# Page:A Dynamical Theory of the Electromagnetic Field.pdf/17

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PROFESSOR CLERK MAXWELL ON THE ELECTROMAGNETIC FIELD.

The total quantity of electricity is ${\displaystyle c{\frac {M}{S}}}$, and the value of ${\displaystyle \int y^{2}dt}$ is ${\displaystyle c^{2}{\tfrac {M^{2}}{2SN}}}$.

The effects on the galvanometer and dynamometer are equal to those of a uniform current ${\displaystyle {\tfrac {1}{2}}c{\tfrac {M}{N}}}$ for a time ${\displaystyle 2{\tfrac {N}{S}}}$.

The heating effect is therefore greater than that of the current on making contact.

(42) If an electromotive force of the form ${\displaystyle \xi =E\cos pt}$ acts on the circuit R, then if the circuit S is removed, the value of ${\displaystyle x}$ will be

${\displaystyle x={\frac {E}{A}}\sin(pt-\alpha )}$

where

${\displaystyle A^{2}=R^{2}+L^{2}p^{2}}$

and

${\displaystyle \tan \alpha ={\frac {Lp}{R}}}$

The effect of the presence of the circuit S in the neighbourhood is to alter the value of A and ${\displaystyle \alpha }$, to that which they would be if R became

${\displaystyle R+p^{2}{\frac {MS}{S^{2}+p^{2}N^{2}}}}$

and L became

${\displaystyle L-p^{2}{\frac {MN}{S^{2}+p^{2}N^{2}}}}$

Hence the effect of the presence of the circuit S is to increase the apparent resistance and diminish the apparent self-induction of the circuit R.

On the Determination of Coefficients of Induction by the Electric Balance.

(43) The electric balance consists of six conductors joining four points, A,C,D,E, two and two. One pair, AC, of these points is connected through the battery B. The opposite pair, DE, is connected through the battery B. The opposite pair, DE, is connected through the galvanometer G. Then if the resistances of the four remaining conductors are represented by P,Q,R,S, and the currents in them by ${\displaystyle x,x-z,y}$, and ${\displaystyle y+z}$, the current through G will be ${\displaystyle z}$. Let the potentials at the four points be A,C,D,E. Then the conditions of steady currents may be found from the equations

 ${\displaystyle \left.{\begin{matrix}Px=A-D,&Q(x-z)=D-C\\Ry=A-E,&S(y+z)=E-C\\Gz=D-E,&B(x+y)=-A+C+F\end{matrix}}\right\}.}$ (21)

Solving these equations for ${\displaystyle z}$, we find

 ${\displaystyle z\left\{{\frac {1}{P}}+{\frac {1}{Q}}+{\frac {1}{R}}+{\frac {1}{S}}+B({\frac {1}{P}}+{\frac {1}{R}})({\frac {1}{Q}}+{\frac {1}{S}})+G({\frac {1}{P}}+{\frac {1}{Q}})({\frac {1}{R}}+{\frac {1}{S}})+{\frac {BG}{PQRS}}(P+Q+R+S)\right\}=F({\frac {1}{PS}}-{\frac {1}{QR}}).}$ (22)