Page:A Treatise on Electricity and Magnetism - Volume 1.djvu/439

$m(A+\beta )-(B+\alpha )={\frac {D'}{E}}\delta ',$

whence

$(m+n)(B-A)={\frac {D}{E}}\delta -{\frac {D'}{E'}}\delta '.$

If $m$ and $n,\,A\,{\mbox{and}}\,B,\,\alpha \,{\mbox{and}}\,\beta \,$ are approximately equal, then

$B-A={\frac {1}{2nE}}(A+\alpha )(A+\alpha +2r)(\delta -\delta ').$

Here $\delta -\delta '$ may be taken to be the smallest observable deflexion of the galvanometer.

If the galvanometer wire be made longer and thinner, retaining the same total mass, then $n$ will vary as the length of the wire and $\alpha$ as the square of the length. Hence there will be a minimum value of ${\frac {(A+\alpha )(A+\alpha +2r)}{n}}$ when

$\alpha ={\frac {1}{3}}(a+r)\left\{2{\sqrt {1-{\frac {3}{4}}{\frac {r^{2}}{(A+r)^{2}}}}}-1\right\}.$

If we suppose $r$, the battery resistance, small compared with $A$, this gives

$\alpha ={\frac {1}{3}}A;$

or, the resistance of each coil of the galvanometer should be one-third of the resistance to be measured.

We then find

$B-A={\frac {8}{9}}{\frac {A^{2}}{nE}}(\delta -\delta ').$

If we allow the current to flow through one only of the coils of the galvanometer, and if the deflexion thereby produced is $\Delta$ (supposing the deflexion strictly proportional to the deflecting force), then

$\Delta ={\frac {mE}{A+\alpha +r}}={\frac {3}{4}}{\frac {nE}{A}}\;{\mbox{if}}\;r=0\;{\mbox{and}}\;a={\frac {1}{3}}A.$

Hence

${\frac {B-A}{A}}={\frac {2}{3}}{\frac {\delta -\delta '}{\Delta }}.$

In the differential galvanometer two currents are made to produce equal and opposite effects on the suspended needle. The force with which either current acts on the needle depends not only on the strength of the current, but on the position of the windings of the wire with respect to the needle. Hence, unless the coil is very carefully wound, the ratio of $m$ to $n$ may change when the position of the needle is changed, and therefore it is necessary to determine this ratio by proper methods during each