Let $B_{P}$ and $B_{Q}$ be the magnetic field produced by the wires P and Q respectively which are placed symmetrically in a rectangular loop ABCD as shown in figure. Current in wire P is I directed inward and in Q is 2I directed outwards.

If $\int_{A}^{B} {\vec{B}}_{Q} \cdot \vec{d l} = 2 μ_{0}$ tesla meter, $\int_{D}^{A} {\vec{B}}_{P} \cdot \vec{d l} = - 2 μ_{0}$ tesla meter
& $\int_{A}^{B} {\vec{B}}_{P} \cdot \vec{d l} = - μ_{0}$ tesla meter, the value of I will be (in amperes)

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answer is 6.

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Detailed Solution

Magnetic field is proportional to current. Applying this idea, and keeping the directions of current in mind, we write $\int \vec{B} \cdot \vec{d l}$ over all 4 segments due to both wires.

$\int \vec{B} \cdot \vec{d l}$	$A \to B$	$B \to C$	$C \to D$	$D \to A$
P	$- μ_{0}$	$- 2 μ_{0}$	$- μ_{0}$	$- 2 μ_{0}$
Q	$2 μ_{0}$	$4 μ_{0}$	$2 μ_{0}$	$4 μ_{0}$