Two wires with current 3A and 1.5A are enclosed in a circular loop P. Third parallel wire with current 1A is situated outside the loop as shown. All the wires are perpendicular to the plane of the circular loop. The value of \(\oint \overset{⃗}{B}\cdot \overset{⃗}{dl}\) around the loop is (\(μ_0 =\) permeability of free space)
Show Hint
Ampere's law counts only the net current passing through the loop. The 1 A wire outside contributes nothing.
Step 1: Understanding the Concept:
Ampere's circuital law states that the line integral of \(\vec B\) around a closed loop equals \(\mu_0\) times the net current that passes through the loop.
Step 2: Key Formula or Approach:
\[ \oint\vec B\cdot d\vec l = \mu_0 I_{\text{enclosed}} \]
Currents in opposite directions through the loop have opposite signs.
Step 3: Detailed Explanation:
In the figure, the loop P encloses two wires. The 3 A wire carries current upward and the 1.5 A wire carries current downward. The third wire, with 1 A, lies outside the loop.
Net enclosed current (taking upward as positive):
\[ I_{\text{enc}} = 3 - 1.5 = 1.5 \text{ A} \]
The wire outside the loop does not contribute to the line integral, even though it does contribute to the field at points on the loop.
\[ \oint\vec B\cdot d\vec l = 1.5\,\mu_0 \]
Option (A) 5.5 adds all three currents. Option (B) 2.5 adds 1.5 and 1, and (D) 0.5 uses 1.5 - 1.
Final Answer:
The value is \(1.5\,\mu_0\), option (C).
\[ \boxed{1.5\,\mu_0 \text{ (C)}} \]
