Page:A Treatise on Electricity and Magnetism - Volume 2.djvu/49

$lK=\iiint {Adxdydz},\quad mK=\iiint {Bdxdydz},\quad nK=\iiint {Cdxdydz};$

(6)

$L=\iiint {Axdxdydz},\quad M=\iiint {Bydxdydz},\quad N=\iiint {Cxdxdydz};$

(7)

$P=\iiint {(Bz+Cy)dxdydz},\quad Q=\iiint {(Cx+Az)dxdydz},\quad R=\iiint {(Ay+Bz)dxdydz}.$

(8)

We thus find for the value of the potential energy of the magnet placed in presence of the unit pole at the point (ξ, η, ζ),

$W=K{\frac {l\xi +m\eta +n\zeta }{r^{3}}}-{\frac {\xi ^{2}(2L-M-N)+\eta ^{2}(2M-N-L)+\zeta ^{2}(2N-L-M)+3(P\eta \zeta +Q\zeta \xi +R\zeta \eta )}{r^{5}}}.$

(9)

This expression may also be regarded as the potential energy of the unit pole in presence of the magnet, or more simply as the potential at the point ξ, η, ζ due to the magnet.

On the Centre of a Magnet and its Primary and Secondary Axes.

392.] This expression may be simplified by altering the directions of the coordinates and the position of the origin. In the first place, we shall make the direction of the axis of x parallel to the axis of the magnet. This is equivalent to making

$l=1,\quad m=0,\quad n=0.$

(10)

If we change the origin of coordinates to the point (x', y', z'), the directions of the axes remaining unchanged, the volume-integrals lK, mK and nK will remain unchanged, but the others will be altered as follows:

$L'=L-lKx',\quad M'=M-mKy',\quad N'=N-nKz';$

(11)

$P'=P-K(mz'+ny'),\,Q'=Q-K(nx'+lz'),\,R'=R-L(ly'+mx').$

(12)

If we now make the direction of the axis of x parallel to the axis of the magnet, and put

$x'={\frac {2L-M-N}{2K}},\quad y'={\frac {R}{K}},\quad z'={\frac {Q}{K}}$

(13)

then for the new axes M and N have their values unchanged, and the value of L' becomes ${\frac {1}{2}}(M+N)$. P remains unchanged, and Q and R vanish. We may therefore write the potential thus,

$K{\frac {\xi }{r^{3}}}+{\frac {{\frac {3}{2}}(\eta ^{2}-\zeta ^{2})(M-N)+3P\eta \zeta }{r^{5}}}.$[1]

(14)

1. The denominator in the second term in the original is r, obviously an error.