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

Daniel Ioan and Irina Munteanu

9.8  Problem 8 - The rectangular current-carrying bar

A very long massive conductor, of rectangular crossection 2 a x 2 b, carries a current of intensity Io. Determine the magnetic field produced by this conductor.

Solution

We will choose a Cartesian coordinate system, with the axis Oz along the conductor.

The current density has the value J=Io/(4ab). Let us consider an elementary conductor, of crossection dx0 x dyo, which will carry a current di = J dx0 dyo. The magnitude of the field produced by this straight infinitely long elementary conductor in a point P of coordinates x, y has the expression obtained in problem 1:

dH = [(di)/(2 p r)],

where r = {(x - x0)2 + (y - y0)2} is the distance from point P to the elementary conductor.

The field produced by the whole conductor is obtained by integration of the elementary field, moving the source-elementary conductor in the rectangular domain of the crossection x0 = - a .. a, y0 = - b .. b. The superposition involved in this integral cannot be made using the magnitude, it must be performed for each component separately. For this reason, we nee the vector elementary field, whose direction is normal on the source-field vector, in the problem's plane:

dH = t dH = (k x r ) [(di)/(2 p r2)],

because the unit vector of the field direction is t = (k x r ) / |r|. The components of the vector product k x r are: [-( y - y0), x - x0, 0].

> restart: with(linalg):

Warning, new definition for norm

Warning, new definition for trace

> dHx := -Io/(4*a*b)/(2*Pi) * (y-y0)/((x-x0)^2+(y-y0)^2);


dHx : = - 1
8
  Io (y - y0)
a b p ((x - x0)2 + (y - y0)2)

> Hxa:=int(dHx,y0=-a..a);


Hxa : =
1
16
  ln(x2 -x x0 + x02 + y2 -y a + a2Io
a b p
- 1
16
  ln(x2 -x x0 + x02 + y2 + 2 y a + a2Io
a b p

> Hx :=int(Hxa,x0=-b..b);


Hx : = - 1
16
Io( -y arctan( x - b
y + a
) + 2 y arctan( x - b
y - a
)
- x ln(x2 -x b + 2 y a + b2 + y2 + a2) + x ln(x2 -x b -y a + b2 + y2 + a2)
- 2 arctan( x - b
y + a
a + ln(x2 -x b + 2 y a + b2 + y2 + a2b
- ln(x2 -x b -y a + b2 + y2 + a2b - 2 arctan( x - b
y - a
a) / (a b p) + 1
16
Io(
-y arctan( x + b
y + a
) + 2 y arctan( x + b
y - a
) - x ln(x2 + 2 x b + 2 y a + b2 + y2 + a2)
+ x ln(x2 + 2 x b -y a + b2 + y2 + a2) - 2 arctan( x + b
y + a
a
- ln(x2 + 2 x b + 2 y a + b2 + y2 + a2b + ln(x2 + 2 x b -y a + b2 + y2 + a2b
- 2 arctan( x + b
y - a
a) / (a b p)

> dHy := Io/(4*a*b)/(2*Pi) * (x-x0)/((x-x0)^2+(y-y0)^2);


dHy : = 1
8
  Io (x - x0)
a b p ((x - x0)2 + (y - y0)2)

> Hya:=int(dHy,y0=-a..a);


Hya : = - 1
8
 
arctan( y - a
x - x0
Io

a b p
+ 1
8
 
arctan( y + a
x - x0
Io

a b p

> Hy :=int(Hya,x0=-b..b);


Hy : = - 1
16
Io(2 arctan( y - a
x - b
b - 2 arctan( y + a
x - b
b + 2 arctan( y + a
x - b
x - 2 arctan( y - a
x - b
x
-y ln(y + a) + y ln(x2 -x b + 2 y a + b2 + y2 + a2)
- y ln(x2 -x b -y a + b2 + y2 + a2) + 2 y ln(y - a) -a ln(y - a) -a ln(y + a)
+ a ln(x2 -x b -y a + b2 + y2 + a2) + a ln(x2 -x b + 2 y a + b2 + y2 + a2)) / (a b p)
- 1
16
Io( - y ln(x2 + 2 x b + 2 y a + b2 + y2 + a2) + 2 y ln(y + a)
+ y ln(x2 + 2 x b -y a + b2 + y2 + a2) -y ln(y - a) + 2 a ln(y - a) - 2 arctan( y + a
x + b
x
+ 2 arctan( y - a
x + b
b - a ln(x2 + 2 x b -y a + b2 + y2 + a2) - 2 arctan( y + a
x + b
b
+ 2 a ln(y + a) + 2 arctan( y - a
x + b
x - a ln(x2 + 2 x b + 2 y a + b2 + y2 + a2)) / (a b p)       

> simplify(Hy); # Impressive expression, difficult to obtain by hand!


1
16
Io( - 2 arctan( y - a
x - b
b + 2 arctan( y + a
x - b
b - 2 arctan( y + a
x - b
x + 2 arctan( y - a
x - b
x
- y ln(x2 -x b + 2 y a + b2 + y2 + a2) + y ln(x2 -x b -y a + b2 + y2 + a2)
- a ln(x2 -x b -y a + b2 + y2 + a2) - a ln(x2 -x b + 2 y a + b2 + y2 + a2)
+ y ln(x2 + 2 x b + 2 y a + b2 + y2 + a2) - y ln(x2 + 2 x b -y a + b2 + y2 + a2)
+ 2 arctan( y + a
x + b
x - 2 arctan( y - a
x + b
b + a ln(x2 + 2 x b -y a + b2 + y2 + a2)
+ 2 arctan( y + a
x + b
b - 2 arctan( y - a
x + b
x + a ln(x2 + 2 x b + 2 y a + b2 + y2 + a2)) / (a b p)

> a:=1; b:=1; Io:=1;f:=fieldplot([Hx,Hy],x=-2..2,y=-2..2):


a : = 1


b : = 1


Io : = 1

> with(plots):

> p:=polygonplot([[a,b],[-a,b],[-a,-b],[a,-b],[ a,a]],color=red):

> display(f,p);

pictures/bsl_en28.gif
> fh:=plot(subs(y=0.1,Hy), x=0.1..3):

> fh0:=plot(1/(2*Pi*x),x=0.8..3,color=blue): # Equivalent filamentary conductor

> display(fh,fh0);

pictures/bsl_en29.gif
> A:=int(Hy, x): contourplot(A,x=-2..2, y=-2..2,color=blue);

pictures/bsl_en30.gif
>

10  Test: Regular polygon

A filamentary conductor shaped as a regular polygon with n branches of length a each, is placed in air and carries the current I. Compute the magnetic flux density in the center of the polygon. Evaluate the approximation error made when a circular wire is modeled by a polygonal line with n vertices.


File translated from TEX by TTH, version 2.62.
On 16 Feb 2000, 02:04.