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JEE Main & Advanced Physics Electro Magnetic Induction Question Bank Static EMI

  • question_answer
    A coil of wire of a certain radius has 600 turns and a self inductance of 108 mH. The self inductance of a 2nd  similar coil of 500 turns will be                                                   [MP PMT 1990]

    A)            74 mH                                     

    B)            75 mH

    C)            76 mH                                     

    D)            77 mH

    Correct Answer: B

    Solution :

               \[\frac{{{L}_{B}}}{{{L}_{A}}}={{\left( \frac{{{n}_{B}}}{{{n}_{A}}} \right)}^{2}}\Rightarrow {{L}_{B}}={{\left( \frac{500}{600} \right)}^{2}}\times 108=75\ mH\]


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