26

MAGNETIC FORCE AND MAGNETIC INDUCTION.

[403.

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

(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,

(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,

(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

or writing *A*, *B*, *C* for the components of magnetization, and *l*, *m*, *n* for the direction-cosines of the normal drawn outwards,

(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),

(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

(18) |

This is the solenoidal condition which is always atisfied by the components of the magnetic induction.