File:Mpl example Helmoltz coils.svg
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Summary
DescriptionMpl example Helmoltz coils.svg |
English: Cross section of B (magnetic field strength) magnitude in a Helmholtz coil (actually consisting of two coils: one at the top, one at the bottom in the plot). The eight contours are for field magnitudes of 0.5 {\displaystyle B_0}, 0.8 {\displaystyle B_0}, 0.9 {\displaystyle B_0}, 0.95 {\displaystyle B_0}, 0.99 {\displaystyle B_0}, 1.01 {\displaystyle B_0}, 1.05 {\displaystyle B_0}, and 1.1 {\displaystyle B_0}, where {\displaystyle B_0} is field strength at center. The large center area has almost uniform field strength. |
Date | |
Source | Own work |
Author | Adrien F. Vincent |
SVG development InfoField | This plot was created with Matplotlib. |
Rationale: this work aims at providing an up-to-date version of the similar work https://commons.wikimedia.org/wiki/File:Helmholtz_coil,_B_magnitude_cross_section.svg, done by Morn.
Source code has been modified into fully object-oriented matplotlib interface. It now uses the "viridis" colormap, instead of "jet" which produces perceptual glitches. Besides, some changes had to be done to work with versions of numpy more recent than the one originally used.
The matplotlib (mpl) version is 1.5.3, with Python 2.7 and numpy 1.10
##########
## Code for the figure
##########
# -*- coding: utf-8 -*-
from __future__ import division
import matplotlib.pyplot as plt
import numpy as np
from matplotlib.cm import viridis as colormap # future default colormap
"""
Setup
"""
r = 1.0
res = 200 # grid resolution. 100 may be enough, resulting in smaller SVG file)
def dist3(a, b, c, d, e, f):
"""Compute the Euclidian distance from (d, e, f) to (a, b, c),
raised to the 3rd power (and with lower boundary `r`).
"""
return np.maximum(r, np.sqrt((a - d)**2 + (b - e)**2 + (c - f)**2))
x = np.linspace(-150, 150, res)
y = np.linspace(-150, 150, res)
X, Y = np.meshgrid(x, y)
F = np.zeros((res, res, 3))
"""
Computing part
"""
# Loop over two coils
for coils in [1.0, -1.0]:
# Sum field contributions from coil in 10-degree steps
for p in np.arange(0, 360, 10):
xc = 100 * np.sin(np.pi * p / 180.0)
yc = 50 * coils
zc = 100 * np.cos(np.pi * p / 180.0)
MAG = 1.0 / ((r + dist3(X, Y, 0.0, xc, yc, zc))**3)
# (We leave out the necessary constants that would be required
# to get proper units because only scaling behavior will be shown
# in the plot. This is also why a sum instead of an integral
# can be used.)
#
# Due to more stringent casting rules in recent Numpy (>=1.10),
# one builds an explicit list of all the vectors (X - xc, Y - yc, -zc)
# instead of relying on broadcasting. One then reshapes the array Z
# (of the cross-product results) as previously expected.
vectors = np.array([[xval - xc, yval - yc, -zc] for (xval, yval)
in zip(X.reshape(-1), Y.reshape(-1))])
Z = np.cross(vectors, (-zc, 0.0, xc))
Z = Z.reshape(res, res, 3)
F += Z * MAG[:,:,np.newaxis]
# Compute the B-field
B = np.sqrt(F[..., 0]**2 + F[..., 1]**2 + F[..., 2]**2)
# Scale field strength by value at center
B = B / B[res // 2, res // 2]
"""
Plotting part
"""
fig_label = "helmoltz_coils"
plt.close(fig_label)
fig, ax = plt.subplots(figsize=(6, 6), num=fig_label, frameon=False)
levels = (0.5, 0.8, 0.9, 0.95, 0.99, 1.01, 1.05, 1.1)
cs = ax.contour(x, y, B, cmap=colormap, levels=levels)
# Add wire symbols
ax.scatter((100, 100, -100, -100), (50, -50, 50, -50), s=400, color="Black")
ax.axis((-130, 130, -130, 130))
ax.set_xticks([])
ax.set_yticks([])
plt.tight_layout()
plt.show()
fig.savefig("Helmholtz_coil,_B_magnitude_cross_section.svg")
##########
Licensing
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- Under the following conditions:
- attribution – You must give appropriate credit, provide a link to the license, and indicate if changes were made. You may do so in any reasonable manner, but not in any way that suggests the licensor endorses you or your use.
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26 September 2016
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Date/Time | Thumbnail | Dimensions | User | Comment | |
---|---|---|---|---|---|
current | 09:52, 27 September 2016 | 540 × 540 (63 KB) | Adrien F. Vincent | User created page with UploadWizard |
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