certain Differential Equations in Quaternions. 63 



This solution gives the magnetic field in the case of a small 

 circular coil, placed with its axis parallel to the axis of the medium, 

 and carrying a rapidly alternating current. 



For the solution of the problem when the vibrator lies with its 

 axis perpendicular to the axis of the medium, physical reasoning 

 shows that it is necessary to find a value of fi (perpendicular to i) 

 which, while otherwise suitable, is composed of two terms which 

 respectively satisfy equations made with the factors of (3) (say 

 /x = (P 1 — P 2 )j, where Pj satisfies (4) and P 2 (6)) but separately 

 give rise to solutions that are infinite or discontinuous along a 

 line through the vibrator parallel to the axis of the medium. On 

 substituting this value of //., we find 



T = _ (f-aV + bVHV) VjV . P 2 - bViVSiVSjV . (P 1 - P 2 ) 

 = _ b [£. + SHV) VjV . P 2 -bVWSiVSjV . (P 1 -P 2 ). 



The first term is of the required form if 



Q—uipr 



g! + #w)p 2 = 



where r is given by (7). We determine P ± from the condition 

 that P x — P 2 is to be finite with its differential coefficients, hence, 

 using Cartesian coordinates x, y, z, 



where R', r' are the values of R, r when £ is substituted for x. 

 The lower limit is taken as — oo in order that the expression may 

 satisfy (3). 



Finally, omitting the factor — b, 



Of course, as many of the steps involved are little more than 

 guesswork, it is necessary to verify this result. To do this we 

 may transform the integral. 



Consider the integral, where, of course, p is a scalar, 



rx e -u>q</p+p* 



f (a!> p) = j_ h dZ-j==cosq(x-& 



- rx e -aqx e -<*q (-l+Vfs+p'-O rx guqxg-toq (?+Vf2+p 2 )" 



dP . + d% 



J-h * VP+o 2 J-h 



Vf 2 + /> 2 J-h " V£ 2 + 



