# Page:A Dynamical Theory of the Electromagnetic Field.pdf/16

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PROFESSOR CLERK MAXWELL ON THE ELECTROMAGNETIC FIELD.

the current, and the impulse it gives to the suspended coil of Weber's dynamometer, depend on the square of the current at every instant during the short time it lasts. Hence we must obtain the solution of the equations, and from the solution we may find the effects both on the galvanometer and dynamometer; and we may then make use of the method of Weber for estimating the intensity and duration of a current uniform while it lasts which would produce the same effects.

(39) Let ${\displaystyle n_{1},n_{2}}$ be the roots of the equation

 ${\displaystyle \left(LN-M^{2}\right)n^{2}+(RN+LS)n+RS=0,}$ (16)

and let the primary coil be acted on by a constant electromotive force ${\displaystyle Rc}$, so that ${\displaystyle c}$ is the constant current it could maintain; then the complete solution of the equations for making contact is

 ${\displaystyle x={\frac {c}{S}}{\frac {n_{1}n_{2}}{n_{1}-n_{2}}}\left\{\left({\frac {S}{n_{1}}}+N\right)e^{n_{1}t}-\left({\frac {S}{n_{2}}}+N\right)e^{n_{2}t}+S{\frac {n_{1}-n_{2}}{n_{1}n_{2}}}\right\}}$ (17)
 ${\displaystyle y={\frac {cM}{S}}{\frac {n_{1}n_{2}}{n_{1}-n_{2}}}\left\{e^{n_{1}t}-e^{n_{2}t}\right\}}$ (18)

From these we obtain for calculating the impulse on the dynamometer,

 ${\displaystyle \int x^{2}dt=c^{2}\left\{t-{\frac {3}{2}}{\frac {L}{R}}-{\frac {1}{2}}{\frac {M^{2}}{RN+LS}}\right\}}$ (19)
 ${\displaystyle \int y^{2}dt=c^{2}{\frac {1}{2}}{\frac {M^{2}R}{S(RN+LS)}}}$ (20)

The effects of the current in the secondary coil on the galvanometer and dynamometer are the same as those of a uniform current

${\displaystyle -{\frac {1}{2}}c{\frac {M^{2}R}{S(RN+LS)}}}$

for a time

${\displaystyle 2\left({\frac {L}{R}}+{\frac {N}{S}}\right)}$

(40) The equation between work and energy may be easily verified. The work done by the electromotive force is

${\displaystyle \xi \int xdt=c^{2}(Rt-L).}$

Work done in overcoming resistance and producing heat,

${\displaystyle R\int x^{2}dt+S\int y^{2}dt=c^{2}(Rt-{\frac {3}{2}}L).}$

Energy remaining in the system,

${\displaystyle ={\frac {1}{2}}c^{2}L}$

(41) If the circuit R is suddenly and completely interrupted while carrying a current ${\displaystyle c}$, then the equation of the current in the secondary coil would be

${\displaystyle y=c{\frac {M}{N}}e^{-{\frac {S}{N}}t}}$

This current begins with a value ${\displaystyle c{\tfrac {M}{N}}}$, and gradually disappears.