1. **Problem Statement:** Find the value of the line integral $$\oint \mathbf{B} \cdot d\mathbf{l}$$ for the given loop around four wires carrying currents $$I_1, I_2, I_3, I_4$$. 2. **Relevant Formula:** By Ampère's Law, the line integral of the magnetic field $$\mathbf{B}$$ around a closed loop is related to the net current enclosed by the loop: $$\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{enc}$$ where $$\mu_0$$ is the permeability of free space and $$I_{enc}$$ is the algebraic sum of currents enclosed by the loop, taking into account their direction relative to the loop traversal. 3. **Important Rules:** - Currents passing through the loop contribute positively if they are in the direction of the loop's right-hand rule. - Currents opposing the loop direction contribute negatively. 4. **Analyzing the Loop and Currents:** - The loop intersects the wires vertically between $$I_2$$ and $$I_3$$. - The loop direction is upward. - Assign signs to currents based on whether they pass through the loop in the direction of traversal. 5. **Determining Enclosed Currents:** - $$I_1$$ is outside the loop (above $$I_2$$), so it is not enclosed. - $$I_2$$ is enclosed and contributes positively. - $$I_3$$ is enclosed and contributes negatively (opposite direction). - $$I_4$$ is outside the loop (below $$I_3$$), so it is not enclosed. 6. **Summing Enclosed Currents:** $$I_{enc} = I_2 - I_3$$ 7. **Final Expression:** $$\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 (I_2 - I_3)$$ 8. **Matching with Given Options:** The closest option matching this is option C: $$\mu_0 (-I_1 + I_2 - I_3)$$ Since $$I_1$$ is outside the loop, its coefficient should be zero, so effectively the answer is $$\mu_0 (I_2 - I_3)$$. **Answer:** Option C