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: Understand the question.
A circular loop encloses two wires carrying $5\ \text{A}$ and $2\ \text{A}$ in opposite directions. A third wire of $3\ \text{A}$ lies outside the loop. We must find the value of $\oint \vec{B}\cdot d\vec{l}$ around the loop.
Step 2: State Ampere's law.
Ampere's circuital law says the line integral of the magnetic field around a closed loop equals $\mu_0$ times the net current passing through the loop:
\[ \oint \vec{B}\cdot d\vec{l} = \mu_0\, I_{\text{enc}} \]
Step 3: Decide which currents count.
Only the currents that pass through the loop matter. The $3\ \text{A}$ wire is outside the loop, so it does not count at all.
Step 4: Add the enclosed currents with signs.
The $5\ \text{A}$ and $2\ \text{A}$ wires point in opposite directions, so we subtract them:
\[ I_{\text{enc}} = 5 - 2 = 3\ \text{A} \]
Step 5: Put into the law.
\[ \oint \vec{B}\cdot d\vec{l} = \mu_0 \times 3 = 3\mu_0 \]
Step 6: Write the answer.
The value of the integral is $3\mu_0$. This matches option (3).
\[ \boxed{\oint \vec{B}\cdot d\vec{l} = 3\mu_0} \]
The ammeter reads 0.9 A. Value of R is