Two wires with currents $2 A$ and $I A$ are enclosed in a circular loop. Another wire with current $3 A$ is situated outside the loop as shown.
The $\oint \to B . d \to l$ around the loop is

$[$ 1 Mark $]$

μ_{0}

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3 μ_{0}

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6 μ_{0}

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2 μ_{0}

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Solution

The correct option is A $μ_{0}$
According to Ampere's circuital law
$\oint \to B . d \to l = μ_{0} I_{enclosed} = μ_{0} (2 - 1) = μ_{0}$

Since, currents $2 A$ and $1 A$ are in the opposite direction.

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Two wires with currents 2 A and I A are enclosed in a circular loop. Another wire with current 3 A is situated outside the loop as shown. The ∮→B.d→l around the loop is ​​​​​​​[1 Mark]

The correct option is A μ0According to Ampere's circuital law ∮→B.d→l=μ0Ienclosed=μ0(2−1)=μ0 Since, currents 2A and 1A are in the opposite direction.

Two wires with currents $2 A$ and $I A$ are enclosed in a circular loop. Another wire with current $3 A$ is situated outside the loop as shown.
The $\oint \to B . d \to l$ around the loop is

$[$ 1 Mark $]$

The correct option is A $μ_{0}$
According to Ampere's circuital law
$\oint \to B . d \to l = μ_{0} I_{enclosed} = μ_{0} (2 - 1) = μ_{0}$

Since, currents $2 A$ and $1 A$ are in the opposite direction.