﻿808 
  Dr. 
  Russell 
  and 
  Mr. 
  Alty: 
  Electromagnetic 
  Method 
  

   This 
  equation 
  can 
  be 
  satisfied 
  either 
  by 
  

  

  fio:) 
  = 
  0, 
  and 
  ,j=0, 
  (9) 
  

  

  or 
  by 
  

  

  /■W-|V"W 
  + 
  f4V'W-.-=o 
  • 
  (10)^ 
  

   ''"'^ 
  /■' 
  (*) 
  - 
  iV'" 
  (^) 
  + 
  J/' 
  i^)-- 
  =0 
  . 
  (11), 
  

  

  The 
  equations 
  (9) 
  obviously 
  give 
  the 
  real 
  roots 
  of 
  the 
  

   equation 
  which 
  must 
  all 
  lie 
  on 
  the 
  line 
  ?/ 
  = 
  0, 
  that 
  is, 
  on 
  the 
  

   axis 
  o£ 
  X. 
  The 
  points 
  o£ 
  intersection 
  o£ 
  the 
  equations 
  (10) 
  

   and 
  (11) 
  determine 
  the 
  imaginary 
  roots. 
  

  

  As 
  equation 
  (11) 
  does 
  not 
  contain 
  the 
  constant 
  term 
  of 
  

   the 
  equation 
  /(j) 
  = 
  0, 
  it 
  follows 
  that 
  it 
  gives 
  the 
  curve 
  locus 
  

   of 
  a 
  series 
  of 
  imaginary 
  roots 
  of 
  the 
  equations 
  formed 
  by 
  

   varying 
  c. 
  Where 
  the 
  curve 
  represented 
  by 
  equation 
  (11) 
  

   cuts 
  the 
  axis 
  of 
  x 
  we 
  have 
  /' 
  {x) 
  = 
  0. 
  At 
  these 
  points 
  there 
  

   are 
  at 
  least 
  two 
  equal 
  roots. 
  

  

  For 
  the 
  quadratic 
  equation 
  x^-{-hx 
  + 
  c=-0^ 
  (11) 
  becomes 
  

   2ct' 
  + 
  & 
  = 
  0, 
  and 
  this 
  is 
  the 
  equation 
  to 
  the 
  line 
  of 
  neutral 
  

   points 
  Ni'PN2' 
  shown 
  in 
  fig. 
  2. 
  

  

  Let 
  us 
  suppose 
  that 
  the 
  equation 
  has 
  been 
  reduced 
  to 
  the 
  

   form 
  

  

  5. 
  Cubic 
  Equation, 
  

   the 
  equation 
  has 
  1 
  

  

  x^-h^x^c=() 
  (12; 
  

  

  We 
  easily 
  find 
  that 
  

  

  x'-h'x 
  + 
  c 
  C 
  20 
  C 
  

  

  x{x^-h'') 
  ~ 
  '^ 
  5(x-b)'~ 
  5x 
  "^ 
  5(^ 
  + 
  6)' 
  ^ 
  ^^ 
  

  

  where 
  C=5Hc/262. 
  

  

  In 
  this 
  case 
  (fig. 
  3) 
  three 
  long 
  vertical 
  wires 
  are 
  used. 
  

   Let 
  them 
  cut 
  the 
  plane 
  of 
  the 
  paper 
  at 
  right 
  angles 
  at 
  Bj, 
  0, 
  

   and 
  B3 
  respectively, 
  and 
  let 
  BiO 
  = 
  063=6. 
  The 
  currents 
  in 
  

   the 
  wires 
  passing 
  through 
  Bj 
  and 
  Bg 
  must 
  each 
  be 
  made 
  

   equal 
  to 
  0, 
  and 
  the 
  current 
  in 
  the 
  wire 
  passing 
  through 
  

   should 
  be 
  the 
  return 
  current 
  20. 
  The 
  arrow-heads 
  indicate 
  

   that 
  the 
  direction 
  of 
  the 
  current 
  flow 
  is 
  out 
  of 
  the 
  paper 
  at 
  

   Bi 
  and 
  B3 
  and 
  into 
  it 
  at 
  0. 
  H 
  denotes 
  the 
  direction 
  of 
  the 
  

   earth's 
  magnetic 
  field. 
  

  

  Let 
  us 
  now 
  suppose 
  that 
  c 
  increases 
  uniformly 
  from 
  zero 
  

   to 
  infinity. 
  When 
  c 
  is 
  small, 
  and 
  therefore 
  the 
  currents 
  are 
  

  

  