Irodov 3.256: Matched Load, No Force Between Wires – Amit & Deepali Sinha Foundation

Problem Irodov 3.256

Two long parallel wires of negligible resistance are joined at one end through a resistor $R$ and at the other end through a dc voltage source. The distance between the wires’ axes is $\eta = 20$ times the radius of each wire. For what value of $R$ is the total force between the wires zero?

The two long parallel wires, joined at one end through the dc source U and at the other through the resistor R. — The two long parallel wires, joined at one end through the dc source $U$ and at the other through the resistor $R$ .

Given

wires of negligible resistance
axes $\eta = 20$ radii apart

Find

the resistance

R

for which the wires do not interact

Solution

Step 1Two forces of opposite sign

Let $U$ be the source voltage, $d$ the distance between the axes and $r$ the radius, so that $d = \eta r$ . The wires have no resistance, so each is an equipotential and the whole voltage $U$ lies across $R$ ; the current out along one wire and back along the other is

I = \frac{U}{R} \tag{1}

The pair seen end on from the source, the positive (upper) wire on the left: the charges λ per unit length and the opposite currents I, with the axes d = η r apart (radius not to scale). — The pair seen end on from the source, the positive (upper) wire on the left: the charges $\pm\lambda$ per unit length and the opposite currents $I$ , with the axes $d = \eta r$ apart (radius not to scale).

Two things therefore act between the wires: opposite currents, which repel, and opposite charges $\pm\lambda$ per unit length, held there by the voltage $U$ , which attract. The force is zero when the two balance.

Step 2The magnetic repulsion

The field of one wire at the axis of the other follows from Ampere’s law on a circle of radius $d$ , round which $B$ is tangential and the same all round:

B\cdot 2\pi d = \mu_0 I

A length of the other wire carrying $I$ across this field is pushed away with the force per unit length

F_m = I B = \frac{\mu_0 I^2}{2\pi d} \tag{2}

Step 3The charge on the wires

Since $r \ll d$ , each wire’s charge acts on the other as a line charge on its axis, whose field follows from Gauss’s law on a coaxial cylinder of radius $x$ , over which $E$ is radial and the same all round:

E\cdot 2\pi x\,\ell = \frac{\lambda\ell}{\varepsilon_0}

Between the wires the two fields add. Integrating from the surface of the positive wire to the surface of the negative one gives the voltage:

U = \int_r^{d - r}\frac{\lambda}{2\pi\varepsilon_0}\left(\frac{1}{x} + \frac{1}{d - x}\right)dx = \frac{\lambda}{\pi\varepsilon_0}\ln\frac{d - r}{r} \approx \frac{\lambda}{\pi\varepsilon_0}\ln\eta

to the leading order in $r/d$ that the line-charge picture itself keeps, so the charge per unit length is

\lambda = \frac{\pi\varepsilon_0 U}{\ln\eta} \tag{3}

Step 4The electric attraction

The charge $\lambda$ per unit length of one wire sits in the field of the other, $E = \lambda/(2\pi\varepsilon_0 d)$ at the distance $d$ , so it is pulled toward it with the force per unit length

F_e = \lambda E = \frac{\lambda^2}{2\pi\varepsilon_0 d} \tag{4}

Step 5The balance

Setting (4) equal to (2), the common factor $1/(2\pi d)$ cancels:

\frac{\lambda^2}{\varepsilon_0} = \mu_0 I^2, \qquad \lambda = \sqrt{\varepsilon_0\mu_0}\,I \tag{5}

Substituting (3) and (1):

\frac{\pi\varepsilon_0 U}{\ln\eta} = \sqrt{\varepsilon_0\mu_0}\,\frac{U}{R}

and $U$ cancels. Solving for $R$ , with $\sqrt{\varepsilon_0\mu_0}/\varepsilon_0 = \sqrt{\mu_0/\varepsilon_0}$ :

R = \frac{\sqrt{\varepsilon_0\mu_0}\,\ln\eta}{\pi\varepsilon_0}

Numerically,

R = \frac{\ln 20}{\pi}\sqrt{\frac{4\pi\times10^{-7}\text{ H/m}}{8.85\times10^{-12}\text{ F/m}}} = \frac{2.996}{\pi}\,(376.8\ \Omega)

R = \frac{\ln\eta}{\pi}\sqrt{\frac{\mu_0}{\varepsilon_0}} \approx 0.36\ \text{k}\Omega

(6)

For a larger $R$ the current is weaker and the wires attract; for a smaller one they repel. The exact treatment of two round wires replaces $\ln\eta$ by $\operatorname{arcosh}(\eta/2) = 2.993$ and the distance in (4) by a slightly shorter one, corrections of relative order $1/\eta^2$ that lower $R$ by about 0.3 per cent.

Check

Units: $\sqrt{\mu_0/\varepsilon_0}$ has the units $\sqrt{(\text{H/m})/(\text{F/m})} = \sqrt{\text{H}/\text{F}} = \sqrt{(\Omega\,\text{s})/(\text{s}/\Omega)} = \Omega$ , a resistance, as it must be. Limits: as $R \to 0$ the current grows without bound while the charge stays fixed by $U$ , so repulsion wins; as $R \to \infty$ the current vanishes and only the attraction of the charges is left; the balance lies between, as found.

Remark

The balance (5) reads $\lambda c = I$ with $c = 1/\sqrt{\varepsilon_0\mu_0}$ : the forces cancel when the current equals the charge per unit length carried at the speed of light. And $(\sqrt{\mu_0/\varepsilon_0}/\pi)\ln\eta$ is the characteristic impedance of a two-wire line, so the forces vanish exactly when the line is terminated in a matched load, the condition under which a signal running down it at the speed of light is absorbed at the end without reflection.