Answer
$R = 3\Omega$
Work Step by Step
The force exerted on one another by parallel wires is given by
$F = \frac{\mu_0 l I_1 I_2}{2\pi d} $
Where $\mu_0 = 4\pi \times 10^{-7} \frac{Tm}{A} $ is the permeability constant, $l$ is the length of the wires, and $d$ is the separation between the wires.
Let $I_1$ represent the current in the circuit on the left, and $I_2$ represent the current in the circuit on the right. To find $R$, we must do 3 things:
1. Find $I_1$ using Kirchhoff's Voltage Law and Ohm's Law.
2. Solve for $I_2$
3. Find $R$ using KVL and Ohm's Law.
1. By Ohm's Law, we know that $\Delta V = IR$ and by KVL, we know the sum of the voltage changes around a loop is equal to zero. So
$9V = \Delta V_{2\Omega} = (2\Omega)I_1$, therefore $I_1 = 4.5A$.
2. Solve for $I_2$. Rearranging our equation for $F$, we get $I_2 = \frac{F2\pi d}{\mu_0 l I_1}$. We are given $F = 5.4\times 10^{-5} N$, $d = .005 m$, $l = .10m$, so we plug in what we know and get:
$I_2 = \frac{(5.4\times 10^{-5} N)(2\pi)(.005m)}{(4\pi\times 10^{-7})(.10m)(4.5A)} = 3A$
3. By KVL and Ohm's law, then, $(3A)R = 9V$ and so $R = 3\Omega$.
