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The part depending on p in the second integral

=-\frac{\mu ep}{2}\int\int\int\left(B\frac{d}{dz}\frac{1}{R}-\frac{Cd}{dy}\frac{1}{R}\right)dx\ dy\ dz,

or (see Maxwell's 'Electricity and Magnetism,' § 405)

=-\frac{\mu ep}{2}F'_{1}.

Adding this to the term epF'_{1} already obtained, we get \tfrac{\mu ep}{2}F'_{1} as the part of the kinetic energy depending on p. We have evidently similar expressions for the parts of the kinetic energy depending on q and r. Hence the part of the kinetic energy with which we are concerned will

=\frac{\mu e}{2}\cdot\left(F'_{1}p+G'_{1}q+H'_{1}r\right).

By Lagrange's equations, the force on the sphere parallel to the axis of x


=\frac{\mu e}{2}\left\{ p\frac{dF'_{1}}{dx}+q\frac{dG'_{1}}{dx}+r\frac{dH'_{1}}{dx}-\frac{dF'_{1}}{dt}\right\}

=\frac{\mu e}{2}\left\{ p\frac{dF'_{1}}{dx}+q\frac{dG'_{1}}{dx}+r\frac{dH'_{1}}{dx}-p\frac{dF'_{1}}{dx}-q\frac{dF'_{1}}{dy}-r\frac{dF'_{1}}{dz}\right\}

=\frac{\mu e}{2}\left\{ q\left(\frac{dG'_{1}}{dx}-\frac{dF'_{1}}{dy}\right)-r\left(\frac{dF'_{1}}{dz}-\frac{dH'_{1}}{dx}\right)\right\}

=\frac{\mu e}{2}\left(qc_{1}-rb_{1}\right).

Similarly, the force parallel to the axis of y

=\frac{\mu e}{2}\left(ra_{1}-pc_{1}\right), (5)

the force parallel to the axis of z

=\frac{\mu e}{2}\left(pb_{1}-qa_{1}\right)

where a1, b1, c1 are the components of magnetic induction at the centre of the sphere due to the external magnet. These forces are the same as would act on unit length of a conductor at the centre of the sphere carrying a current whose components are \tfrac{\mu ep}{2}, \tfrac{\mu eq}{2}, \tfrac{\mu er}{2} . The resultant force is perpendicular