Page:A Treatise on Electricity and Magnetism - Volume 2.djvu/58

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 k², 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.