Let's solve the problem step by step: --- ## **Step 1: Find the force per unit length between two parallel wires** The magnetic force per unit length between two parallel wires carrying currents \( I_1 \) and \( I_2 \), separated by distance \( r \), is given by: \[ \frac{F}{L} = \frac{\mu_}{2\pi} \frac{I_1 I_2}{r} \] where - \( \mu_ = 4\pi \times 10^{-7} \, \text{T}\cdot\text{m}/\text{A} \) - \( I_1 = I_2 = 10\,\text{A} \) - \( r = .02\,\text{m} \) (since 2. cm = .02 m) **Explanation:** This equation tells us how much force each meter of one wire experiences due to the other, depending on their currents and separation. --- ## **Step 2: Calculate net force on the second wire** The second wire (middle) is affected by both the top and bottom wires: - **Wire 1 (top):** Current is to the right; distance = .02 m above wire 2. - **Wire 3 (bottom):** Current is to the right; distance = .02 m below wire 2. Since the currents in wire 1 and wire 3 are both to the right, and wire 2 is to the left: - The force from each is **attractive** (opposite current directions), so both forces act in the **same direction** (toward the other wires). Calculate force from one wire: \[ F_{12} = \left(\frac{\mu_}{2\pi}\right) \frac{I_1 I_2}{r} \cdot L = \left(\frac{4\pi \times 10^{-7}}{2\pi}\right) \frac{10 \times 10}{.02} \cdot .5 \] \[ = (2 \times 10^{-7}) \frac{100}{.02} \cdot .5 \] \[ = (2 \times 10^{-7}) \times 500 \times .5 \] \[ = (2 \times 10^{-7}) \times 250 \] \[ = 5 \times 10^{-4}\ \text{N} \] Since both wires exert the same magnitude of force on wire 2 and in the same direction: \[ F_{net} = 2 \times 5 \times 10^{-4} = 1 \times 10^{-3}\ \text{N} \] **Explanation:** Both adjacent wires pull wire 2 with equal force, so the total magnitude is double the single-wire value. --- ## **Final Answer** **The magnitude of the net force on the second wire is:** \[ \boxed{1 \times 10^{-3}\ \text{N}} \] --- If you also need the answer for the third wire, let me know! The process is almost identical.