MHTCET 18 Apr 2026 Shift 2 Paper

Concept:Ampere's circuital law states that the line integral of the magnetic field around a closed loop equals $\mu_{0}$ times the net current enclosed by that loop.Explanation:According to Ampere's law,$$\oint \vec{B} \cdot \overrightarrow{d l} = \mu_{0} I_{\text{enclosed}}$$Here, the circular loop $P$ encloses only two wires.These two wires carry currents of $3\,\text{A}$ and $1.5\,\text{A}$ respectively.The third wire carrying $1\,\text{A}$ lies outside the loop $P$, so it does not contribute to $I_{\text{enclosed}}$ and must be ignored.The two enclosed currents flow in opposite directions through the loop.Taking the $3\,\text{A}$ current as positive and the $1.5\,\text{A}$ current as negative, the net enclosed current is:$$I_{\text{enclosed}} = 3 - 1.5 = 1.5\,\text{A}$$Substituting this into Ampere's law gives:$$\oint \vec{B} \cdot \overrightarrow{d l} = \mu_{0} \times 1.5 = 1.5\,\mu_{0}$$Answer:$\oint \vec{B} \cdot \overrightarrow{d l} = 1.5\,\mu_{0}$Therefore, the correct option is C. $1.5\,\mu_{0}$.