Question

Two long parallel wires of negligible resistance are connected at one end to a resistance $R$ and at the other end to a constant voltage source of voltage $V$. The distance between the axes of the wires is $\eta$ times greater than the cross-sectional radius of each wire. At what value of resistance $R$, does the resultant force of interaction between the wires will become zero?

Answer 1

$\frac{\mu_0 c}{\pi} \cosh^{-1}(\eta/2)$

Answer 2

The circuit consists of a voltage source $V$, two long parallel wires of negligible resistance, and a resistance $R$ connected in series. The current flowing through the circuit is $I = V/R$, assuming the wire resistance is zero. The current flows in one wire and returns in the other, so the currents in the two parallel wires are in opposite directions.

Two long parallel wires carrying currents $I_1$ and $I_2$ separated by a distance $d$ exert a magnetic force on each other. The force per unit length is given by $F_B/L = \frac{\mu_0 I_1 I_2}{2\pi d}$. Since the currents are in opposite directions ($I_1 = I, I_2 = -I$), the force is repulsive, and its magnitude per unit length is $F_B/L = \frac{\mu_0 I^2}{2\pi d}$. In this problem, the distance between the axes of the wires is $d = \eta a$, where $a$ is the radius of each wire. So, $F_B/L = \frac{\mu_0 I^2}{2\pi \eta a}$.

The two parallel wires also form a capacitor, as there is a potential difference between them. Since the wires have negligible resistance and are connected to a constant voltage source $V$ at one end, the potential difference between the wires is $V$ along their entire length. The capacitance per unit length of two long parallel wires of radius $a$ separated by a distance $d$ between their centers is given by $C/L = \frac{\pi \epsilon_0}{\cosh^{-1}(d/2a)}$. With $d = \eta a$, the capacitance per unit length is $C/L = \frac{\pi \epsilon_0}{\cosh^{-1}(\eta a/2a)} = \frac{\pi \epsilon_0}{\cosh^{-1}(\eta/2)}$.

The charge per unit length on the wires is related to the potential difference by $\lambda = (C/L) V$. One wire has charge density $+\lambda$ and the other $-\lambda$. These charged wires exert an electrostatic force on each other. The force per unit length between two long parallel lines of charge with linear charge densities $\lambda_1$ and $\lambda_2$ separated by a distance $d$ is $F_E/L = \frac{\lambda_1 \lambda_2}{2\pi \epsilon_0 d}$. Since $\lambda_1 = +\lambda$ and $\lambda_2 = -\lambda$, the force is attractive, and its magnitude per unit length is $F_E/L = \frac{\lambda^2}{2\pi \epsilon_0 d}$. Substituting $d = \eta a$ and $\lambda = (C/L)V$, we get $F_E/L = \frac{((C/L)V)^2}{2\pi \epsilon_0 \eta a} = \frac{1}{2\pi \epsilon_0 \eta a} \left(\frac{\pi \epsilon_0 V}{\cosh^{-1}(\eta/2)}\right)^2 = \frac{\pi (\epsilon_0 V)^2}{2\eta a (\cosh^{-1}(\eta/2))^2}$.

The resultant force of interaction between the wires is the sum of the magnetic force (repulsive) and the electrostatic force (attractive). The resultant force is zero when the magnitudes of these forces are equal: $F_B/L = F_E/L$ $\frac{\mu_0 I^2}{2\pi \eta a} = \frac{\lambda^2}{2\pi \epsilon_0 \eta a}$ $\mu_0 I^2 = \frac{\lambda^2}{\epsilon_0}$ $\mu_0 \epsilon_0 I^2 = \lambda^2$

We know that $\mu_0 \epsilon_0 = 1/c^2$, where $c$ is the speed of light in vacuum. $\frac{I^2}{c^2} = \lambda^2 \implies \lambda = I/c$. (Taking the positive root for magnitude).

Substitute $\lambda = (C/L)V$: $I/c = (C/L)V$ Substitute $I = V/R$: $(V/R)/c = (C/L)V$ $1/(Rc) = C/L$ $R = \frac{1}{c (C/L)}$

Substitute the expression for $C/L$: $R = \frac{1}{c \frac{\pi \epsilon_0}{\cosh^{-1}(\eta/2)}} = \frac{\cosh^{-1}(\eta/2)}{\pi c \epsilon_0}$

Using $\epsilon_0 = \frac{1}{\mu_0 c^2}$: $R = \frac{\cosh^{-1}(\eta/2)}{\pi c (1/(\mu_0 c^2))} = \frac{\mu_0 c^2 \cosh^{-1}(\eta/2)}{\pi c} = \frac{\mu_0 c}{\pi} \cosh^{-1}(\eta/2)$.

The value of resistance $R$ at which the resultant force of interaction between the wires is zero is $R = \frac{\mu_0 c}{\pi} \cosh^{-1}(\eta/2)$.