﻿766 
  Mr. 
  W. 
  A. 
  Jenkins 
  : 
  New 
  Method 
  of 
  Determining 
  the 
  

  

  magnet 
  takes 
  place, 
  i. 
  e., 
  it 
  is 
  placed 
  making 
  an 
  angle 
  a 
  

   with 
  the 
  earth's 
  field. 
  

  

  Consider 
  now 
  what 
  it 
  is 
  that 
  we 
  actually 
  measure 
  in 
  the 
  

   second 
  part 
  of 
  the 
  experiment. 
  The 
  magnet 
  vibrates 
  under 
  

   a 
  resultant 
  field 
  of 
  H 
  cos 
  a 
  in 
  both 
  parts 
  of 
  the 
  experiment. 
  

   Let 
  Z 
  be 
  the 
  field 
  actually 
  produced 
  by 
  the 
  helix. 
  

  

  a 
  is 
  the 
  angle 
  between 
  the 
  zero 
  position 
  of 
  the 
  magnet 
  

   and 
  the 
  earth's 
  field, 
  but 
  the 
  coil 
  is 
  placed 
  so 
  that 
  its 
  field 
  

   does 
  not 
  disturb 
  the 
  zero 
  position 
  of 
  the 
  magnet. 
  The 
  angle 
  

   between 
  the 
  direction 
  of 
  H 
  and 
  the 
  magnetic 
  axis 
  of 
  the 
  

   tube 
  is 
  therefore 
  a. 
  

  

  Suppose 
  the 
  magnet 
  reversed 
  and 
  the 
  extra 
  torsion 
  taken 
  

   out 
  of 
  the 
  fibre. 
  It 
  comes 
  to 
  rest 
  under 
  the 
  forces 
  due 
  to 
  

   the 
  earth's 
  field, 
  the 
  field 
  of 
  the 
  solenoid, 
  and 
  the 
  torsion 
  

   of 
  the 
  fibre. 
  Let 
  ft 
  be 
  the 
  angle 
  between 
  this 
  zero 
  position 
  

   of 
  the 
  magnet 
  and 
  the 
  magnetic 
  axis 
  of 
  the 
  solenoid. 
  

  

  The 
  resultant 
  field 
  giving 
  the 
  same 
  time 
  of 
  swing 
  as 
  in 
  

   the 
  first 
  part 
  of 
  the 
  experiment 
  = 
  H 
  cos 
  a. 
  

  

  Therefore 
  we 
  get 
  

  

  Hcosa 
  = 
  Zcos/3— 
  Hcos(a 
  + 
  /3). 
  . 
  . 
  . 
  (B) 
  

  

  Also 
  the 
  sum 
  of 
  the 
  moments 
  of 
  the 
  forces 
  acting 
  on 
  the 
  

   magnet 
  at 
  the 
  centre 
  of 
  the 
  magnet 
  when 
  it 
  is 
  in 
  its 
  zero 
  

   position 
  is 
  zero. 
  Hence 
  we 
  get 
  

  

  n(6 
  -a-ft) 
  + 
  2Rl 
  sin 
  (u-t 
  ft) 
  =2Zl 
  sin 
  ft 
  . 
  . 
  (C) 
  

  

  where 
  21 
  is 
  the 
  length 
  of 
  the 
  magnet. 
  

  

  Angles 
  a 
  and 
  ft 
  are 
  small 
  and 
  we 
  may 
  substitute 
  the 
  

   angles 
  for 
  the 
  sine*. 
  

  

  Eliminating 
  n 
  and 
  ft 
  from 
  equations 
  A, 
  B, 
  and 
  C 
  we 
  get 
  

   the 
  equation 
  

  

  „ 
  2*(0-ct) 
  „ 
  , 
  tt 
  Sa0-2ot 
  2 
  

  

  Z 
  cos 
  3 
  — 
  = 
  xi 
  cos 
  a 
  + 
  J± 
  cos 
  -= 
  — 
  > 
  

  

  o 
  o 
  

  

  which 
  to 
  a 
  first 
  approximation 
  gives 
  

  

  „ 
  Hcos 
  a 
  + 
  H 
  cos 
  doc. 
  

   n— 
  5 
  

  

  COS 
  ZOC 
  

  

  = 
  2Hcos 
  a. 
  

  

  Hence, 
  instead 
  of 
  measuring 
  H 
  we 
  are 
  measuring 
  some- 
  

   thing 
  very 
  nearly 
  H 
  cos 
  a. 
  

  

  Cosine 
  «. 
  must 
  therefore 
  be 
  1 
  within 
  the 
  limits 
  of 
  accuracy 
  

   of 
  determination 
  of 
  the 
  other 
  quantities. 
  As 
  described 
  

   previously, 
  the 
  torsion 
  was 
  eliminated 
  as 
  far 
  as 
  possible 
  by 
  

   allowing 
  the 
  system 
  to 
  come 
  to 
  rest 
  before 
  inserting 
  the 
  

  

  