Let's break down the problem and solve it step by step: ## **Step 1: Understanding the Problem** Three parallel wires (A, B, C), each carrying current **upward**. Given: - Current in wire A, \( I_A = 25 \) A - Current in wire B, \( I_B = 22 \) A - Length of segment of wire B, \( L = 30 \) cm \( = 0.30 \) m - Distance from A to B, \( r_1 = 20 \) cm \( = 0.20 \) m - Distance from C to B, \( r_2 = 6.0 \) cm \( = 0.060 \) m - Current in wire C, \( I_C = 13 \) A We need to find the **net force** on wire B due to wires A and C. --- ## **Step 2: Formula for Force Between Two Parallel Currents** The force per unit length between two parallel currents is: \[ F = \frac{\mu_0}{2\pi} \frac{I_1 I_2}{r} L \] Where: - \( \mu_0 = 4\pi \times 10^{-7} \) T·m/A - \( I_1 \) and \( I_2 \) are the currents - \( r \) is the separation - \( L \) is the length --- ## **Step 3: Find the Individual Forces** ### **Force by A on B (\( F_{AB} \)):** \[ F_{AB} = \frac{\mu_0}{2\pi} \frac{I_A I_B}{r_1} L \] Substitute: - \( I_A = 25 \) A - \( I_B = 22 \) A - \( r_1 = 0.20 \) m - \( L = 0.30 \) m \[ F_{AB} = \frac{4\pi \times 10^{-7}}{2\pi} \cdot \frac{25 \cdot 22}{0.20} \cdot 0.30 \] \[ = 2 \times 10^{-7} \cdot \frac{550}{0.20} \cdot 0.30 \] \[ = 2 \times 10^{-7} \cdot 2750 \cdot 0.30 \] \[ = 2 \times 10^{-7} \cdot 825 \] \[ = 1.65 \times 10^{-4} \, \text{N} \] --- ### **Force by C on B (\( F_{CB} \)):** \[ F_{CB} = \frac{\mu_0}{2\pi} \frac{I_C I_B}{r_2} L \] - \( I_C = 13 \) A - \( r_2 = 0.060 \) m \[ F_{CB} = 2 \times 10^{-7} \cdot \frac{13 \cdot 22}{0.060} \cdot 0.30 \] \[ = 2 \times 10^{-7} \cdot \frac{286}{0.060} \cdot 0.30 \] \[ = 2 \times 10^{-7} \cdot 4766.67 \cdot 0.30 \] \[ = 2 \times 10^{-7} \cdot 1430 \] \[ = 2.86 \times 10^{-4} \, \text{N} \] --- ## **Step 4: Direction of Forces** - **Currents are in the same direction:** The wires attract each other. - **Assume A is to the left, C is to the right of B:** - Force by A on B: leftward (B attracted towards A) - Force by C on B: rightward (B attracted towards C) --- ## **Step 5: Net Force on B** Positive = rightward, Negative = leftward. \[ F_\text{net} = F_{CB} - F_{AB} \] \[ F_\text{net} = 2.86 \times 10^{-4} - 1.65 \times 10^{-4} \] \[ = 1.21 \times 10^{-4} \, \text{N} \] --- ## **Step 6: Final Answer** \[ \boxed{F = 1.2 \times 10^{-4}~\text{N}} \] (Rounded to 2 significant figures, rightward) --- **Summary Table:** | Source | Magnitude (\(N\)) | Direction | |-------------|--------------------------|---------------------| | A on B | \(1.65 \times 10^{-4}\) | Leftward (−) | | C on B | \(2.86 \times 10^{-4}\) | Rightward (+) | | **Net** | \(\mathbf{1.2 \times 10^{-4}}\) | **Rightward (+)** | **Units:** Newtons (N) --- ### **Final Answer:** \[ \boxed{F = 1.2 \times 10^{-4}~\text{N}} \] (Rightward, since positive)