1. **Problem statement:** Two long parallel wires A and B carry currents $I_A = 8.0$ A and $I_B = 5.0$ A in the same direction, separated by distance $d = 4.0$ cm = 0.04 m. Find the force on a 10 cm = 0.10 m section of wire A due to wire B. 2. **Formula used:** The magnetic force per unit length between two parallel currents is given by $$F/L = \frac{\mu_0 I_A I_B}{2 \pi d}$$ where $\mu_0 = 4\pi \times 10^{-7}$ T·m/A is the permeability of free space. 3. **Calculate force per unit length:** $$F/L = \frac{4\pi \times 10^{-7} \times 8.0 \times 5.0}{2 \pi \times 0.04} = \frac{4\pi \times 10^{-7} \times 40}{2 \pi \times 0.04}$$ Simplify numerator and denominator: $$F/L = \frac{4\pi \times 10^{-7} \times 40}{2 \pi \times 0.04} = \frac{160\pi \times 10^{-7}}{2 \pi \times 0.04}$$ Cancel $\pi$: $$F/L = \frac{160 \times 10^{-7}}{2 \times 0.04} = \frac{160 \times 10^{-7}}{0.08}$$ Calculate: $$F/L = 2.0 \times 10^{-4} \text{ N/m}$$ 4. **Calculate force on 10 cm section:** $$F = (F/L) \times L = 2.0 \times 10^{-4} \times 0.10 = 2.0 \times 10^{-5} \text{ N}$$ 5. **Direction:** Since currents are in the same direction, the force is attractive, so wire A is pulled toward wire B. **Final answer:** $$\boxed{2.0 \times 10^{-5} \text{ N (attractive force on wire A)}}$$