• # question_answer Let ${{B}_{P}}$ and ${{B}_{Q}}$ be the magnetic field produced by the wire P and Q 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}{{{B}_{Q}}\cdot dl=+2{{\mu }_{0}}}$$T-m$,  $\int_{D}^{A}{{{B}_{P}}\cdot dl=-2{{\mu }_{0}}}$ T-m  and $\int_{A}^{B}{{{B}_{P}}\cdot dl=-{{\mu }_{0}}}$ T-m, the value of $I$ will be A)  8 A                              B)  4 A                C)  5 A                              D)  6 A

From diagram, $\int_{A}^{B}{{{B}_{P}}dI+}\int_{B}^{C}{{{B}_{P}}di+}\int_{C}^{D}{{{B}_{P}}dI}+\int_{D}^{A}{{{B}_{P}}dI}$ $+\int_{A}^{B}{{{B}_{Q}}dI+}\int_{B}^{C}{{{B}_{Q}}dI}+\int_{C}^{D}{{{B}_{Q}}dI}+\int_{D}^{A}{{{B}_{Q}}dI}={{\mu }_{0}}\,(2I-I)$$(-{{\mu }_{0}}-2{{\mu }_{0}}-{{\mu }_{0}}-2{{\mu }_{0}})\,+(2{{\mu }_{0}}+4{{\mu }_{0}}+2{{\mu }_{0}}+4{{\mu }_{0}})$$={{\mu }_{0}}I$ $\Rightarrow$            $I=12-6=6A$