# CHAPTER II.
## MAGNETIC FORCE AND MAGNETIC INDUCTION.

395.] We have already (Art. 386) determined the magnetic potential at a given point due to a magnet, the magnetization of which is given at every point of its substance, and we have shewn that the mathematical result may be expressed either in terms of the actual magnetization of every element of the magnet, or in terms of an imaginary distribution of magnetic matter, partly condensed on the surface of the magnet and partly diffused through out its substance.
The magnetic potential, as thus defined, is found by the same mathematical process, whether the given point is outside the magnet or within it. The force exerted on a unit magnetic pole placed at any point outside the magnet is deduced from the potential by the same process of differentiation as in the corresponding electrical problem. If the components of this force are α, β, γ,
| $\alpha =-{\frac {dV}{dx}},\quad \beta =-{\frac {dV}{dy}},\quad \gamma =-{\frac {dV}{dz}}.$ | (1) |
To determine by experiment the magnetic force at a point within the magnet we must begin by removing part of the magnetized substance, so as to form a cavity within which we are to place the magnetic pole. The force acting on the pole will depend, in general, in the form of this cavity, and on the inclination of the walls of the cavity to the direction of magnetization. Hence it is necessary, in order to avoid ambiguity in speaking of the magnetic force within a magnet, to specify the form and position of the cavity within which the force is to be measured. It is manifest that when the form and position of the cavity is specified, the point within it at which the magnetic pole is placed must be regarded as ​no longer within the substance of the magnet, and therefore the ordinary methods of determining the force become at once applicable.
396.] Let us now consider a portion of a magnet in which the direction and intensity of the magnetization are uniform. Within this portion let a cavity be hollowed out in the form of a cylinder, the axis of which is parallel to the direction of magnetization, and let a magnetic pole of unit strength be placed at the middle point of the axis.
Since the generating lines of this cylinder are in the direction of magnetization, there will be no superficial distribution of magnetism on the curved surface, and since the circular ends of the cylinder are perpendicular to the direction of magnetization, there will be a uniform superficial distribution, of which the surface- density is Ifor the negative end, and –I for the positive end.
Let the length of the axis of the cylinder be 2b, and its radius a. Then the force arising from this superficial distribution on a magnetic pole placed at the middle point of the axis is that due to the attraction of the disk on the positive side, and the repulsion of the disk on the negative side. These two forces are equal and in the same direction, and their sum is
| $R=4\pi I\left(1-{\frac {b}{\sqrt {a^{2}+b^{2}}}}\right).$ | (2) |
From this expression it appears that the force depends, not on the absolute dimensions of the cavity, but on the ratio of the length to the diameter of the cylinder. Hence, however small we make the cavity, the force arising from the surface distribution on its walls will remain, in general, finite.
397.] We have hitherto supposed the magnetization to be uniform and in the same direction throughout the whole of the portion of the magnet from which the cylinder is hollowed out. When the magnetization is not thus restricted, there will in general be a distribution of imaginary magnetic matter through the substance of the magnet. The cutting out of the cylinder will remove part of this distribution, but since in similar solid figures the forces at corresponding points are proportional to the linear dimensions of the figures, the alteration of the force on the magnetic pole due to the volume-density of magnetic matter will diminish indefinitely as the size of the cavity is diminished, while the effect due to the surface-density on the walls of the cavity remains, in general, finite.
If, therefore, we assume the dimensions of the cylinder so small ​that the magnetization of the part removed may be regarded as everywhere parallel to the axis of the cylinder, and of constant magnitude I, the force on a magnetic pole placed at the middle point of the axis of the cylindrical hollow will be compounded of two forces. The first of these is that due to the distribution of magnetic matter on the outer surface of the magnet, and throughout its interior, exclusive of the portion hollowed out. The components of this force are α, β, γ, derived from the potential by equations (1). The second is the force R, acting along the axis of the cylinder in the direction of magnetization. The value of this force depends on the ratio of the length to the diameter of the cylindric cavity.
398.] Case I. Let this ratio be very great, or let the diameter of the cylinder be small compared with its length. Expanding the expression for R in terms of $|frac{a}{b}$, it becomes
| $R=4\pi I\left\{{\frac {1}{2}}{\frac {a^{2}}{b^{2}}}-{\frac {3}{8}}{\frac {a^{4}}{b^{4}}}+{\And }c.\right\},$ | (3) |
a quantity which vanishes when the ratio of b to a is made infinite. Hence, when the cavity is a very narrow cylinder with its axis parallel to the direction of magnetization, the magnetic force within the cavity is not affected by the surface distribution on the ends of the cylinder, and the components of this force are simply α, β, γ, where
| $\alpha =-{\frac {dV}{dx}},\quad \beta =-{\frac {dV}{dy}},\quad \gamma =-{\frac {dV}{dz}}.$ | (4) |
We shall define the force within a cavity of this form as the magnetic force within the magnet. Sir William Thomson has called this the Polar definition of magnetic force. When we have occasion to consider this force as a vector we shall denote it by ${\mathfrak {A}}$.
399.] Case II. Let the length of the cylinder be very small compared with its diameter, so that the cylinder becomes a thin disk. Expanding the expression for R in terms of ${\frac {b}{a}}$, it becomes
| $R=4\pi I\left\{1-{\frac {b}{a}}+{\frac {1}{2}}{\frac {b^{3}}{a^{3}}}-\And \!{c}.\right\},$ | (5) |
the ultimate value of which, when the ratio of a to b is made $4\pi I$.
Hence, when the cavity is in the form of a thin disk, whose plane is normal to the direction of magnetization, a unit magnetic pole ​placed at the middle of the axis experiences a force $4\pi I$ in the direction of magnetization arising from the superficial magnetism on the circular surfaces of the disk. Since the components of I are A, B and C, the components of this force are 4πA, 4πB and 4πC. This must be compounded with the force whose components are α, β, γ.
400.] Let the actual force on the unit pole be denoted by the vector ${\mathfrak {B}}$, and its components by a, b and c, then
| ${\begin{matrix}a=\alpha +4\pi A,\\b=\beta +4\pi B,\\c=\gamma +4\pi C.\end{matrix}}$ | (6) |
We shall define the force within a hollow disk, whose plane sides are normal to the direction of magnetization, as the Magnetic Induction within the magnet. Sir William Thomson has called this the Electromagnetic definition of magnetic force.
The three vectors, the magnetization ${\mathfrak {J}}$, the magnetic force ${\mathfrak {H}}$, and the magnetic induction ${\mathfrak {B}}$ are connected by the vector equation
| ${\mathfrak {B}}={\mathfrak {H}}+4\pi {\mathfrak {J}}.$ | (7) |

### Line-Integral of Magnetic Force.

401.] Since the magnetic force, as defined in Art. 398, is that due to the distribution of free magnetism on the surface and through the interior of the magnet, and is not affected by the surface-magnetism of the cavity, it may be derived directly from the general expression for the potential of the magnet, and the line-integral of the magnetic force taken along any curve from the point A to the point B is
| $\int _{A}^{B}{\left(\alpha {\frac {dx}{ds}}+\beta {\frac {dy}{ds}}+\gamma {\frac {dz}{ds}}\right)ds}=V_{a}-V_{B},$ | (8) |

### Surface-Integral of Magnetic Induction.

402.] The magnetic induction through the surface S is defined as the value of the integral
| $Q=\iint {{\mathfrak {B}}\cos \epsilon \,dS},$ | (9) |
where ${\mathfrak {B}}$ denotes the magnitude of the magnetic induction at the element of surface dS, and ε the angle between the direction of the induction and the normal to the element of surface, and the integration is to be extended over the whole surface, which may be either closed or bounded by a closed curve.
If a, b, c denote the components of the magnetic induction, and l, m, n the direction-cosines of the normal, the surface-integral may be written
| $Q=\iint {(la+mb+nc)dS}.$ | (10) |
If we substitute for the components of the magnetic induction their values in terms of those of the magnetic force, and the magnetization as given in Art. 400, we find
| $Q=\iint {(l\alpha +m\beta +n\gamma )dS}+4\pi \iint {(lA+mB+nC)dS}.$ | (11) |
We shall now suppose that the surface over which the integration extends is a closed one, and we shall investigate the value of the two terms on the right-hand side of this equation.
Since the mathematical form of the relation between magnetic force and free magnetism is the same as that between electric force and free electricity, we may apply the result given in Art. 77 to the first term in the value of Q by substituting α, β, γ, the components of magnetic force, for X, Y, Z, the components of electric force in Art. 77, and M, the algebraic sum of the free magnetism within the closed surface, for e, the algebraic sum of the free electricity.
We thus obtain the equation
| $\iint {(l\alpha +m\beta +n\gamma )dS}=4\pi M.$ | (12) |
Since every magnetic particle has two poles, which are equal in numerical magnitude but of opposite signs, the algebraic sum of the magnetism of the particle is zero. Hence, those particles which are entirely within the closed surface S can contribute nothing to the algebraic sum of the magnetism within S. The ​value of M must therefore depend only on those magnetic particles which are cut by the surface S.
Consider a small element of the magnet of length s and transverse section k$^2$, magnetized in the direction of its length, so that the strength of its poles is m. The moment of this small magnet will be ms, and the intensity of its magnetization, being the ratio of the magnetic moment to the volume, will be
| $I={\frac {m}{k^{2}}}.$ | (13) |
Let this small magnet be cut by the surface S, so that the direction of magnetization makes an angle ε' with the normal drawn outwards from the surface, then if dS denotes the area of the section,
| $k^{2}=dS\cos \epsilon '.\,$ | (14) |
The negative pole –m of this magnet lies within the surface S.
Hence, if we denote by dM the part of the free magnetism within S which is contributed by this little magnet,
| ${\begin{aligned}dM=-m&=-Ik^{2},\\&=-I\cos \epsilon '\,dS.\end{aligned}}$ | (15) |
To find M, the algebraic sum of the free magnetism within the closed surface S, we must integrate this expression over the closed surface, so that
* $M=-\iint {I\cos \epsilon '\,dS},$
or writing A, B, C for the components of magnetization, and l, m, n for the direction-cosines of the normal drawn outwards,
| $M=-\iint {(lA+mB+nC)dS}.$ | (16) |
This gives us the value of the integral in the second term of equation (11). The value of Q in that equation may therefore be found in terms of equations (12) and (16),
| $Q=4\pi M-4\pi M=0,\,$ | (17) |
or, the surface-integral of the magnetic induction through any closed surface is zero.
403.] If we assume as the closed surface that of the differential element of volume dxdydz, we obtain the equation
| ${\frac {da}{dx}}+{\frac {db}{dy}}+{\frac {dc}{dz}}=0.$ | (18) |
This is the solenoidal condition which is always atisfied by the components of the magnetic induction.
​Since the distribution of magnetic induction is solenoidal, the induction through any surface bounded by a closed curve depends only on the form and position of the closed curve, and not on that of the surface itself.
404.] Surfaces at every point of which
| $la+mb+nc=0\,$ | (19) |
are called Surfaces of no induction, and the intersection of two such surfaces is called a Line of induction. The conditions that a curve, s, may be a line of induction are
| ${\frac {1}{a}}{\frac {dx}{ds}}={\frac {1}{b}}{\frac {dy}{ds}}={\frac {1}{c}}{\frac {dz}{ds}}.$ | (20) |
A system of lines of induction drawn through every point of a closed curve forms a tubular surface called a Tube of induction.
The induction across any section of such a tube is the same. If the induction is unity the tube is called a Unit tube of induction.
All that Faraday says about lines of magnetic force and magnetic sphondyloids is mathematically true, if understood of the lines and tubes of magnetic induction.
The magnetic force and the magnetic induction are identical outside the magnet, but within the substance of the magnet they must be carefully distinguished. In a straight uniformly magnetized bar the magnetic force due to the magnet itself is from the end which points north, which we call the positive pole, towards the south end or negative pole, both within the magnet and in the space without.
The magnetic induction, on the other hand, is from the positive pole to the negative outside the magnet, and from the negative pole to the positive within the magnet, so that the lines and tubes of induction are re-entering or cyclic figures.
The importance of the magnetic induction as a physical quantity will be more clearly seen when we study electromagnetic phenomena. When the magnetic field is explored by a moving wire, as in Faraday's Exp. Res. 3076, it is the magnetic induction and not the magnetic force which is directly measured.

### The Vector-Potential of Magnetic Induction.

405.] Since, as we have shewn in Art. 403, the magnetic induction through a surface bounded by a closed curve depends on ​the closed curve, and not on the form of the surface which is bounded by it, it must be possible to determine the induction through a closed curve by a process depending only on the nature of that curve, and not involving the construction of a surface forming a diaphragm of the curve.
This may be done by finding a vector ${\mathfrak {A}}$ related to ${\mathfrak {B}}$, the magnetic induction, in such a way that the line-integral of ${\mathfrak {A}}$, extended round the closed curve, is equal to the surface-integral of ${\mathfrak {B}}$, extended over a surface bounded by the closed curve.
If, in Art. 24, we write F, G, H for the components of ${\mathfrak {A}}$, and a, b, c for the components of ${\mathfrak {B}}$, we find for the relation between these components
| $a={\frac {dH}{dy}}-{\frac {dG}{dz}},\quad b={\frac {dF}{dz}}-{\frac {dH}{dx}},\quad c={\frac {dG}{dx}}-{\frac {dF}{dy}}$ | (21) |
The vector ${\mathfrak {A}}$, whose components are F, G, H, is called the vector-potential of magnetic induction. The vector-potential at a given point, due to a magnetized particle placed at the origin, is numerically equal to the magnetic moment of the particle divided by the square of the radius vector and multiplied by the sine of the angle between the axis of magnetization and the radius vector, and the direction of the vector-potential is perpendicular to the plane of the axis of magnetization and the radius vector, and is such that to an eye looking in the positive direction along the axis of magnetization the vector-potential is drawn in the direction of rotation of the hands of a watch.
Hence, for a magnet of any form in which A, B, C are the components of magnetization at the point x y z, the components of the vector-potential at the point ξ η ζ, are
| ${\begin{matrix}F=\iiint {\left(B{\frac {dp}{dz}}-C{\frac {dp}{dy}}\right)dxdydz},\\G=\iiint {\left(C{\frac {dp}{dx}}-A{\frac {dp}{dz}}\right)dxdydz},\\H=\iiint {\left(A{\frac {dp}{dy}}-B{\frac {dp}{dx}}\right)dxdydz};\end{matrix}}$ | (22) |
where p is put, for conciseness, for the reciprocal of the distance between the points (ξ η ζ,) and (x, y, z), and the integrations are extended over the space occupied by the magnet.
406.] The scalar, or ordinary, potential of magnetic force, Art. 386, becomes when expressed in the same notation,
| $V=\iiint {\left(A{\frac {dp}{dx}}+B{\frac {dp}{dy}}+C{\frac {dp}{dz}}\right)dxdydz}.$ | (23) |
​Remembering that ${\frac {dp}{dx}}=-{\frac {dp}{d\xi }}$, and that the integral
* $\iiint {A\left({\frac {d^{2}p}{dx^{2}}}+{\frac {d^{2}p}{dy^{2}}}+{\frac {d^{2}p}{dz^{2}}}\right)dxdydz}$
has the value –4π(A) when the point (ξ, η, ζ) is included within the limits of integration, and is zero when it is not so included, (A) being the value of (A) at the point (ξ, η, ζ), we find for the value of the x-component of the magnetic induction,
| ${\begin{aligned}a&={\frac {dH}{d\eta }}-{\frac {dG}{d\zeta }}\\&=\iiint {\left\{A\left({\frac {d^{2}p}{dyd\eta }}+{\frac {d^{2}p}{dzd\zeta }}\right)-B{\frac {d^{2}p}{dxd\eta }}-C{\frac {d^{2}p}{dxd\zeta }}\right\}dxdydz}\\&=-{\frac {d}{d\xi }}\iiint {\left\{A{\frac {dp}{dx}}+B{\frac {dp}{dy}}+C{\frac {dp}{dz}}\right\}dxdydz}\\&\quad \quad -\iiint {A\left({\frac {d^{2}p}{dx^{2}}}+{\frac {d^{2}p}{dy^{2}}}+{\frac {d^{2}p}{dz^{2}}}\right)dxdydz}.\end{aligned}}$ | (24) |
The first term of this expression is evidently $-{\frac {dV}{d\xi }}$, or α, the component of the magnetic force.
The quantity under the integral sign in the second term is zero for every element of volume except that in which the point (ξ, η, ζ) is included. If the value of A at the point (ξ, η, ζ) is (A), the value of the second term is 4π(A), where (A) is evidently zero at all points outside the magnet.
We may now write the value of the x-component of the magnetic induction
| $a=\alpha +4\pi (A)\,$ | (25) |
an equation which is identical with the first of those given in Art. 400. The equations for b and c will also agree with those of Art. 400.
We have already seen that the magnetic force ${\mathfrak {H}}$ is derived from the scalar magnetic potential V by the application of Hamilton's operator $\nabla$, so that we may write, as in Art. 17,
| ${\mathfrak {H}}=-\nabla V,$ | (26) |
and that this equation is true both without and within the magnet.
It appears from the present investigation that the magnetic induction ${\mathfrak {B}}$ is derived from the vector-potential ${\mathfrak {A}}$ by the application of the same operator, and that the result is true within the magnet as well as without it.
The application of this operator to a vector-function produces, ​in general, a scalar quantity as well as a vector. The scalar part, however, which we have called the convergence of the vector-function, vanishes when the vector-function satisfies the solenoidal condition
| ${\frac {dF}{d\xi }}+{\frac {dG}{d\eta }}+{\frac {dH}{d\zeta }}=0.$ | (27) |
By differentiating the expressions for F, G, H in equations (22), we find that this equation is satisfied by these quantities.
We may therefore write the relation between the magnetic induction and its vector-potential
* ${\mathfrak {B}}=\nabla {\mathfrak {A}},$
which may be expressed in words by saying that the magnetic induction is the curl of its vector-potential. See Art. 25.
* ↑   On the force within cavities of other forms. 1. Any narrow crevasse. The force arising from the surface-magnetism is $4\pi I\cos \epsilon$ in the direction of the normal to the plane of the crevasse, where ε is the angle between this normal and the direction of magnetization. When the crevasse is parallel to the direction of magnetization the force is the magnetic force ${\mathfrak {H}}$; when the crevasse is perpendicular to the direction of magnetization the force is the magnetic induction ${\mathfrak {B}}$. 2. In an elongated cylinder, the axis of which makes an angle ε with the direction of magnetization, the force arising from the surface-magnetism is $2\pi I\sin \epsilon$, perpendicular to the axis in the plane containing the axis and the direction of magnetization. 3. In a sphere the force arising from surface-magnetism is ${\frac {4}{3}}\pi I$ in the direction of magnetization.
* ↑ * Exp. Res., series xxviii.