Python：使用比奥-萨伐尔定律计算电线磁场
我想用比奥-萨伐尔定律来计算一根电线的磁场。有些人建议我使用numpy数组。最开始我用vpython做的，效果还不错。但现在我想用Matplotlib来可视化，所以我需要数组，对吧？可是我现在卡住了。
我也把这个问题发到codereview上，但他们让我来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()
编辑 No. 7：我删除了其他的编辑。我不想让大家困惑。输出结果不正确。请看下一个编辑。
# 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：我最终没有使用numpy就解决了。下面这个版本可以正常工作。
# 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()
4 个回答
1
你可以这样做：
bfield3 = 0
也许你应该试试这样：
bfield3 = np.zeros((len(...),len(...)))
或者说，bfield3可能已经分配过了？你只是想把所有的值都设为零？那就这样做：
bfield3[:,:] = 0
2
这不是对你最初问题的回答，只是给你一个关于如何使用 numpy 数组的小提示：
In []: 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)]
In []: coord= np.array(coord).T
In []: coord
Out[]:
array([[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 9, 9, 9, 9, 9, 9, 9, 9],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 3, 4, 5, 6, 7, 8]])
In [170]: wiremiddle= (coord[:, 1:]- coord[:, :-1])/ 2.
In []: wiremiddle
Out[]:
array([[ 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0. , 0. , 0. , 0. , 0. , 0. , 0. , 0. ],
[ 0. , 0. , 0. , 0. , 0. , 0. , 0. , 0. , 0. , 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5]])
希望这能帮助你重写你的代码。
2
更改
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
为
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，所以这个没有经过测试。简单来说，就是把你的元组转换成Numpy数组，使用np.array这个函数。
你也可以不动dl，直接把bfield2改成用np.array(dl)来代替dl。