Page:A Treatise on Electricity and Magnetism - Volume 2.djvu/310

If $N$ is the whole number of windings, and if $\gamma$ is the strength of the current in each winding,

$\phi ={\frac {1}{2}}N\gamma \cos \theta$.

Hence the magnetic force within the coil is

$M={\frac {4\pi }{3}}{\frac {N\gamma }{a}}$.

673.] Let us next find the method of coiling the wire in order to produce within the sphere a magnetic potential of the form of a solid zonal harmonic of the second degree,

12= jl*(f cos 2 0-i).

Here = ~A (f cos 2 0-i).

i .

If the whole number of windings is $N$ the number between the pole and the polar distance $\theta$ is ${\frac {1}{2}}N\sin ^{2}\theta$.

The windings are closest at latitude 45°. At the equator the direction of winding changes, and in the other hemisphere the windings are in the contrary direction.

Let $\gamma$ be the strength of the current in the wire, then within the shell

$\Omega ={\frac {4\pi }{5}}N\gamma {\frac {r^{2}}{a^{2}}}\left({\frac {3}{2}}\cos ^{2}\theta -{\frac {1}{2}}\right)$.

Let us now consider a Conductor in the form of a plane closed curve placed anywhere within the shell with its plane perpendicular to the axis. To determine its coefficient of induction we have to find the surface-integral of ${\frac {d\Omega }{dz}}$ over the plane bounded by the curve, putting $\gamma =1$.

Now 12 = ~ 2 N (z 2 - i (x* 4 /)),

O d

d& STT ,

and - = 2 Nz. dz 5 a 2

Hence, if $S$ is the area of the closed curve, its coefficient of induction is

$M={\frac {8\pi }{5a^{2}}}NSz$.

If the current in this conductor is $\gamma ^{\prime }$, there will be, by Art. 583, a force $Z$, urging it in the direction of $z$, where

$Z=\gamma \gamma ^{\prime }{\frac {dM}{dz}}={\frac {8\pi }{5a^{2}}}NS\gamma \gamma ^{\prime }$,

and, since this is independent of $x$, $y$, $z$, the force is the same in whatever part of the shell the circuit is placed.

674.] The method given by Poisson, and described in Art. 437,