Magnetic Field of a Hollow Cylinder

© 1998 Waterloo Maple Inc.

NOTE: This worksheet shows how Maple can be used to determine the magnetic screening provided by a hollow cylinder.

Introduction

restart;

A ferromagnetic tube can be used to shield its interior from static magnetic fields. To calculate the screening constant, the tube is presumed to be set up so that its axis is perpendicular to the direction of an outer homogeneous magnetic field H0 [A/m]. The outer radius of the tube is Ro [m], and the inner radius is Ri [m].

Theory

Under the influence of the outside magnetic field, the ferromagnetic material is magnetized. The resulting magnetic field can be calculated as a superposition of the outside magnetic field and the fields of magnetic line dipoles. The free constants in the equations are calculated with the help of boundary conditions.
The magnetic scalar potential V [A] is defined by H =-grad( V ).
The scalar magnetic potential of a homogeneous magnetic field in polar coordinates (r [m], [rad]) is given by
(characteristic constants A [A], B [A/m]).

The magnetic field of a line dipole is given by
(characteristic constant C [A*m]).

Solution

There are three equations that represent the superposition of a homogeneous magnetic scalar potential and the scalar potential of a line dipole: one for the outside, a second for the wall and a third for the inside of the tube.

The permeabitiy of the material is defined as

( : permeability of vacuum [H/m], mur: relative permeability [-])

Outside:

Wall:

Inside:

On the inside, the potential must be finite so that C3 = 0. On the axis where r = 0 and where , the potential has the value V3 = V2 = V1 = V0 = A1 = A2 = A3.

Choose V0 = 0. The result on the inside is a homogeneous magnetic field with a field intensity of Hi, so that B3 = Hi.

A3:=0: A2:=0: A1:=0: C3:=0: B3:=Hi:

On the outside, the magnetic field in very big distances is undisturbed by the magnetic field caused by the tube, so that

~ .

This results in B1 = H0.

B1:=H0:

The conditions at the interfaces are that the tangential components of the vector H and the normal components of the vector B (B = *H) are continuous.

From in cylinder coordinates, one obtains the following equations:

with(linalg):
Warning, new definition for norm
Warning, new definition for trace

h1:=grad(V1, v, coords=cylindrical);

h2:=grad(V2, v, coords=cylindrical);

h3:=grad(V3,v,coords=cylindrical);

At r = Ri, the tangential components of H must be continuous (component of h1, h2 and h3 with 'd/d( )').

eqHi:=simplify(evalf(subs(r=Ri,h2[2]))=evalf(subs(r=Ri,h3[2])));

At r = Ro, the tangential components of H must be continuous.

eqHa:=simplify(evalf(subs(r=Ro,h1[2]))=evalf(subs(r=Ro,h2[2])));

At r = Ri, the normal components of B must be continuous (component of h1, h2 and h3 with 'd/d(r)').

eqBi:=simplify(evalf(subs(r=Ri,mu*h2[1]))=evalf(subs(r=Ri,mu0*h3[1])));

At r = Ro, the normal components of B must be continuous.

eqBa:=simplify(evalf(subs(r=Ro,mu0*h1[1]))=evalf(subs(r=Ro,mu*h2[1])));

Solving these equations for C1, C2, B2 and Hi gives

sol:=simplify(solve({eqHi,eqHa,eqBi,eqBa},{C1,C2,B2,Hi}));

assign(sol);

The result for the screening factor 'Screen' (= Hi/H0) is then given by

Screen:=simplify(Hi/H0);

Graphical Display of the Solution

For the plots, values are assigned to the parameters. The relative permeability for ferromagnetic materials is a very small value of mur = 6.

Ri:=1: Ro:=2: H0:=10: mur:=6:

For the contour plot, the plotted function must be written in cartesian coordinates so the radius r and the angle phi are substituted by x and y.

The plot a piecewise function is defined.

Now the equipotential lines can be plotted. In addition, two circles are included so that the location of the tube is indicated.

with(plottools):
c1 := circle([0,0], Ri, color=red): c2 := circle([0,0], Ro, color=red): p:=plots[contourplot](Pot,x=-3..3,y=-3..3,scaling=CONSTRAINED, color=blue,numpoints=4000):
plots[display](c1,c2,p);

In this case, the screening factor is about 0.56, so that the magnetic field inside the tube is still more than half of the value on the outside.

Screen;

Summary

For some applications, magnetic fields need to be shielded. In the preceeding calculation, an illustrative example is given on how this can be accomplished. A ferromagnetic tube (with a very low relative permeability) is placed in a static magnetic field and the screening factor is calculated.