# CHAPTER V.
## PARTICULAR PROBLEMS IN MAGNETIC INDUCTION.
### A Hollow Spherical Shell.

431.] The first example of the complete solution of a problem in magnetic induction was that given by Poisson for the case of a hollow spherical shell acted on by any magnetic forces whatever.
For simplicity we shall suppose the origin of the magnetic forces to be in the space outside the shell.
If V denotes the potential due to the external magnetic system, we may expand V in a series of solid harmonics of the form
| $V=C_{0}S_{0}+C_{1}S_{1}r+{\And }\!{c.}+C_{i}S_{i}r^{i}.$ | (1) |
wherer is the distance from the centre of the shell, S$_i$ is a surface harmonic of order i, and C$_i$ is a coefficient.
This series will be convergent provided r is less than the distance of the nearest magnet of the system which produces this potential. Hence, for the hollow spherical shell and the space within it, this expansion is convergent.
Let the external radius of the shell be a$_2$ and the inner radius a$_1$ and let the potential due to its induced magnetism be Ω. The form of the function Ω will in general be different in the hollow space, in the substance of the shell, and in the space beyond. If we expand these functions in harmonic series, then, confining our attention to those terms which involve the surface harmonic S$_i$, we shall find that if Ω$_1$ is that which corresponds to the hollow space within the shell, the expansion of Ω$_1$ must be in positive harmonics of the form A$_1$S$_i$r$^i$, because the potential must not become infinite within the sphere whose radius is aa$_1$.
In the substance of the shell, where r$_1$ lies between a$_1$ and a$_2$, the series may contain both positive and negative powers of r, of the form
* $A_{2}S_{i}r^{i}+B_{2}S_{i}r^{-(i+1)}.\,$
Outside the shell, where r is greater than a$_2$, since the series ​must be convergent however great r may be, we must have only negative powers of r, of the form
* $B_{3}S_{i}r^{-(i+1)}.\,$
The conditions which must be satisfied by the function Ω, are: It must be (1) finite, and (2) continuous, and (3) must vanish at an infinite distance, and it must (4) everywhere satisfy Laplace's equation.
On account of (1) B$_1$ = 0.
On account of (2) when r = a$_1$,
| $(A_{1}-A_{2})a_{1}^{2i+1}-B_{2}=0,\,$ | (2) |
and when r = a$_2$,
| $(A_{2}-A_{3})a_{2}^{2i+1}+B_{2}-B_{3}=0.\,$ | (3) |
On account of (3) A$_3$ = 0, and the condition (4) is satisfied everywhere, since the functions are harmonic.
But, besides these, there are other conditions to be satisfied at the inner and outer surface in virtue of equation (10), Art. 427.
At the inner surface where r = a$_1$,
| $(1+4\pi \kappa ){\frac {d\Omega _{2}}{dr}}-{\frac {d\Omega _{1}}{dr}}+4\pi \kappa {\frac {dV}{dr}}=0,$ | (4) |
and at the outer surface where r = a$_2$,
| $-(1+4\pi \kappa ){\frac {d\Omega _{2}}{dr}}+{\frac {d\Omega _{3}}{dr}}-4\pi \kappa {\frac {dV}{dr}}=0.$ | (5) |
From these conditions we obtain the equations
| $(1+4\pi \kappa )(iA_{2}a_{1}^{2i+1}-(i+1)B_{2})-iA_{1}a_{1}^{2i+1}+4\pi \kappa iC_{i}a_{1}^{2i+1}=0,$ | (6) |
| $(1+4\pi \kappa )(iA_{2}a_{2}^{2i+1}-(i+1)B_{2})+(i+1)B_{3}+4\pi \kappa iC_{i}a_{2}^{2i+1}=0,$ | (7) |
and if we put
| $N_{i}={\frac {1}{(1+4\pi \kappa )(2i+1)^{2}+(4\pi \kappa )^{2}i(i+1)\left(1-\left({\frac {a_{1}}{a_{2}}}\right)^{2i+1}\right)}},$ | (8) |
we find
| $A_{1}=-(4\pi \kappa )^{2}i(i+1)\left(1-\left({\frac {a_{1}}{a_{2}}}\right)^{2i+1}\right)N_{i}C_{i},$ | (9) |
| $A_{2}=-4\pi \kappa i\left[2i+1+4\pi \kappa (i+1)\left(1-\left({\frac {a_{1}}{a_{2}}}\right)^{2i+1}\right)\right]N_{i}C_{i},$ | (10) |
| $B_{2}=4\pi \kappa i(2i+2)a_{1}^{2i+1}N_{i}C_{i},$ | (11) |
| $B_{3}=4\pi \kappa i(2i+1+4\pi \kappa (i+1))(a_{2}^{2i+1}-a_{1}^{2i+1})N_{i}C_{i}.$ | (12) |
These quantities being substituted in the harmonic expansions give the part of the potential due to the magnetization of the shell. The quantity N$_i$ is always positive, since 1+4πκ can never be negative. Hence A$_1$ is always negative, or in other words, the ​action of the magnetized shell on a point within it is always opposed to that of the external magnetic force whether the shell be paramagnetic or diamagnetic. The actual value of the resultant potential within the shell is
* $(c_{i}+A_{1})S_{i}r^{i},\,$
| or | $(1+4\pi \kappa )(2i+1)^{3}N_{i}C_{i}S_{i}r.\,$ | (13) |
432.] When κ is a large number, as it is in the case of soft iron, then, unless the shell is very thin, the magnetic force within it is but a small fraction of the external force.
In this way Sir W. Thomson has rendered his marine galvanometer independent of external magnetic force by enclosing it in a tube of soft iron.
433.] The case of greatest practical importance is that in which i = 1. In this case
| $N_{1}={\frac {1}{9(1+4\pi \kappa )+2(4\pi \kappa )^{2}\left(1-\left({\frac {a_{1}}{a_{2}}}\right)^{3}\right)}},$ | (14) |
| ${\begin{aligned}A_{1}&=-2(4\pi \kappa )^{2}\left(1-\left({\frac {a_{1}}{a_{2}}}\right)^{3}\right)N_{1}C_{1},\\A_{2}&=-4\pi \kappa \left[3+8\pi \kappa \left(1-\left({\frac {a_{1}}{a_{2}}}\right)^{3}\right)\right]N_{1}C_{1},\\B_{2}&=12\pi \kappa a_{1}^{3}N_{1}C_{1},\\B_{3}&=4\pi \kappa (3+8\pi \kappa )(a_{2}^{3}-a_{1}^{3})N_{1}C_{1}.\end{aligned}}$ | (15) |
The magnetic force within the hollow shell is in this case uniform and equal to
| $C_{1}+A_{1}={\frac {9(1+4\pi \kappa )}{9(1+4\pi \kappa )+2(4\pi \kappa )^{2}\left(1-\left({\frac {a_{1}}{a_{2}}}\right)^{3}\right)}}C_{1}.$ | (16) |
If we wish to determine κ by measuring the magnetic force within a hollow shell and comparing it with the external magnetic force, the best value of the thickness of the shell may be found from the equation
| $1-{\frac {a_{1}^{3}}{a_{2}^{3}}}={\frac {9}{2}}{\frac {1+4\pi \kappa }{(4\pi \kappa )^{2}}}.$ | (17) |
The magnetic force inside the shell is then half of its value outside.
Since, in the case of iron, κ is a number between 20 and 30, the thickness of the shell ought to be about the hundredth part of its radius. This method is applicable only when the value of κ is large. When it is very small the value of A$_l$ becomes insensible, since it depends on the square of κ. ​For a nearly solid sphere with a very small spherical hollow,
| ${\begin{aligned}A_{1}&=-{\frac {2(4\pi \kappa )^{2}}{(3+4\pi \kappa )(3+8\pi \kappa )}}C_{1},\\A_{2}&=-{\frac {4\pi \kappa }{3+4\pi \kappa }}C_{1},\\B_{3}&={\frac {4\pi \kappa }{3+4\pi \kappa }}C_{1}a_{2}^{3}.\end{aligned}}$ | (18) |
The whole of this investigation might have been deduced directly from that of conduction through a spherical shell, as given in Art. 312, by putting k$_1$ = (1 + 4πκ)k$_2$ in the expressions there given, remembering that A$_l$ and A$_2$ in the problem of conduction are equivalent to C$_l$ + A$_l$ and C$_l$ + A$_2$ in the problem of magnetic induction.
434.] The corresponding solution in two dimensions is graphically represented in Fig. XV, at the end of this volume. The lines of induction, which at a distance from the centre of the figure are nearly horizontal, are represented as disturbed by a cylindric rod magnetized transversely and placed in its position of stable equilibrium. The lines which cut this system at right angles represent the equipotential surfaces, one of which is a cylinder. The large dotted circle represents the section of a cylinder of a paramagnetic substance, and the dotted horizontal straight lines within it, which are continuous with the external lines of induction, represent the lines of induction within the substance. The dotted vertical lines represent the internal equipotential surfaces, and are continuous with the external system. It will be observed that the lines of induction are drawn nearer together within the substance, and the equipotential surfaces are separated farther apart by the paramagnetic cylinder, which, in the language of Faraday, conducts the lines of induction better than the surrounding medium.
If we consider the system of vertical lines as lines of induction, and the horizontal system as equipotential surfaces, we have, in the first place, the case of a cylinder magnetized transversely and placed in the position of unstable equilibrium among the lines of force, which it causes to diverge. In the second place, considering the large dotted circle as the section of a diamagnetic cylinder, the dotted straight lines within it, together with the lines external to it, represent the effect of a diamagnetic substance in separating

### Case of a Sphere in which the Coefficients of Magnetization are Different in Different Directions.

435.] Let α, β, γ be the components of magnetic force, andA, B, C those of the magnetization at any point, then the most general linear relation between these quantities is given by the equations
| ${\begin{aligned}A&=r_{1}\alpha +p_{3}\beta +q_{2}\gamma ,\\B&=q_{3}\alpha +r_{2}\beta +p_{1}\gamma ,\\C&=p_{2}\alpha +q_{1}\beta +r_{3}\gamma ,\\\end{aligned}}$ | (1) |
where the coefficients r, p, q are the nine coefficients of magnetization.
Let us now suppose that these are the conditions of magnetization within a sphere of radius a, and that the magnetization at every point of the substance is uniform and in the same direction, having the components A, B, C.
Let us also suppose that the external magnetizing force is also uniform and parallel to one direction, and has for its components X, Y, Z.
The value of V is therefore
| $V=-(Xx+Yy+Zz),\,$ | (2) |
and that of Ω' the potential of the magnetization outside the sphere is
| $\Omega '=(Ax+By+Cz){\frac {4\pi a^{3}}{3r^{3}}}.$ | (3) |
The value of Ω, the potential of the magnetization within the sphere, is
| $\Omega ={\frac {4\pi }{3}}(Ax+By+Cz).$ | (4) |
The actual potential within the sphere is V + Ω, so that we shall have for the components of the magnetic force within the sphere
| ${\begin{matrix}\alpha =X-{\frac {4}{3}}\pi A,\\\beta =Y-{\frac {4}{3}}\pi B,\\\gamma =Z-{\frac {4}{3}}\pi C.\end{matrix}}$ | (5) |
Hence
| ${\begin{aligned}(1+{\frac {4}{3}}\pi r_{1})A+{\frac {4}{3}}\pi p_{3}B+{\frac {4}{3}}\pi q_{2}C&=r_{1}X+p_{3}Y+q_{2}Z,\\{\frac {4}{3}}\pi q_{3}A+(1+{\frac {4}{3}}\pi r_{2})B+{\frac {4}{3}}\pi p_{1}C&=q_{3}X+r_{2}Y+p_{1}Z,\\{\frac {4}{3}}\pi p_{2})A+{\frac {4}{3}}\pi q_{1}B+(1+{\frac {4}{3}}\pi r_{3}C&=p_{2}X+q_{1}Y+r_{3}Z.\end{aligned}}$ | (6) |
Solving these equations, we find
| ${\begin{aligned}A=r_{1}'X+p_{3}'Y+q_{2}'Z,\,\\B=q_{3}'X+r_{2}'Y+p_{1}'Z,\,\\C=p_{2}'X+q_{1}'Y+r_{3}'Z,\,\end{aligned}}$ | (7) |
​where
| ${\begin{aligned}&D'r_{1}'=r_{1}+{\frac {4}{3}}\pi (r_{3}r_{1}-p_{2}q_{2}+r_{1}r_{2}-p_{3}q_{3})+\left({\frac {4}{3}}\pi \right)^{2}D,\\&D'p_{1}'=p_{1}-{\frac {4}{3}}\pi (q_{2}q_{3}-p_{1}r_{1}),\\&D'q_{1}'=q_{1}-{\frac {4}{3}}\pi (p_{2}p_{3}-q_{1}r_{1}),\\&{\And }c.,\end{aligned}}$ | (8) |
where D is the determinant of the coefficients on the right side of equations (6), and D' that of the coefficients on the left.
The new system of coefficients p', q', r' will be symmetrical only when the system p, q, r is symmetrical, that is, when the coefficients of the form p are equal to the corresponding ones of the form q.
436.] The moment of the couple tending to turn the sphere about the axis of x from y towards z is
| ${\begin{aligned}L&={\frac {4}{3}}\pi a^{3}(ZB-YC)\\&={\frac {4}{3}}\pi a^{3}\left[p_{1}'Z^{2}-q_{1}'Y^{2}+(r_{2}'-r_{3}')YZ+X(q_{3}'Z-p_{2}'Y)\right].\end{aligned}}$ | (9) |
If we make
* $X=0,\quad Y=F\cos \theta ,\quad Z=F\sin \theta ,$
this corresponds to a magnetic force F in the plane of yz, and inclined to y at an angle θ. If we now turn the sphere while this force remains constant the work done in turning the sphere will be $\int _{0}^{2\pi }{L\,d\theta }$ in each complete revolution. But this is equal to
| ${\frac {2}{3}}\pi a^{3}F^{2}(p_{1}'-q_{1}').$ | )10 |
Hence, in order that the revolving sphere may not become an inexhaustible source of energy, p$_1$' = q$_1$', and similarly q$_1$' = q$_1$'.
These conditions shew that in the original equations the coefficient of B in the third equation is equal to that of C in the second, and so on. Hence, the system of equations is symmetrical, and the equations become when referred to the principal axes of magnetization,
| ${\begin{aligned}A&={\frac {r_{1}}{1+{\frac {4}{3}}\pi r_{1}}}X,\\B&={\frac {r_{2}}{1+{\frac {4}{3}}\pi r_{2}}}Y,\\C&={\frac {r_{3}}{1+{\frac {4}{3}}\pi r_{3}}}Z.\end{aligned}}$ | (11) |
The moment of the couple tending to turn the sphere round the axis of x is
| $L={\frac {4}{3}}\pi a^{3}{\frac {r_{2}-r_{3}}{(1+{\frac {4}{3}}\pi r_{2})(1+{\frac {4}{3}}\pi r_{3})}}YZ.$ | (12) |
​In most cases the differences between the coefficients of magnet ization in different directions are very small, so that we may put
| $L={\frac {4}{3}}\pi a^{3}{\frac {r_{2}-r_{3}}{(1+{\frac {4}{3}}\pi r)^{2}}}F^{2}\sin 2\theta .$ | (13) |
This is the force tending to turn a crystalline sphere about the axis of x from y towards z. It always tends to place the axis of greatest magnetic coefficient (or least diamagnetic coefficient) parallel to the line of magnetic force.
The corresponding case in two dimensions is represented in Fig. XVI.
If we suppose the upper side of the figure to be towards the north, the figure represents the lines of force and equipotential surfaces as disturbed by a transversely magnetized cylinder placed with the north side eastwards. The resultant force tends to turn the cylinder from east to north. The large dotted circle represents a section of a cylinder of a crystalline substance which has a larger coefficient of induction along an axis from north-east to south-west than along an axis from north-west to south-east. The dotted lines within the circle represent the lines of induction and the equipotential surfaces, which in this case are not at right angles to each other. The resultant force on the cylinder is evidently to turn it from east to north.
437.] The case of an ellipsoid placed in a field of uniform and parallel magnetic force has been solved in a very ingenious manner by Poisson.
If V is the potential at the point (x, y, z) due to the gravitation of a body of any form of uniform density ρ, then $-{\frac {dV}{dx}}$ is the potential of the magnetism of the same body if uniformly magnetized in the direction of x with the intensity I = ρ.
For the value of $-{\frac {dV}{dx}}\delta x$ at any point is the excess of the value of V, the potential of the body, above V', the value of the potential when the body is moved –δx in the direction of x.
If we supposed the body shifted through the distance –δx, and its density changed from ρ to –ρ (that is to say, made of repulsive instead of attractive matter,) then $-{\frac {dV}{dx}}\delta x$ would be the potential due to the two bodies.
Now consider any elementary portion of the body containing a volume δv. Its quantity is ρ δv, and corresponding to it there is ​an element of the shifted body whose quantity is –ρ δv at a distance –δx. The effect of these two elements is equivalent to that of a magnet of strength –r and length –δx. The intensity of magnetization is found by dividing the magnetic moment of an element by its volume. The result is δx.
Hence $-{\frac {dV}{dx}}\delta x$ is the magnetic potential of the body magnetized with the intensity δx in the direction of x, and $-{\frac {dV}{dx}}$ is that of the body magnetized with intensity ρ.
This potential may be also considered in another light. The body was shifted through the distance –δx and made of density –ρ. Throughout that part of space common to the body in its two positions the density is zero, for, as far as attraction is con cerned, the two equal and opposite densities annihilate each other. There remains therefore a shell of positive matter on one side and of negative matter on the other, and we may regard the resultant potential as due to these. The thickness of the shell at a point where the normal drawn outwards makes an angle e with the axis of x is δx cos ε and its density is ρ cos ε. The surface-density is therefore ρ δx cos ε, and, in the case in which the potential is $-{\frac {dV}{dx}}$, the surface-density is ρ cos ε.
In this way we can find the magnetic potential of any body uniformly magnetized parallel to a given direction. Now if this uniform magnetization is due to magnetic induction, the magnetizing force at all points within the body must also be uniform and parallel.
This force consists of two parts, one due to external causes, and the other due to the magnetization of the body. If therefore the external magnetic force is uniform and parallel, the magnetic force due to the magnetization must also be uniform and parallel for all points within the body.
Hence, in order that this method may lead to a solution of the problem of magnetic induction, $-{\frac {dV}{dx}}$ must be a linear function of the coordinates x, y, z within the body, and therefore V must be a quadratic function of the coordinates.
Now the only cases with which we are acquainted in which V is a quadratic function of the coordinates within the body are those in which the body is bounded by a complete surface of the second degree, and the only case in which such a body is of finite ​dimensions, is when it is an ellipsoid. We shall therefore apply the method to the case of an ellipsoid.
| Let | ${\frac {x^{2}}{a^{2}}}+{\frac {y^{2}}{b^{2}}}+{\frac {z^{2}}{c^{2}}}=1$ | (1) |
be the equation of the ellipsoid, and let Φ$_0$ denote the definite integral
| $\int _{0}^{\infty }{\frac {d(\phi ^{2})}{\sqrt {(a^{2}+\phi ^{2})(b^{2}+\phi ^{2})(c^{2}+\phi ^{2})}}}.$ | (2) |
Then if we make
| $L=2\pi abc{\frac {d\Phi _{0}}{d(a^{2})}},\quad M=2\pi abc{\frac {d\Phi _{0}}{d(b^{2})}},\quad N=2\pi abc{\frac {d\Phi _{0}}{d(c^{2})}},$ | (3) |
the value of the potential within the ellipsoid will be
| $V_{0}=-{\frac {\rho }{2}}(Lx^{2}+My^{2}+Nz^{2})+{\text{const.}}$ | (4) |
If the ellipsoid is magnetized with uniform intensity I in a direction making angles whose cosines are l, m, n with the axes of x, y, z, so that the components of magnetization are
* $A=Il,\quad B=Im,\quad C=In,$
the potential due to this magnetization within the ellipsoid will be
| $\Omega =-I(Llx+Mmy+Nnz).\,$ | (5) |
If the external magnetizing force is ${\mathfrak {H}}$, and if its components are α, β, γ, its potential will be
| $V=Xx+Yy+Zz.\,$ | (6) |
The components of the actual magnetizing force at any point within the body are therefore
| $X-AL,\quad Y-BM,\quad Z-CN.\,$ | (7) |
The most general relations between the magnetization and the magnetizing force are given by three linear equations, involving nine coefficients. It is necessary, however, in order to fulfil the condition of the conservation of energy, that in the case of magnetic induction three of these should be equal respectively to other three, so that we should have
| ${\begin{aligned}A&=K_{1}(X-AL)+K_{3}'(Y-BM)+K_{2}'(Z-CN),\\B&=K_{3}'(X-AL)+K_{2}(Y-BM)+K_{1}'(Z-CN),\\C&=K_{2}'(X-AL)+K_{1}'(Y-BM)+K_{3}(Z-CN).\end{aligned}}$ | (8) |
From these equations we may determine A, B and C in terms of X, Y. Z, and this will give the most general solution of the problem.
The potential outside the ellipsoid will then be that due to the ​magnetization of the ellipsoid together with that due to the external magnetic force.
438.] The only case of practical importance is that in which
| $\kappa '_{1}=\kappa '_{2}=\kappa '_{3}=0.$ | (9) |
We have then
| ${\begin{aligned}A={\frac {\kappa _{1}}{1+\kappa _{1}L}}X,\\B={\frac {\kappa _{2}}{1+\kappa _{2}M}}Y,\\C={\frac {\kappa _{3}}{1+\kappa _{3}N}}Z.\end{aligned}}$ | (10) |
If the ellipsoid has two axes equal, and is of the planetary or flattened form,
| $b=c={\frac {a}{\sqrt {1-e^{2}}}};$ | (11) |
| ${\begin{aligned}L&=4\pi \left({\frac {1}{e^{2}}}-{\frac {\sqrt {1-e^{2}}}{e^{3}}}\sin ^{-1}e\right),\\M=N&=2\pi \left({\frac {\sqrt {1-e^{2}}}{e^{3}}}\sin ^{-1}e-{\frac {1-e^{2}}{e^{2}}}\right).\end{aligned}}$ | (12) |
If the ellipsoid is of the ovary or elongated form
| $a=b={\sqrt {1-e^{2}}}c;\,$ | (13) |
| ${\begin{aligned}L=M&=2\pi \left({\frac {1}{e^{2}}}-{\frac {1-e^{2}}{2e^{2}}}\log {\frac {1+e}{1-e}}\right),\\N&=4\pi \left({\frac {1}{e^{2}}}-1\right)\left({\frac {1}{2e}}\log {\frac {1+e}{1-e}}-1\right).\end{aligned}}$ | (14) |
In the case of a sphere, when e = 0,
| $L=M=N={\frac {4}{3}}\pi .$ | (15) |
In the case of a very flattened planetoid L becomes in the limit equal to 4π, and M and N become $\pi ^{2}{\frac {a}{c}}$.
In the case of a very elongated ovoid L and M approximate to the value 2π, while N approximates to the form
* $4\pi {\frac {a^{2}}{c^{2}}}\left(\log {\frac {2c}{a}}-1\right)$
and vanishes when e = 1.
It appears from these results that—
(1) When κ, the coefficient of magnetization, is very small, whether positive or negative, the induced magnetization is nearly equal to the magnetizing force multiplied by κ, and is almost independent of the form of the body. ​(2) When κ is a large positive quantity, the magnetization depends principally on the form of the body, and is almost independent of the precise value of κ, except in the case of a longitudinal force acting on an ovoid so elongated that Nκ is a small quantity though κ is large.
(3) If the value of κ could be negative and equal to ${\frac {1}{4\pi }}$ we should have an infinite value of the magnetization in the case of a magnetizing force acting normally to a flat plate or disk. The absurdity of this result confirms what we said in Art. 428.
Hence, experiments to determine the value of κ may be made on bodies of any form provided κ is very small, as it is in the case of all diamagnetic bodies, and all magnetic bodies except iron, nickel, and cobalt.
If, however, as in the case of iron, κ is a large number, experiments made on spheres or flattened figures are not suitable to determine κ; for instance, in the case of a sphere the ratio of the magnetization to the magnetizing force is as 1 to 4.22 if κ = 30, as it is in some kinds of iron, and if K were infinite the ratio would be as 1 to 4.19, so that a very small error in the determination of the magnetization would introduce a very large one in the value of κ.
But if we make use of a piece of iron in the form of a very elongated ovoid, then, as long as Nκ is of moderate value com pared with unity, we may deduce the value of κ from a determination of the magnetization, and the smaller the value of N the more accurate will be the value of κ.
In fact, if Nκ be made small enough, a small error in the value of N itself will not introduce much error, so that we may use any elongated body, such as a wire or long rod, instead of an ovoid.
We must remember, however, that it is only when the product Nκ is small compared with unity that this substitution is allowable. In fact the distribution of magnetism on a long cylinder with flat ends does not resemble that on a long ovoid, for the free mag netism is very much concentrated towards the ends of the cylinder, whereas it varies directly as the distance from the equator in the case of the ovoid.
The distribution of electricity on a cylinder, however, is really comparable with that on an ovoid, as we have already seen, Art. 152.
​These results also enable us to understand why the magnetic moment of a permanent magnet can be made so much greater when the magnet has an elongated form. If we were to magnetize a disk with intensityI in a direction normal to its surface, and then leave it to itself, the interior particles would experience a constant demagnetizing force equal to 4πI, and this, if not sufficient of itself to destroy part of the magnetization, would soon do so if aided by vibrations or changes of temperature.
If we were to magnetize a cylinder transversely the demagnetizing force would be only 2πI.
If the magnet were a sphere the demagnetizing force would be ${\frac {4}{3}}\pi I$.
In a disk magnetized transversely the demagnetizing force is $\pi ^{2}{\frac {a}{c}}I$, and in an elongated ovoid magnetized longitudinally it is least of all, being $4\pi {\frac {a^{2}}{c^{2}}}I\log {\frac {2c}{a}}$.
Hence an elongated magnet is less likely to lose its magnetism than a short thick one.
The moment of the force acting on an ellipsoid having different magnetic coefficients for the three axes which tends to turn it about the axis of x, is
* ${\frac {4}{3}}\pi abc(BZ-CY)={\frac {4}{3}}\pi abcYZ{\frac {\kappa _{3}-\kappa _{2}+\kappa _{2}\kappa _{3}(M-N)}{(1-\kappa _{2}M)(1-\kappa _{3}N)}}.$
Hence, if κ$_2$ and κ$_3$ are small, this force will depend principally on the crystalline quality of the body and not on its shape, provided its dimensions are not very unequal, but if κ$_2$ and κ$_3$ are considerable, as in the case of iron, the force will depend principally on the shape of the body, and it will turn so as to set its longer axis parallel to the lines of force.
If a sufficiently strong, yet uniform, field of magnetic force could be obtained, an elongated isotropic diamagnetic body would also set itself with its longest dimension parallel to the lines of magnetic force.
439.] The question of the distribution of the magnetization of an ellipsoid of revolution under the action of any magnetic forces has been investigated by J. Neumann. Kirchhoff has extended the method to the case of a cylinder of infinite length acted on by any force.
​Green, in the 17th section of his Essay, has given an investigation of the distribution of magnetism in a cylinder of finite length acted on by a uniform external force parallel to its axis. Though some of the steps of this investigation are not very rigorous, it is probable that the result represents roughly the actual magnetization in this most important case. It certainly expresses very fairly the transition from the case of a cylinder for which κ is a large number to that in which it is very small, but it fails entirely in the case in which K is negative, as in diamagnetic substances.
Green finds that the linear density of free magnetism at a distance as from the middle of a cylinder whose radius is a and whose length is 2l, is
* $\lambda =\pi \kappa Xpa{\frac {e^{\frac {px}{a}}-e^{-{\frac {px}{a}}}}{e^{\frac {px}{a}}+e^{-{\frac {px}{a}}}}},$
where p is a numerical quantity to be found from the equation
* $0.231863-2\log _{e}p+2p={\frac {1}{\pi \kappa p^{2}}}.$
The following are a few of the corresponding values of p and κ.
| κ | p | κ | p |
| ∞ | 0 | 1.802 | 0.07 |
| 336.4 | 0.01 | 9.139 | 0.08 |
| 62.02 | 0.02 | 7.517 | 0.09 |
| 48.416 | 0.03 | 6.319 | 0.10 |
| 29.475 | 0.04 | 0.1427 | 1.00 |
| 20.185 | 0.05 | 0.0002 | 10.00 |
| 14.794 | 0.06 | 0.0000 | ∞ |
| | | negative | imaginary |
When the length of the cylinder is great compared with its radius, the whole quantity of free magnetism on either side of the middle of the cylinder is, as it ought to be,
* $M=\pi ^{2}a\kappa X.\,$
Of this ½pM is on the flat end of the cylinder, and the distance of the centre of gravity of the whole quantity M from the end of the cylinder is a/p.
When κ is very small p is large, and nearly the whole free magnetism is on the ends of the cylinder. As κ increases p diminishes, and the free magnetism is spread over a greater distance ​from the ends. When κ is infinite the free magnetism at any point of the cylinder is simply proportional to its distance from the middle point, the distribution being- similar to that of free electricity on a conductor in a field of uniform force.
440.] In all substances except iron, nickel, and cobalt, the coefficient of magnetization is so small that the induced magnetization of the body produces only a very slight alteration of the forces in the magnetic field. We may therefore assume, as a first approximation, that the actual magnetic force within the body is the same as if the body had not been there. The superficial magnetization of the body is therefore, as a first approximation, $-\kappa {\frac {dV}{d\nu }}$, where ${\frac {dV}{d\nu }}$ is the rate of increase of the magnetic potential due to the external magnet along a normal to the surface drawn inwards. If we now calculate the potential due to this superficial distribution, we may use it in proceeding to a second approximation.
To find the mechanical energy due to the distribution of magnetism on this first approximation we must find the surface-integral
* $E=\iint {\kappa V{\frac {dV}{d\nu }}dS}$
taken over the whole surface of the body. Now we have shewn in Art. 100 that this is equal to the volume-integral
* $E=-\iiint {\kappa \left({\frac {\overline {dV}}{dx}}{\Bigg |}^{2}+{\frac {\overline {dV}}{dy}}{\Bigg |}^{2}+{\frac {\overline {dV}}{dz}}{\Bigg |}^{2}\right)dxdydz},$
taken through the whole space occupied by the body, or, if R is the resultant magnetic force,
* $E=-\iiint {\kappa R^{2}dxdydz}.$
Now since the work done by the magnetic force on the body during a displacement δx is Xδx where X is the mechanical force in the direction of x, and since
* $\int {X\delta x}+E={\text{constant}},$
* $X=-{\frac {dE}{dx}}={\frac {d}{dx}}\iiint {\kappa R^{2}dxdydz}=\iiint {\kappa {\frac {d\cdot R^{2}}{dx}}R^{2}dxdydz}$
which shews that the force acting on the body is as if every part of it tended to move from places where R$^2$ is less to places where it is greater with a force which on every unit of volume is
* $\kappa {\frac {d\cdot R^{2}}{dx}}.$
​If κ is negative, as in diamagnetic bodies, this force is, as Faraday first shewed, from stronger to weaker parts of the magnetic field. Most of the actions observed in the case of diamagnetic bodies depend on this property.

### On Ship's Magnetism.

441.] Almost every part of magnetic science finds its use in navigation. The directive action of the earth s magnetism on the compass needle is the only method of ascertaining the ship's course when the sun and stars are hid. The declination of the needle from the true meridian seemed at first to be a hindrance to the application of the compass to navigation, but after this difficulty had been overcome by the construction of magnetic charts it appeared likely that the declination itself would assist the mariner in determining his ship s place.
The greatest difficulty in navigation had always been to ascertain the longitude; but since the declination is different at different points on the same parallel of latitude, an observation of the de clination together with a knowledge of the latitude would enable the mariner to find his position on the magnetic chart.
But in recent times iron is so largely used in the construction of ships that it has become impossible to use the compass at all without taking into account the action of the ship, as a magnetic body, on the needle.
To determine the distribution of magnetism in a mass of iron of any form under the influence of the earth s magnetic force, even though not subjected to mechanical strain or other disturbances, is, as we have seen, a very difficult problem.
In this case, however, the problem is simplified by the following considerations.
The compass is supposed to be placed with its centre at a fixed point of the ship, and so far from any iron that the magnetism of the needle does not induce any perceptible magnetism in the ship. The size of the compass needle is supposed so small that we may regard the magnetic force at any point of the needle as the same.
The iron of the ship is supposed to be of two kinds only.
(1) Hard iron, magnetized in a constant manner.
(2) Soft iron, the magnetization of which is induced by the earth or other magnets.
In strictness we must admit that the hardest iron is not only ​capable of induction but that it may lose part of its so-called permanent magnetization in various ways.
The softest iron is capable of retaining what is called residual magnetization. The actual properties of iron cannot be accurately represented by supposing it compounded of the hard iron and the soft iron above defined. But it has been found that when a ship is acted on only by the earth's magnetic force, and not subjected to any extraordinary stress of weather, the supposition that the magnetism of the ship is due partly to permanent magnetization and partly to induction leads to sufficiently accurate results when applied to the correction of the compass.
The equations on which the theory of the variation of the compass is founded were given by Poisson in the fifth volume of the Mémoires de l'Istitnt, p. 533 (1824).
The only assumption relative to induced magnetism which is involved in these equations is, that if a magnetic force X due to external magnetism produces in the iron of the ship an induced magnetization, and if this induced magnetization exerts on the compass needle a disturbing force whose components are X', Y', Z', then, if the external magnetic force is altered in a given ratio, the components of the disturbing force will be altered in the same ratio.
It is true that when the magnetic force acting on iron is very great the induced magnetization is no longer proportional to the external magnetic force, but this want of proportionality is quite insensible for magnetic forces of the magnitude of those due to the earth's action.
Hence, in practice we may assume that if a magnetic force whose value is unity produces through the intervention of the iron of the ship a disturbing force at the compass needle whose components are a in the direction of x, d in that of y, and g in that of z, the components of the disturbing force due to a force X in the direction of x will be a X, dX, and gX.
If therefore we assume axes fixed in the ship, so that x is towards the ship's head, y to the starboard side, and z towards the keel, and if X, Y, Z represent the components of the earth's magnetic force in these directions, and X', Y', Z' the components of the combined magnetic force of the earth and ship on the compass needle,
| ${\begin{aligned}X'=X+aX+bY+cZ+P,\\Y'=Y+dX+eY+fZ+Q,\\Z'=Z+gX+hY+kZ+R.\end{aligned}}$ | (1) |
​In these equations a, b, c, d, e, f, g, h, k are nine constant coefficients depending on the amount, the arrangement, and the capacity for induction of the soft iron of the ship.
P, Q, and R are constant quantities depending on the permanent magnetization of the ship.
It is evident that these equations are sufficiently general if magnetic induction is a linear function of magnetic force, for they are neither more nor less than the most general expression of a vector as a linear function of another vector.
It may also be shewn that they are not too general, for, by a proper arrangement of iron, any one of the coefficients may be made to vary independently of the others.
Thus, a long thin rod of iron under the action of a longitudinal magnetic force acquires poles, the strength of each of which is numerically equal to the cross section of the rod multiplied by the magnetizing force and by the coefficient of induced magnetization. A magnetic force transverse to the rod produces a much feebler magnetization, the effect of which is almost insensible at a distance of a few diameters.
If a long iron rod be placed fore and aft with one end at a distance x from the compass needle, measured towards the ship's head, then, if the section of the rod is A, and its coefficient of magnetization κ, the strength of the pole will be AκX, and, if $A={\frac {ax^{2}}{\kappa }}$, the force exerted by this pole on the compass needle will be aX. The rod may be supposed so long that the effect of the other pole on the compass may be neglected.
We have thus obtained the means of giving any required value to the coefficient a.
If we place another rod of section B with one extremity at the same point, distant x from the compass toward the head of the vessel, and extending to starboard to such a distance that the distant pole produces no sensible effect on the compass, the disturbing force due to this rod will be in the direction of x, and equal to ${\frac {B\kappa Y}{x^{2}}}$, or if $B={\frac {bx^{2}}{\kappa }}$, the force will be bY.
This rod therefore introduces the coefficient b.
A third rod extending downwards from the same point will introduce the coefficient c.
The coefficients d, e, f may be produced by three rods extending to head, to starboard, and downward from a point to starboard of ​the compass, and g, h, k by three rods in parallel directions from a point below the compass.
Hence each of the nine coefficients can be separately varied by means of iron rods properly placed.
The quantities P, Q, R are simply the components of the force on the compass arising from the permanent magnetization of the ship together with that part of the induced magnetization which is due to the action of this permanent magnetization.
A complete discussion of the equations (1), and of the relation between the true magnetic course of the ship and the course as indicated by the compass, is given by Mr. Archibald Smith in the Admiralty Manual of the Deviation of the Compass.
A valuable graphic method of investigating the problem is there given. Taking a fixed point as origin, a line is drawn from this point representing in direction and magnitude the horizontal part of the actual magnetic force on the compass-needle. As the ship is swung round so as to bring her head into different azimuths in succession, the extremity of this line describes a curve, each point of which corresponds to a particular azimuth.
Such a curve, by means of which the direction and magnitude of the force on the compass is given in terms of the magnetic course of the ship, is called a Dygogram.
There are two varieties of the Dygogram. In the first, the curve is traced on a plane fixed in space as the ship turns round. In the second kind, the curve is traced on a plane fixed with respect to the ship.
The dygogram of the first kind is the Limaçon of Pascal, that of the second kind is an ellipse. For the construction and use of these curves, and for many theorems as interesting to the mathematician as they are important to the navigator, the reader is referred to the Admiralty Manual of the Deviation of the Compass.
* ↑ See Thomson and Tait's Natural Philosophy, §522.
* ↑ Crelle, bd. xxxvii (1848).
* ↑ Crelle, bd. xlviii (1854.