# copyright 2000 Oregon State University Capstones in
# Physics: Electromagnetism Philip J. Siemens
#
# MAGNETIC FIELD OF A RING
#
# This Maplesheet is about the magnetic field of a localized static
# current distribution.
#
# In the class notes, we calculate the field far away from the sources.
#
# Here we use numerical methods to view the pattern of the field
# everywhere.
#
> restart: with (linalg): with (plots):
#
# BIOT SAVART LAW
#
# The simplest finite current distribution is a loop of current I
# circulating around a circle of radius R.
#
# From the principle of superposition we know that the magnetic field
# for the loop is just the vector sum of the magnetic fields from each
# part of the ring.
#
# For a current I, the constant of proportionality for the magnetic
# field is k = mu[0]*I/4/pi.
#
# Then the magnetic field dB at a position r due to a segment dL of the
# ring at position R is
#
# dB = k (dL x (r-R))/(r-R)^3
>
#
# SYMMETRY CONSIDERATIONS
#
# By symmetry,
#
# the total magnetic field B depends only on
#
# the distance r of the observer from the center of the ring
# and
# the angle theta of the observer from the ring's axis of symmetry
#
# the direction of B is in the plane which contains the axis
# and r
#
# We will need to use these symmetry considerations to simplify the
# computation.
#
# Otherwise,
#
# the math expressions would get really complicated
# and
# it would take too much time for Maple to calculate the fields
#
# Due to the symmetry of the ring, it is convenient to choose a
# coordinate system in which
#
# the center of the ring is the origin
# and
# the axis of the ring is the z-axis
#
# We will assume that, in the first quadrant, the current is flowing
# from the x-axis toward the y-axis.
#
#
# It will be convenient to switch back and forth between cartesian and
# spherical coordinates.
#
#
# We can start by finding the field at a point r in the x - z plane.
#
# Then we will find the field elsewhere by applying the symmetry
# arguments.
#
# To set up the integral we choose cartesian coordinates,
# because then the unit vectors do not depend on the coordinates.
#
# The observer's position is
#
> r:=[x,0,z];
#
# The position of the current segment on the loop is given by its
# angular coordinate phi:
#
> R:=[a*cos(phi),a*sin(phi),0];
#
#
# With these substitutions, the square of the distance between the
# source and observer is
#
> Rr2:=dotprod(r-R,r-R);
#
# This can be simplified:
#
> Rr2:=simplify(Rr2);
#
#
# The differential segment dL on the loop is proportional to the
# differential dphi ;
# the ratio of proportionality is the vector
#
> dLoverdphi:=a*[-sin(phi),cos(phi),0];
#
#
# The numerator in the Biot-Savart law is, except for a factor dphi ,
#
> numerator:=crossprod(dLoverdphi,(r-R));
#
# This can be simplified:
#
> numerator:=[numerator[1],numerator[2],simplify(numerator[3])];
#
#
# We are now ready to do the integrals.
#
#
# NUMERICAL INTEGRATIONS
#
# The integrals can, after some manipulation, be converted to elliptic
# integrals.
#
# Instead, we will do them numerically.
#
#
# We first define the integrands.
#
> dBx:=k*numerator[1]/Rr2^(3/2);
#
#
> dBz:=k*numerator[3]/Rr2^(3/2);
#
# There is no point in doing the integral for B[y] , because it is zero
# by symmetry.
#
#
# We need to give values to the ring's radius and current. Let's take
# a=1,k=1
#
> dBx:=subs(a=1,k=1,dBx); dBz:=subs(a=1,k=1,dBz);
#
#
#
# Next, we set up the integrals, using the integration routine int.
#
# If we use int in its usual format, Maple will try to do the integral
# analytically.
#
# We can save it this step by capitalizing the first letter.
#
# We can also save time by halving the range of integration from 0 to pi
# , then doubling the answer by doubling the integrand:
#
#
> Bx:=Int(2*dBx,phi=0..Pi); Bz:=Int(2*dBz,phi=0..Pi);
#
# These integrals will take some time for Maple to evaluate numerically.
#
# To save time, we will plot them in the x - z plane. We can do this
# because:
#
# For y=0, the magnetic field only has x and z components.
#
# We can find the fields elsewhere by symmetry.
#
#
#
# To plot the fields, we have to introduce procedures defining the
# integrals.
#
# We will also save time by limiting the integration procedure to 0.1% =
# 10^(-3)
#
#
# ->
#
> Bxp:=(x,z) ->
> evalf(Int(2*cos(phi)*z/(-2*cos(phi)*x+x^2+1+z^2)^(3/2),phi = 0 ..
> Pi),3);
#
> Bzp:=(x,z) ->
> evalf(Int(2*(1-cos(phi)*x)/(-2*cos(phi)*x+x^2+1+z^2)^(3/2),phi = 0 ..
> Pi),3);
#
#
#
# We can save time by plotting the result in just one quadrant. The
# others follow from symmetry.
#
> fieldplot([Bxp,Bzp],0.5..2.5,0.5..2.5,grid=[5,5],arrows=slim);
#
#
>
#
# QUESTIONS FOR HOMEWORK
#
# Question 1.
#
# Sketch the extension of the fieldplot to all 4 quadrants of the xz
# plane.
#
# Question 2.
#
# What goes wrong if we change the limits of the plot to include the x
# and z axes, for example to 0..2,0..2 ? Explain, both physically and
# numerically.
#
# Question 3.
#
# Choose a larger scale, 0.5..5.5,0.5..5.5 .
# Multiply the fields by an appropriate power of r so that the
# magnitudes of the field vectors aren't falling off with increasing r.
# What is the power you need? Explain. Show an output graph which
# illustrates your answer. What is happening at small r? Why is this
# okay?
#
#
#
#
# version 17 October 2001