# CHAPTER XIII.
## PARALLEL CURRENTS.

Cylindrical Conductors.
682.] In a very important class of electrical arrangements the current is conducted through round wires of nearly uniform section, and either straight, or such that the radius of curvature of the axis of the wire is very great compared with the radius of the transverse section of the wire. In order to be prepared to deal mathematically with such arrangements, we shall begin with the case in which the circuit consists of two very long parallel conductors, with two pieces joining their ends, and we shall confine our attention to a part of the circuit which is so far from the ends of the conductors that the fact of their not being infinitely long does not introduce any sensible change in the distribution of force.
We shall take the axis of $z$ parallel to the direction of the conductors, then, from the symmetry of the arrangements in the part of the field considered, everything will depend on $H$, the component of the vector-potential parallel to $z$.
The components of magnetic induction become, by equations (A),
| $a={\frac {dH}{dy}}$, | (1) |
| $b=-{\frac {dH}{dx}}$, | (2) |
| $c=0$. |
For the sake of generality we shall suppose the coefficient of magnetic induction to be $\mu$, so that $a=\mu \alpha$, $b=\mu \beta$, where $\alpha$ and $\beta$ are the components of the magnetic force.
$u=0$
$v=0$
$4\pi w={\frac {d\beta }{dx}}-{\frac {d\alpha }{dy}}$
$r$
$z$
$x=r\cos \theta$
$y=r\sin \theta$
$\beta$
$\theta$
$z$
$4\pi w={\frac {d\beta }{dr}}+{\frac {1}{r}}\beta ={\frac {1}{r}}{\frac {d}{dr}}(\beta r)$
$C$
$xy$
$r$
$C=\int _{0}^{r}2\pi rw\,dr={\frac {1}{2}}\beta r$
It appears, therefore, that the magnetic force at a given point due to a current arranged in cylindrical strata, whose common axis is the axis of $z$, depends only on the total strength of the current flowing through the strata which lie between the given point and the axis, and not on the distribution of the current among the different cylindrical strata.
For instance, let the conductor be a uniform wire of radius $a$, and let the total current through it be $C$, then, if the current is uniformly distributed through all parts of the section, $w$ will be
$C=\pi wa^{2}$
The current flowing through a circular section of radius $r$, $r$ being less than $a$, is $C^{\prime }=\pi wr^{2}$. Hence at any point within the wire,
| | $\beta ={\frac {2C^{\prime }}{r}}=2C{\frac {r}{a^{2}}}$. | (8) |
| Outside the wire | $\beta =2{\frac {C}{r}}$. | (9) |
In the substance of the wire there is no magnetic potential, for within a conductor carrying an electric current the magnetic force does not fulfil the condition of having a potential.
$\Omega =2C\theta$
$a_{1}$
$a_{2}$
$C$
$C=\pi w(a_{1}^{2}-a_{2}^{2})$
$r$
$a_{1}$
$a_{2}$
$\beta =2C{\frac {1}{a_{1}^{2}-a_{2}^{2}}}\left(r-a{\frac {a_{2}^{2}}{r}}\right)$
$\beta =2{\frac {C}{r}}$
684.] The magnetic induction at any point is $b=\mu \beta$, and since, by equation (2),
| $b$ | $=-{\frac {dH}{dr}}$, | (14) |
| $H$ | $=-\int \mu \beta \,dr$. | (15) |
$H$
$A-2\mu _{0}C\log r$
$\mu _{0}$
$\mu$
$A$
$H=A-2\mu _{0}C\log a_{1}+{\frac {\mu C}{a_{1}^{2}-a_{2}^{2}}}\left(a_{1}^{2}-r^{2}+2a_{2}^{2}\log {\frac {r}{a_{1}}}\right)$
$H$
$H=A-2\mu _{0}C\log a_{1}+\mu C\left(1+{\frac {2a_{2}^{2}}{a_{1}^{2}-a_{2}^{2}}}\log {\frac {a_{2}}{a_{1}}}\right)$
$b$
$T={\frac {1}{2}}\iiint Hw\,dx\,dy\,dz$
$l$
$T={\frac {1}{2}}l\iint Hw\,dx\,dy$
${\frac {2T}{l}}=\iint Hw^{\prime }\,dx^{\prime }\,dy^{\prime }+\iint H^{\prime }w\,dx\,dy+\iint Hw\,dx\,dy+\iint H^{\prime }w^{\prime }\,dx^{\prime }\,dy^{\prime }$
Since the action of the current on any point outside the tube is the same as if the same current had been concentrated at the axis of the tube, the mean value of $H$ for the section of the return current is $A-2\mu _{0}C\log b$, and the mean value of $H^{\prime }$ for the section of the positive current is $A^{\prime }-2\mu _{0}C^{\prime }\log b$.
$T$
$AC^{\prime }-2\mu _{0}CC^{\prime }\log b$ and $A^{\prime }C-2\mu _{0}CC^{\prime }\log b$.
Integrating the two latter terms in the ordinary way, and adding the results, remembering that $C+C^{\prime }=0$, we obtain the value of the kinetic energy $T$. Writing this ${\frac {1}{2}}LC^{2}$, where $L$ is the coefficient of self-induction of the system of two conductors, we find as the value of $L$ for unit of length of the system
| ${\frac {L}{l}}=2\mu _{0}\log {\frac {b^{2}}{a_{1}a_{1}^{\prime }}}$ | $+{\frac {1}{2}}{\frac {a_{1}^{2}-3a_{2}^{2}}{a_{1}^{2}-a_{2}^{2}}}+{\frac {4a_{2}^{4}}{(a_{1}^{2}-a_{2}^{2})^{2}}}\log {\frac {a_{1}}{a_{2}}}$ | |
| | $+{\frac {1}{2}}\mu ^{\prime }{\frac {a_{1}^{\prime 2}-3a_{2}^{\prime 2}}{a_{1}^{\prime 2}-a_{2}^{\prime ^{2}}}}+{\frac {4a_{2}^{\prime 4}}{(a_{1}^{\prime 2}-a_{2}^{\prime 2})^{2}}}\log {\frac {a_{1}^{\prime }}{a_{2}^{\prime }}}$. | (22) |
If the conductors are solid wires, $a_{2}$ and $a_{2}^{\prime }$ are zero, and
| ${\frac {L}{l}}=2\mu _{o}\log {\frac {b^{2}}{a_{1}a_{1}^{\prime }}}+{\frac {1}{2}}(\mu +\mu ^{\prime })$. | (23) |
It is only in the case of iron wires that we need take account of the magnetic induction in calculating their self-induction. In other cases we may make $\mu _{0}$, $\mu$, and $\mu ^{\prime }$ all equal to unity. The smaller the radii of the wires, and the greater the distance between them, the greater is the self-induction.
To find the Repulsion, $X$, between the Two Portions of Wire.
686.] By Art. 580 we obtain for the force tending to increase $b$,
| $X$ $={\frac {1}{2}}{\frac {dL}{db}}C^{2}$, $=2\mu _{0}{\frac {l}{b}}C^{2}$, | (24) |
which agrees with Ampère's formula, when $\mu _{0}=1$, as in air.
687.] If the length of the wires is great compared with the distance between them, we may use the coefficient of self-induction to determine the tension of the wires arising from the action of the current.
If $Z$ is this tension,
| $Z$ | $={\frac {1}{2}}{\frac {dL}{dl}}C^{2}$, | |
| | $=C^{2}\left\{\mu _{0}\log {\frac {c^{2}}{a_{1}a_{1}^{\prime }}}+{\frac {\mu }{2}}\right\}$. | (25) |
In one of Ampère's experiments the parallel conductors consist of two troughs of mercury connected with each other by a floating bridge of wire. When a current is made to enter at the extremity of one of the troughs, to flow along it till it reaches one extremity ​of the floating wire, to pass into the other trough through the floating bridge, and so to return along the second trough, the floating bridge moves along the troughs so as to lengthen the part of the mercury traversed by the current.
Professor Tait has simplified the electrical conditions of this experiment by substituting for the wire a floating siphon of glass filled with mercury, so that the current flows in mercury throughout its course.
Fig. 40.
This experiment is sometimes adduced to prove that two elements of a current in the same straight line repel one another, and thus to shew that Ampère's formula, which indicates such a repulsion of collinear elements, is more correct than that of Grassmann, which gives no action between two elements in the same straight line; Art. 526.
But it is manifest that since the formulae both of Ampère and of Grassmann give the same results for closed circuits, and since we have in the experiment only a closed circuit, no result of the experiment can favour one more than the other of these theories.
In fact, both formulae lead to the very same value of the repulsion as that already given, in which it appears that $b$, the distance between the parallel conductors is an important element.
When the length of the conductors is not very great compared with their distance apart, the form of the value of $L$ becomes somewhat more complicated.
$L$
$b=a_{1}+a_{2}$
$L=2l\log \left({\frac {(a_{1}+a_{2})^{2}}{a_{1}a_{2}}}+{\frac {1}{2}}\right)$
​This is a minimum when $a_{1}=a_{2}$, and then
| $L$ | $=2l(\log 4+{\frac {1}{2}})$, | |
| | $=2l(1.8863)$, |
| | $=3.7726l$. | (27) |
This is the smallest value of the self-induction of a round wire doubled on itself, the whole length of the wire being $2l$.
Since the two parts of the wire must be insulated from each other, the self-induction can never actually reach this limiting value. By using broad flat strips of metal instead of round wires the self-induction may be diminished indefinitely.
On the Electromotive Force required to produce a Current of Varying Intensity along a Cylindrical Conductor.
689.] When the current in a wire is of varying intensity, the electromotive force arising from the induction of the current on itself is different in different parts of the section of the wire, being in general a function of the distance from the axis of the wire as well as of the time. If we suppose the cylindrical conductor to consist of a bundle of wires all forming part of the same circuit, so that the current is compelled to be of uniform strength in every part of the section of the bundle, the method of calculation which we have hitherto used would be strictly applicable. If, however, we consider the cylindrical conductor as a solid mass in which electric currents are free to flow in obedience to electromotive force, the intensity of the current will not be the same at different distances from the axis of the cylinder, and the electromotive forces themselves will depend on the distribution of the current in the different cylindric strata of the wire.
The vector-potential $H$, the density of the current $w$, and the electromotive force at any point, must be considered as functions of the time and of the distance from the axis of the wire.
The total current, $C$, through the section of the wire, and the total electromotive force, $E$, acting round the circuit, are to be regarded as the variables, the relation between which we have to find.
$H$
$H=S+T_{0}+T_{1}r^{2}+\mathrm {\&c.} +T_{n}r^{2n}$
$S$
$T_{0}$
$T_{1}$
${\frac {d^{2}H}{dr^{2}}}+{\frac {1}{r}}{\frac {dH}{dr}}=-4\pi w$
$-\pi w=T_{1}+\mathrm {\&c} +n^{2}T_{n}r^{2n-2}$
$\rho$
$\rho w$
$H$
$\rho w=-{\frac {d\Psi }{dz}}-{\frac {dH}{dt}}$
$-\rho w={\frac {d\Psi }{dz}}+{\frac {dS}{dt}}+{\frac {dT_{0}}{dt}}+{\frac {dT_{1}}{dt}}r^{2}+\mathrm {\&c.} +{\frac {dT_{n}}{dt}}r^{2n}$
Comparing the coefficients of like powers of $r$ in equations (3) and (5),
| $T_{1}$ | $={\frac {\pi }{\rho }}\left({\frac {d\Psi }{dz}}+{\frac {dS}{dt}}+{\frac {dT_{0}}{dt}}\right)$, | (6) |
| $T_{2}$ | $={\frac {\pi }{\rho }}{\frac {dT_{1}}{dt}}$, | (7) |
| $T_{n}$ | $={\frac {\pi }{\rho }}{\frac {1}{n^{2}}}{\frac {dT_{n-1}}{dt}}$. | (8) |
${\frac {dS}{dt}}=-{\frac {d\Psi }{dz}}$
$T_{0}=T$
$T_{1}={\frac {\pi }{\rho }}{\frac {dT}{dt}}$
$T_{n}={\frac {\pi ^{n}}{\rho ^{n}}}{\frac {1}{(|{\underline {n}})^{2}}}{\frac {d^{n}T}{dt^{n}}}$
$C$
$w$
$a$
$C=2\pi \int _{0}^{a}wr\,dr$
$\pi w$
$C=-(T_{1}a^{2}+\mathrm {\&c.} +nT_{n}a^{2n})$
$H$
$C$
$H$
$AC$
$A$
$H=AC$
$r=a$
$AC=S+T_{0}+T_{1}a^{2}+\mathrm {\&c.} +T_{n}a^{2n}$
If we now write ${\frac {\pi a^{2}}{\rho }}=\alpha$, $\alpha$ is the value of the conductivity of unit of length of the wire, and we have
| $C$ | $=-\left(\alpha {\frac {dT}{dt}}+{\frac {2\alpha ^{2}}{1^{2}2^{2}}}{\frac {d^{2}T}{dt^{2}}}+\mathrm {\&c.} +{\frac {n\alpha ^{n}}{(|{\underline {n}})^{2}}}{\frac {d^{n}T}{dt^{n}}}+\mathrm {\&c.} \right)$, | (14) |
| $AC-S$ | $=T+\alpha {\frac {dT}{dt}}+{\frac {\alpha ^{2}}{1^{2}2^{2}}}{\frac {d^{2}T}{dt^{2}}}+\mathrm {\&c.} +{\frac {\alpha ^{n}}{(|{\underline {n}})^{2}}}{\frac {d^{n}T}{dt^{n}}}+\mathrm {\&c.}$ | (15) |
$T$
| $\alpha \left(A{\frac {dC}{dt}}-{\frac {dS}{dt}}\right)+C+{\frac {1}{2}}\alpha {\frac {dC}{dt}}-{\frac {1}{12}}\alpha ^{2}$ | ${\frac {d^{2}C}{dt^{2}}}+{\frac {1}{48}}\alpha ^{3}{\frac {d^{3}C}{dt^{3}}}$ | |
| | $-{\frac {1}{160}}\alpha ^{4}{\frac {d^{4}C}{dt^{4}}}+\mathrm {\&c.} =0$. | (16) |
$l$
$R$
$E$
${\frac {dS}{dt}}=-{\frac {E}{l}}$
$\alpha ={\frac {l}{R}}$
$E=RC+l(A+{\frac {1}{2}}){\frac {dC}{dt}}-{\frac {1}{12}}{\frac {l^{2}}{R}}{\frac {d^{2}C}{dt^{2}}}+{\frac {1}{48}}{\frac {l^{3}}{R^{2}}}{\frac {d^{3}C}{dt^{3}}}-{\frac {1}{180}}{\frac {l^{4}}{R^{3}}}{\frac {d^{4}C}{dt^{4}}}+\mathrm {\&c}$
The first term, $RC$, of the right-hand member of this equation expresses the electromotive force required to overcome the resistance according to Ohm's law.
The second term, $l(A+{\frac {1}{2}}){\frac {dC}{dt}}$, expresses the electromotive force which would be employed in increasing the electrokinetic momentum of the circuit, on the hypothesis that the current is of uniform strength at every point of the section of the wire.
The remaining terms express the correction of this value, arising from the fact that the current is not of uniform strength at different distances from the axis of the wire. The actual system of currents has a greater degree of freedom than the hypothetical system, in which the current is constrained to be of uniform strength throughout the section. Hence the electromotive force required to produce a rapid change in the strength of the current is somewhat less than it would be on this hypothesis.
$\int E\,dt=R\int C\,dt+l(A+{\frac {1}{2}})C-{\frac {1}{12}}{\frac {l^{2}}{R}}{\frac {dC}{dt}}+\mathrm {\&c}$
$C_{0}$
$C_{1}$
$C$
$\int E\,dt=R\int C\,dt+l(A+{\frac {1}{2}})(C_{1}-C_{0})$
On the Geometrical Mean Distance of Two Figures in a Plane.
$\iiiint \log r\;dx\,dy\,dx^{\prime }\,dy^{\prime }$
$dx\,dy$
$dx^{\prime }\,dy^{\prime }$
$r$
$R$
$A_{1}A_{2}\log R$
$A_{1}$
$A_{2}$
$R$
$R$
$R$
$R$
$r$
$R_{A}$
$R_{B}$
$A$
$B$
$C$
$R_{A+B}$
$C$
$(A+B)\log R_{A+B}=A\log R_{A}+B\log R_{B}$
By means of this relation we can determine $R$ for a compound figure when we know $R$ for the parts of the figure.
$R$
$O$
$AB$
$OP$
$AB$
$AB(\log R+1)=AP\log OA+PB\log OB+OP\;{\widehat {AOB}}$
Fig. 41.
​(2) For two lines (Fig. 42) of lengths $a$ and $b$ drawn perpendicular to the extremities of a line of length $c$ and on the same side of it.
| $ab(2\log R+3)=$ | $(c^{2}-(a-b)^{2})\log {\sqrt {c^{2}+(a-b)^{2}}}+c^{2}\log c$ |
| | $+(a^{2}-c^{2})\log {\sqrt {a^{2}+c^{2}}}+(b^{2}-c^{2})\log {\sqrt {b^{2}+c^{2}}}$ |
| | $-c(a-b)\tan ^{-1}{\frac {a-b}{c}}+ac\tan ^{-1}{\frac {a}{c}}+bc\tan ^{-1}{\frac {b}{c}}$. |
Fig. 42.
(3) For two lines, $PQ$ and $RS$ (Fig. 43), whose directions intersect at $O$.
* $=\log PR(2OP.OR\sin ^{2}O-PR^{2}\cos O)$
* $+\log QS(2OQ.OS\sin ^{2}O-QS^{2}\cos O)$
* $-\log PS(2OP.OS\sin ^{2}O-PS^{2}\cos O)$
* $-\log QR(2OQ.OR\sin ^{2}O-QR^{2}\cos O)$
Fig. 43.
(4) For a point $O$ and a rectangle $ABCD$ (Fig. 44). Let $OP$, $OQ$, $OR$, $OS$, be perpendiculars on the sides, then
* $=2.OP.OQ\log OA+2.OQ.OR\log OB$
* $+2.OR.OS\log OC+2.OS.OP\log OD$
* $+OP^{2}.{\widehat {DOA}}+OQ^{2}.{\widehat {AOB}}$
* $+OR^{2}.{\widehat {BOC}}+OS^{2}.{\widehat {COD}}$.
Fig. 44.
​(5) It is not necessary that the two figures should be different, for we may find the geometric mean of the distances between every pair of points in the same figure. Thus, for a straight line of length $a$,
| | $\log R$ | $=\log a-{\frac {3}{2}}$, |
| or | $R$ | $=ae^{-{\frac {3}{2}}}$, |
| | $R$ | $=0.22313a$. |
(6) For a rectangle whose sides are $a$ and $b$,
| $\log R=\log {\sqrt {a^{2}+b^{2}}}-{\frac {1}{6}}{\frac {a^{2}}{b^{2}}}\log$ | ${\sqrt {1+{\frac {b^{2}}{a^{2}}}}}-{\frac {1}{6}}{\frac {b^{2}}{a^{2}}}\log {\sqrt {1+{\frac {a^{2}}{b^{2}}}}}$ |
| | $+{\frac {2}{3}}{\frac {a}{b}}\tan ^{-1}{\frac {b}{a}}+{\frac {2}{3}}{\frac {b}{a}}\tan ^{-1}{\frac {a}{b}}-{\frac {25}{12}}$. |
When the rectangle is a square, whose side is $a$,
| $\log R$ | $=\log a+{\frac {1}{3}}\log 2+{\frac {\pi }{3}}-{\frac {25}{12}}$, |
| $R$ | $=0.44705a$. |
(7) The geometric mean distance of a point from a circular line is equal to the greater of the two quantities, its distance from the centre of the circle, and the radius of the circle.
$\log R={\frac {a_{1}^{2}\log a_{1}-a_{2}^{2}\log a_{2}}{a_{1}^{2}-a_{2}^{2}}}-{\frac {1}{2}}$,
$a_{1}$
$a_{2}$
$R$
(9) The geometric mean distance of all pairs of points in the ring is found from the equation
$\log R=\log a_{1}-{\frac {a_{2}^{4}}{(a_{1}^{2}-a_{2}^{2})^{2}}}\log {\frac {a_{1}}{a_{2}}}+{\frac {1}{4}}{\frac {3a_{2}^{2}-a_{1}^{2}}{a_{1}^{2}-a_{2}^{2}}}$.
For a circular area of radius $a$, this becomes
| | $\log R$ | $=\log a-{\frac {1}{4}}$, |
| or | $R$ | $=ae^{-{\frac {1}{4}}}$, |
| | $R$ | $=0.7788a$. |
$R=a$.
693.] In calculating the coefficient of self-induction of a coil of uniform section, the radius of curvature being great compared with ​the dimensions of the transverse section, we first determine the geometric mean of the distances of every pair of points of the section by the method already described, and then we calculate the coefficient of mutual induction between two linear conductors of the given form, placed at this distance apart.
This will be the coefficient of self-induction when the total current in the coil is unity, and the current is uniform at all points of the section.
But if there are $n$ windings in the coil we must multiply the coefficient already obtained by $n^{2}$, and thus we shall obtain the coefficient of self-induction on the supposition that the windings of the conducting wire fill the whole section of the coil.
But the wire is cylindric, and is covered with insulating material, so that the current, instead of being uniformly distributed over the section, is concentrated in certain parts of it, and this increases the coefficient of self-induction. Besides this, the currents in the neighbouring wires have not the same action on the current in a given wire as a uniformly distributed current.
The corrections arising from these considerations may be determined by the method of the geometric mean distance. They are proportional to the length of the whole wire of the coil, and may be expressed as numerical quantities, by which we must multiply the length of the wire in order to obtain the correction of the coefficient of self-induction.
Fig. 45. Let the diameter of the wire be $d$. It is covered with insulating material, and wound into a coil. We shall suppose that the sections of the wires are in square order, as in Fig. 45, and that the distance between the axis of each wire and that of the next is $D$, whether in the direction of the breadth or the depth of the coil. $D$ is evidently greater than $d$.
We have first to determine the excess of self-induction of unit of length of a cylindric wire of diameter $d$ over that of unit of length of a square wire of side $D$, or
* $2\log {\frac {R\;\;\mathrm {for\;the\;square} }{R\;\;\mathrm {for\;the\;circle} }}$
* $=2\left(\log {\frac {D}{d}}+{\frac {4}{3}}\log 2+{\frac {\pi }{3}}-{\frac {11}{6}}\right)$
* $=2\left(\log {\frac {D}{d}}+0.1380606\right)$.
​The inductive action of the eight nearest round wires on the wire under consideration is less than that of the corresponding eight square wires on the square wire in the middle by 2 × (.01971).
$2(\log _{\epsilon }{\frac {D}{d}}+0.11835)$.
$L=n^{2}M+2l(\log _{\epsilon }{\frac {D}{d}}+0.11835)$,
where $n$ is the number of windings, and $l$ the length of the wire, $M$ the mutual induction of two circuits of the form of the mean wire of the coil placed at a distance $R$ from each other, where $R$ is the mean geometric distance between pairs of points of the section. $D$ is the distance between consecutive wires, and $d$ the diameter of the wire.
* ↑ Trans. R. S. Edin., 1871–2.