the Earth's Magnetic Field. 99 



Xow we can readily show that 



~d . ~d cos a c) 

 5-= sin*. =- + .^- . 



0.** oy y oa 



Carrying out this operation of differentiation wo have 



AH=2wl!feft»T |^. P„(0) . P„'(") • sin 4 ^ 



sin a. cos a y n+1 _ //1X ^ [" . „ -0 // \~1"| 



It readily follows that if n is the number of doublets per c.c, 

 the magnetic potential due to the ring of cross-section ydu . dy 

 is obtained by replacing N in the above expression by 

 nydxdy. Making this substitution, and then integrating 

 with respect to a, from to it, and with respect to y from 

 to a, we obtain, without much difficulty, the magnetic 

 potential due to the sphere of radius a in the form 



where /n = cos *. 



After some rather laborious work the integral becomes 

 evaluated as a function of n. We need only concern our- 

 selves with the first three terms, which lead to 



H 



= -i|^ = ^{^sm^r0-46sin3^ + 0-09sin^^ 



r ov 4 U L Jr° 



+ [o*112 sin 50 + 0-04 sin 30 + 0-01 sin 0~]- 7 + ...~j. 



At the equator, and on the surface, H amounts to 

 1*75 nelicoa. 



(5) The electrostatic field due to the sphere considered 

 in the last problem. 



In the notation of the last problem the electrostatic potential 

 due to the charge elements of the same kind in the ring of 

 radius x—y sin a is, at the point r . 6, 



SV =2TrNey sin «T -^. P„(a)P n (0). 



n — l r 



H 2 



