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

  • question_answer
    The energy stored in a 50 mH inductor carrying a current of 4 A will be                                    [MP PET 1999]

    A)            0.4 J                                         

    B)            4.0 J

    C)            0.8 J                                         

    D)            0.04 J

    Correct Answer: A

    Solution :

               \[U=\frac{1}{2}L{{i}^{2}}=\frac{1}{2}\times (50\times {{10}^{-3}})\times {{(4)}^{2}}\]\[=400\times {{10}^{-3}}=0.4J\]


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