348 Prof. J. J. Thomson on 
the magnet, and we are left with the momentum through the 
fixed point, 2. €. we get the same result as if we replace the 
magnet by its equivalent currents, 
The following illustration will show the difference in the 
physical effects ‘produced by the differences in the distribu- 
tion of momentum just considered. Suppose AB is a mag- 
netized piece of soft iron, P the charged point, then we have 
seen that the momentum in the field has for resultant a 
momentum I through P perpendicular to the plane containing 
AB and P. Suppose. that AB is demagnetized by tapping, 
there will then be no momentum in the field; hence by the 
principle of the conservation of momentum the momentum in 
the field before the tapping must be transferred to the material 
system in the field. Thus the charged point will acquire 
momentum I,but the momentum of AB will remain unchanged. 
If, however, the distribution of momentum in the field were 
equivalent to I at P and —I at the magnet, then, when the 
magnet was tapped I would gain momentum as before ; but in 
this case the magnet would gain an equal momentum in the 
opposite direction. Thus in this case the system would 
behave as if the point and magnet were acted on by equal 
and opposite impulses, while in the former case, which is that 
given by the rule we have used for finding the momentum, 
the charged particle would alone be acted upon by an impulse, 
the magnet remaining undisturbed. 
Connewion between the Momentum and the Vector Potential — 
From the investigation on page 344, it follows that U, V, W, 
the components of the momentum due to magnetic force pro- 
duced by an element of current ids and the electric field due 
to a charge é at P, are given by the equations 
Fab Piece dxvd?R dy PR | dz oe 
ds da? " ds dyrde * dsdedzJ °* 
_. .,b d@t pan oe Aas 
= pep 7. ds — ghe't en pds 
with similar expressions for V and W ; here Ris the distance 
between the current element and the charged point. Now if 
we find the momentum at P due to a closed cireuit from this 
expression by integration, we see that the second term 
vanishes, and U the momentum parallel to « acting through 
P is equal to 
pe’ io ds pocpout fine 
