Two wires with currents 3 text{ A} and 1.5 te | vedclass

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(C) According to Ampere's Circuital Law, the line integral of the magnetic field $\vec{B}$ around a closed loop is given by $\oint \vec{B} \cdot d\vec

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$1.5$

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(C) According to Ampere's Circuital Law, the line integral of the magnetic field $\vec{B}$ around a closed loop is given by $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{	ext{enclosed}}$.Here, $I_{	ext{enclosed}}$ is the net current passing through the area enclosed by the loop $P$.The currents inside the loop are $I_1 = 3 	ext{ A}$ (upwards) and $I_2 = 1.5 	ext{ A}$ (downwards).Taking the upward direction as positive, the net enclosed current is $I_{	ext{enclosed}} = 3 	ext{ A} - 1.5 	ext{ A} = 1.5 	ext{ A}$.The current outside the loop does not contribute to the line integral.Therefore, $\oint \vec{B} \cdot d\vec{l} = \mu_0 (1.5 	ext{ A}) = 1.5 \mu_0$.

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