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AIIMS AIIMS Solved Paper-2015

  • question_answer
    The self inductance of a coil having 500 turns is\[50\text{ }mH\]. The magnetic flux through the cross-sectional area of the coil while current through it is \[8\text{ }mA\] is found to be

    A) \[50\mu A\]      

    B) \[114\mu A\]

    C) \[{{l}_{1}}\]                        

    D) \[{{T}_{1}}\]

    Correct Answer: A

    Solution :

                     Given, n = 500 \[\left( \because a=\frac{dv}{dt} \right)\] and     \[F'=ilB=C{{B}^{2}}{{I}^{2}}a\] The magnetic flux linked with the coil is \[F-F'=ma\Rightarrow F=ma+C{{B}^{2}}{{I}^{2}}a\] \[a=\frac{F}{m+C{{B}^{2}}{{I}^{2}}}\].                


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