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JEE Main & Advanced Sample Paper JEE Main Sample Paper-47

  • 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

    Correct Answer: D

    Solution :

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


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