Python:使用生物萨伐尔定律计算电线的磁场

Question

我想使用生物萨伐尔定律计算电线的磁场。有些人建议使用 numpy 数组。起初我用 vpython 做了它并且它有效。但知道我想使用 Matplotlib 进行可视化。因此我需要数组，对吗？但我现在卡住了。

我也将这个问题发布到了 codereview，但他们将我发送到了 stackoverflow。

问题出在这一行 --> bfield2 = konstante*I*cross(dl, (rx,ry,rz))/r**3

import matplotlib
import numpy as np
import matplotlib.cm as cm
import matplotlib.mlab as mlab
import matplotlib.pyplot as plt
from visual import *

I = 1
mu0 = 1e-7
konstante = mu0/(4*np.pi)

# wire elements; always lenght one
coord = [(0,0), (1,0), (2,0), (3,0), (4,0), (5,0), (6,0), (7,0), (8,0), (9,0), (9,1),
         (9,2), (9,3), (9,4), (9,5), (9,6), (9,7), (9,8)]
# draw the wires
#for i in range(len(coord)-1):
#    wire = curve(pos=(coord[i],coord[i+1]), radius=0.2)

# calculate the b-field
def bfield(x,y,z):
    bfield3 = 0
    # number of wire elements
    for i in range(len(coord)-1):
        # center of the wire element
        wiremiddlex = coord[i][0]+(coord[i+1][0]-coord[i][0])/2.0
        wiremiddley = coord[i][1]+(coord[i+1][1]-coord[i][1])/2.0
        wiremiddlez = 0
        rx = x-wiremiddlex
        ry = y-wiremiddley
        rz = 0
        r = (rx**2+ry**2+rz**2)**0.5
        dl = ((coord[i+1][0]-coord[i][0]), (coord[i+1][1]-coord[i][1]), 0)
        bfield2 = konstante*I*cross(dl, (rx,ry,rz))/r**3 # i have to use numpy arrays
        bfield3 += (bfield2[0]**2 + bfield2[1]**2 + bfield2[2]**2)**0.5
    return bfield3

# visualize
xwidth=10
ywidth=10
delta = 1
x = np.arange(0, xwidth, delta)
y = np.arange(0, ywidth, delta)
X, Y = np.meshgrid(x, y)
slicee = 3
Z = bfield(X,Y,slicee)

plt.figure()
CS = plt.contour(X, Y, Z)
plt.clabel(CS, inline=1, fontsize=10)
plt.title('Simplest default with labels')
plt.show()

第 7 条编辑:我删除了其他编辑。我不想混淆。输出不正确。请参阅下一个编辑。

# Calculation of a magnetic field of a wire
# later I want to to it three dimensional

import matplotlib
import numpy as np
import matplotlib.cm as cm
import matplotlib.mlab as mlab
import matplotlib.pyplot as plt
from pylab import *

I = 10000000000
mu0 = 1e-7
constant = mu0/(4*np.pi)

# wire elements; always lenght one
coord = [(0,0), (1,0), (2,0), (3,0), (4,0), (5,0), (6,0), (7,0), (8,0),
         (9,0), (9,1), (9,2), (9,3), (9,4), (9,5), (9,6), (9,7), (9,8),
         (8,8), (7,8), (6,8), (5,8)]

# calculate the b-field
def bfield(x,y,z):
    b2 = np.zeros((xwidth,ywidth))
    for x in range(xwidth):
        for y in range(ywidth):
            # number of wire elements
            for i in range(21):
                rx = (coord[i][0]+coord[i+1][0])/2. - x
                ry = (coord[i][1]+coord[i+1][1])/2. - y
                rz = z * 1.0 # = z-0
                r = (rx**2+ry**2+rz**2)**0.5 # distance r between field and middle of the wire
                dl = np.array([(coord[i+1][0]-coord[i][0]), (coord[i+1][1]-coord[i][1]), 0])
                b = np.cross(dl, np.array([rx,ry,rz]))
                e = constant*I*b/r**3
                b2[y][x] += e[2] # why not x y?
    return b2

xwidth = 15 
ywidth = 15
delay = 1
x = np.arange(0, xwidth, delay)
y = np.arange(0, ywidth, delay)
X, Y = np.meshgrid(x, y)
slicee = 0.1
Z = bfield(X,Y,slicee)

# visualize
plt.figure()
CS = plt.contour(X, Y, Z)
plt.clabel(CS, inline=1, fontsize=10)
x1 = array([0,1,2,3,4,5,6,7,8,9,9,9,9,9,9,9,9,9,8,7,6,5])
y1 = array([0,0,0,0,0,0,0,0,0,0,1,2,3,4,5,6,7,8,8,8,8,8])
plot(x1,y1)
plt.title('magnetic field')
plt.show()

最后编辑: 最后我在没有 numpy 的情况下做到了。 以下版本有效。

# Calculation of a magnetic field of a wire
# later I want to to it three dimensional

import matplotlib
import numpy as np
import matplotlib.cm as cm
import matplotlib.mlab as mlab
import matplotlib.pyplot as plt
from pylab import *

# constant
I = 10000000000
mu0 = 1e-7
constant = mu0/(4*np.pi)

# wire position
coord = [(10,10), (20,10), (20,20), (10,20), (10,25)]
coord2 = []

# devide path of the wire in parts of length one
parts = 0
for n in range(len(coord)-1):
    lengthx = coord[n+1][0] - coord[n][0]
    lengthy = coord[n+1][1] - coord[n][1]
    length = (lengthx**2 + lengthy**2)**.5
    for m in range(int(length)):
        coord2.append((coord[n][0]+lengthx/length*m, coord[n][1]+lengthy/length*m))
        parts += 1

# calculate the b-field
def bfield(x,y,z):
    b = 0
    for i in range(parts-1):
        dlx = coord2[i+1][0]-coord2[i][0]
        dly = coord2[i+1][1]-coord2[i][1] 
        dlz = 0
        dl = np.array([dlx,dly,dlz])
        rspace_minus_rwire_x = x - (coord2[i][0]+dlx)
        rspace_minus_rwire_y = y - (coord2[i][1]+dly)
        rspace_minus_rwire_z = z - 0
        rspace_minus_rwire = np.array([rspace_minus_rwire_x, rspace_minus_rwire_y, rspace_minus_rwire_z])
        absr = (rspace_minus_rwire_x**2 + rspace_minus_rwire_y**2 + rspace_minus_rwire_z**2)**0.5
        a = constant * I * np.cross(dl, rspace_minus_rwire) / absr**3
        b += (a[0]**2 + a[1]**2 + a[2]**2)**0.5
    return b

xwidth = 26
ywidth = 26
z = 1
bmatrix = np.zeros((xwidth,ywidth))
for x in range(xwidth):
    for y in range(ywidth):
        bmatrix[x][y] = bfield(x,y,z)

# visualize
plt.figure()
x = range(xwidth)
y = range(ywidth)
z = bmatrix[x][y].T
contour(x,y,z,35)
plt.show()

Answer 1

改变

dl = ((coord[i+1][0]-coord[i][0]), (coord[i+1][1]-coord[i][1]), 0)
bfield2 = konstante*I*cross(dl, (rx,ry,rz))/r**3 # i have to use numpy arrays

至

dl = np.array([(coord[i+1][0]-coord[i][0]), (coord[i+1][1]-coord[i][1]), 0])
bfield2 = konstante*I*cross(dl, np.array([rx,ry,rz]))/r**3 # i have to use numpy arrays

我这台机器上没有 Numpy，因此未经测试。基本上，使用 np.array 将元组更改为 numpy 数组.

您也可以单独保留 dl 并将 bfield2 更改为使用 np.array(dl) 而不是 dl.