r2 z8 + 9175040 r2 z10 + 11468800 r2 z12 + 7208960 r2 z14
+ 2375680 r2 z16 - 358400 r4 z12 - 9072000 r6 z8 + 8279040 r8 z10 - 1182720 r8 z12
- 131072 + 7714560 r6 z4 + 3494400 r6 z6 - 716800 r4 z2 + 2652160 r4 z4
- 1638400 r2 z2 - 55495440 r10 z4 + 63423360 r10 z6 - 21141120 r10 z8 + 6597360 r8 z2
- 3686760 r8 z4 - 18849600 r8 z6 + 1478400 r8 z8 + 1881600 r6 z2 + 13873860 r10 z2) /
(1 + z2)25/2

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


Bzc : = - 1
131072
(5160960 r4 z6 + 1064448 r10 z10 - 43659 r10 - 16515072 z10
- 13762560 z8 - 7864320 z6 - 2949120 z4 + 196608 r2 z4 + 196608 r2 z18 + 645120 r4 z10
+ 573440 r6 z14 - 4085760 r6 z10 - 44100 r8 - 44800 r6 - 46080 r4 - 655360 z2
- 49152 r2 - 65536 z20 - 1658880 r4 z14 - 368640 r4 z16 - 655360 z18 - 2949120 z16
+ 143360 r6 z12 - 7864320 z14 - 13762560 z12 + 4838400 r4 z8 + 2752512 r2 z6
+ 7569408 r2 z8 + 11010048 r2 z10 + 9633792 r2 z12 + 5111808 r2 z14 + 1523712 r2 z16
- 2257920 r4 z12 - 65536 - 6352640 r6 z8 + 4032000 r8 z10 - 806400 r8 z12
+ 1881600 r6 z4 - 1756160 r6 z6 + 276480 r4 z2 + 2257920 r4 z4 - 196608 r2 z2
- 17463600 r10 z4 + 27941760 r10 z6 - 11975040 r10 z8 + 1675800 r8 z2 - 3572100 r8 z4
- 6703200 r8 z6 + 3427200 r8 z8 + 896000 r6 z2 + 2619540 r10 z2) / (1 + z2)23/2

> with(plots):fieldplot(subs(i=1,mu=1,a=1,[Brc, Bzc]),r=0..1,z=-2..2);

pictures/bsl_en20.gif
>


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On 16 Feb 2000, 02:04.