Two long straight wires $$P$$ and $$Q$$ carrying equal current 10 A each were kept parallel to each other at 5 cm distance. Magnitude of magnetic force experienced by 10 cm length of wire $$P$$ is $$F_1$$. If distance between wires is halved and currents on them are doubled, force $$F_2$$ on 10 cm length of wire $$P$$ will be:

Two long parallel wires $$P$$ and $$Q$$ carry equal current $$I = 10$$ A, separated by distance $$d = 5$$ cm. We need to find the new force when the distance is halved and the currents are doubled. The magnetic force per unit length between two parallel wires is $$\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}$$ and for a length $$L = 10$$ cm $$= 0.1$$ m of wire $$P$$ this becomes $$F = \frac{\mu_0 I_1 I_2 L}{2\pi d}$$.

For the original configuration, substituting $$I_1 = I_2 = 10$$ A and $$d = 0.05$$ m gives $$F_1 = \frac{\mu_0 \times 10 \times 10 \times 0.1}{2\pi \times 0.05} = \frac{10\mu_0}{\pi \times 0.1} = \frac{100\mu_0}{\pi}.$$

In the new configuration where the distance is halved ($$d' = 2.5$$ cm) and currents are doubled ($$I' = 20$$ A), we have $$F_2 = \frac{\mu_0 \times 20 \times 20 \times 0.1}{2\pi \times 0.025} = \frac{40\mu_0}{\pi \times 0.05} = \frac{800\mu_0}{\pi}.$$

The ratio of the forces is $$\frac{F_2}{F_1} = \frac{(I')^2 / d'}{I^2 / d} = \frac{(20)^2 \times 0.05}{(10)^2 \times 0.025} = \frac{400 \times 0.05}{100 \times 0.025} = \frac{20}{2.5} = 8.$$ Alternatively, doubling each current increases the force by a factor of $$2 \times 2 = 4$$ and halving the distance increases it by a factor of $$2$$, giving a total factor of $$4 \times 2 = 8$$. Hence $$F_2 = 8F_1.$$

Answer: Option (1): $$\boxed{8F_1}$$.