JEE Main & Advanced Sample Paper JEE Main - Mock Test - 6

  • question_answer
    A capacitor of capacitance C is having a charge\[{{Q}_{0}}\]. It is connected to a pure inductor of inductance L. The inductor is a solenoid having N turns. Find the magnitude of magnetic flux through each of the N turns in the coil at the instant charge on the capacitor becomes\[\frac{{{Q}_{0}}}{2}\].

    A) \[\frac{{{Q}_{0}}}{2N}\sqrt{\frac{3L}{C}}\]     

    B) \[\frac{{{Q}_{0}}}{N}\sqrt{\frac{3L}{C}}\]

    C) \[\frac{2{{Q}_{0}}}{N}\sqrt{\frac{L}{C}}\]                  

    D) \[\frac{{{Q}_{0}}}{N}\sqrt{\frac{L}{C}}\]

    Correct Answer: A

    Solution :

    [a] \[Q={{Q}_{0}}\cos (\omega t)\] where \[\omega =\frac{1}{\sqrt{LC}}\] When   \[Q=\frac{{{Q}_{0}}}{1};\]  \[\cos \,(\omega t)=\frac{1}{2}\] Flux linked to the coil is \[\phi =LI=L\frac{dQ}{dt}\] \[\therefore \,\,\,\,\,\,\,\,\,|\phi |=L{{Q}_{0}}\omega |\sin (\omega t)|\] When   \[\cos \,(\omega t)=\frac{1}{2};\,\,\sin (\omega t)=\frac{\sqrt{3}}{2}\] \[\therefore \,\,\,\,\,|\phi |=\frac{\sqrt{3}}{2}L{{Q}_{0}}\omega \] Flux through each turn \[\frac{|\phi |}{N}=\frac{\sqrt{3}L{{Q}_{0}}\omega }{2N}=\frac{1}{2}\sqrt{\frac{3L}{C}}\frac{{{Q}_{0}}}{N}\]

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