the Earth? s Magnetic Field. 97 



(2) The magnetic field due to a rotating sphere of radius a in 

 which currents circulate round the axis, the current density 

 a- being proportional to the distance from the axis of 

 rotation and to the square of the angular velocity. 



Consider any spherical shell of radius y, and thickness dy . 

 The current density in this shell may be written in the form 



cr = a co 2 y sin a. 



The current in the element yda . dy is 



cr co 2 y 2 sin a . da . dy. 



so that the problem of: finding the magnetic potential AH, 

 due to the shell, is the same as that worked out in problem (1) , 

 and we have 



4 v 4 



AQ = T ira co 2 ^ 2 cos 6 . dy. 



The magnetic potential O, due to the whole solid sphere, is 



n 4 co 2 a C a 4 . 4 co 2 , 



12= 4 j7r o-q-y cos " ] y dy= — Tro-Q-ycrcos 6. 



The horizontal component H of the field is 



dft 4 a) 2 5 . , 

 rod lo u r j 



At the equator on the surface of the sphere we have 



4 



H^n 9 9 



= — ira Q wa". 



(3) The magnetic field due to a sphere of radius a, rotating 

 with angular velocity co, the current density being merely 

 a function of co. 



If o- = F(o)), the current in the ring of cross-section 

 yd* . dy, lying on the sphere of radius y, is y¥{co)da . dy. 



The magnetic potential due to this ring of current at the 

 point r . 6 is 



Sft = 2,n, dy F(») sin 2 "T-^-.gJ.P^fV) . Pn(ff)d«. 



?l=l ft T J- i 



The potential due to the whole shell of thickness dy is 



An = 27rF(aOT— ^ J^>?n(d)dy \ ">»'(«) Bin* *.<fe. 

 jRWt il%. S. 6. Vol. 24. No. 139, July 1912. H 



