Two wires carrying currents $5\ \text{A}$ and $2\ \text{A}$ are enclosed in a circular loop as shown in the figure. Another wire carrying a current of $3\ \text{A}$ is situated outside the loop. The value of $\oint \vec{B} \cdot d\vec{l}$ around the loop is ($\mu_0 = $ permeability of free space, $d\vec{l}$ is the length of the element on the Amperion loop)
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Ampere's Law only cares about the net current piercing the enclosed area. You can completely ignore any complex currents happening outside the loop, as their net contribution to the closed line integral perfectly cancels out to zero.
Step 1: Understanding the Question:
We need to calculate the line integral of the magnetic field ($\oint \vec{B} \cdot d\vec{l}$) around a closed Amperian loop containing multiple current-carrying wires.
Step 2: Key Formula or Approach:
Ampere's Circuital Law states that the line integral of the magnetic field around a closed loop is equal to $\mu_0$ times the net current enclosed by the loop:
$$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}}$$
Currents outside the loop do not contribute to this integral.
Step 3: Detailed Explanation:
There are three wires described:
1. A $5\ \text{A}$ wire inside the loop.
2. A $2\ \text{A}$ wire inside the loop (pointing in the opposite direction).
3. A $3\ \text{A}$ wire outside the loop.
The wire outside the loop ($3\ \text{A}$) is ignored because it is not enclosed by the Amperian surface.
The net enclosed current ($I_{\text{enc}}$) is the algebraic sum of the currents inside the loop. Assuming standard convention where opposite directions subtract:
$$I_{\text{enc}} = 5\ \text{A} - 2\ \text{A} = 3\ \text{A}$$
Apply Ampere's Law:
$$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}} = \mu_0 (3)$$
$$\oint \vec{B} \cdot d\vec{l} = 3\mu_0$$
Step 4: Final Answer:
The value of the integral is $3\mu_0$, matching option (C).
