value of the potential will be m -V ( — ) . This quantity is the potential of a 



r 



magnetic particle or dipole of magnetic moment m. 



If we think of a magnetized body as being built up of dipole elements, 

 then the magnetic moment of the body will be the vector sum of the magnetic 

 moments of the individual dipoles. Let the magnetic moment per unit volume 

 or intensity of magnetization be designated by I. The magnetic moment of a 

 volume element dudvdw would then be \dudvdw and its magnetic potential 



I • V ( — ) dudvdw. The potential of the entire body would then be given by 

 r 



V = ff f | . V (— ) dudvdw (40) 



or, if (l,m,n) be the direction cosines of I, by 



V = I f f f | / — (— ) + m — (— ) + n — (— ) dudvdw. (41) 

 |_ 3" r 'dv r ^w r 



The components of I will be in general functions of the source point coordinates 

 (u,v,w) . In what follows we shall suppose I to remain constant. 



Vertical Component of Field of Infinite Horizontal Cylinder 



As in the problem of computing the vertical component of the attraction of 

 the horizontal cylinder, let the coordinate axes be chosen such that the z axis 

 is vertically downward and the y axis is parallel to the axis of the cylinder. 

 The x,y plane is supposed to coincide with the surface of the ground. Let the 

 direction cosines of I in such a system be (1, m, n) . 



The potential V at a point whose coordinates are (x,y,z) will be given by 



V 



= 'J JT 



I —— ( — ) + n — — ( — ) 

 qu r qw r 



dudvdw , (42) 



since the term containing m as a factor has the value zero. The limits — oo and 

 co refer to integration with respect to v. When this integration is performed 

 (42) becomes 



7/ _ 9 r C C f I (* ~ u) + n (z - w)) 



V - 21 )} (x - u)* + (z- w)*~ dudW ■ (43) 



dV 

 The field component H e — — , so that at the point (x,y,z,) 



?}* 



601 



