Two $$10$$ cm long, straight wires, each carrying a current of $$5$$ A are kept parallel to each other. If each wire experienced a force of $$10^{-5}$$ N, then separation between the wires is ______ cm.

We need to find the separation between two parallel current-carrying wires.

Since the length of each wire is $$L = 10$$ cm = $$0.1$$ m, each carries a current of $$I_1 = I_2 = 5$$ A, and the force experienced by each wire is $$F = 10^{-5}$$ N, we apply the expression for the force between two parallel currents. The formula is $$F = \frac{\mu_0 I_1 I_2 L}{2\pi d}$$ where $$d$$ is the separation between the wires and $$\mu_0 = 4\pi \times 10^{-7}$$ T m/A.

Rearranging this expression for the distance gives

$$d = \frac{\mu_0 I_1 I_2 L}{2\pi F}$$

Substituting the known values yields

$$d = \frac{4\pi \times 10^{-7} \times 5 \times 5 \times 0.1}{2\pi \times 10^{-5}}$$

$$= \frac{4\pi \times 10^{-7} \times 2.5}{2\pi \times 10^{-5}}$$

$$= \frac{4 \times 2.5 \times 10^{-7}}{2 \times 10^{-5}}$$

$$= \frac{10 \times 10^{-7}}{2 \times 10^{-5}}$$

$$= \frac{10^{-6}}{2 \times 10^{-5}}$$

$$= \frac{1}{20} = 0.05$$ m = $$5$$ cm

Therefore, the separation between the wires is 5 cm.