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

  • question_answer
    The inductance of a closed-packed coil of 400 turns is 8 mH. A current of 5 mA is passed through it. The magnetic flux through each turn of the coil is        [Roorkee 2000]

    A)            \[\frac{1}{4\pi }\,{{\mu }_{0}}Wb\]                                

    B)            \[\frac{1}{2\pi }\] m0Wb

    C)            \[\frac{1}{3\pi }{{\mu }_{0}}Wb\]                                   

    D)            \[0.4\,{{\mu }_{0}}Wb\]

    Correct Answer: A

    Solution :

               \[N\varphi =Li\Rightarrow \varphi =\frac{Li}{N}=\frac{8\times {{10}^{-3}}\times 5\times {{10}^{-3}}}{400}={{10}^{-7}}=\frac{{{\mu }_{0}}}{4\pi }wb\]


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