Let B_P and B_Q be the magnetic fields 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 ${\int }_{\mathrm{A}}^{\mathrm{B}} {\overrightarrow{\mathrm{B}}}_{\mathrm{Q}}\cdot \overrightarrow{\mathrm{d\ell }}=2{\mathrm{\mu }}_0$ tesla meter , ${\int }_{\mathrm{D}}^{\mathrm{A}} {\overrightarrow{\mathrm{B}}}_{\mathrm{P}}\cdot \overrightarrow{\mathrm{d\ell }}=-2{\mathrm{\mu }}_0$ tesla meter & ${\int }_{\mathrm{A}}^{\mathrm{B}} {\overrightarrow{\mathrm{B}}}_{\mathrm{P}}\cdot \overrightarrow{\mathrm{d\ell }}=-{\mathrm{\mu }}_0$  tesla meter the value of "I"  will be :