Am
ds lvii? dv'
dY
and
dZ
become infin small of higher order of R and are to be
neglected.
=
Li

2~
sin
?
cos
imd?
dX
1+2Rs].n10c0sA
2ir
r
=
J0
R
÷
2~
sin
cos
ci?
The second term vanishes, so that we have
2irim
ft
Now we
formulate the torque of the electromagnetic force.
The
torque of both forces is
4111w
2AX
cos
u
cos
ii.
I?
[Fig.]
The
quantity
m.2A
is called the magnetic
moment
M.
Then
we
have
Torque
=
cos a.
In addition, however, the magnet is also acted
upon by
the
terrestrial magnetic force.
As
the field of force of strength
1
we
define the field which exerts force
1
in absolute units on the unit
magnetic mass.
The
horizontal component of the terrestrial magnetic
force, which is the only
one
to exert an effect in our case,
we
denote
by
H,
and imagine that it has been measured absolutely by a procedure
which will
be
described later.
We
further imagine that the apparatus
is oriented such that the plane current circuit coincides with the
magnetic meridian.
The
force acting
upon
the north
end
is
[Fig.]
mH.
Hence its torque is
mHy
sin
u.
Thus, the total torque arising
from
the terrestrial magnetic force,
which acts in the same or the other direction than the electromagnetic
force of the current circuit, is
Torque
=
2mHA
sin
u
=
2
MAH
sin u.
120