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

  • question_answer
    An e.m.f. of 5 volt is produced by a self inductance, when the current changes at a steady rate from 3 A to 2 A in 1 millisecond. The value of self inductance is [CPMT 1982; MP PMT 1991; CBSE PMT 1993; AFMC 2002]

    A)            Zero                                         

    B)            5 H

    C)            5000 H                                     

    D)            5 mH

    Correct Answer: D

    Solution :

               \[L=\frac{e}{di/dt}=\frac{5}{(3-2)/{{10}^{-3}}}=\frac{5}{1}\times {{10}^{-3}}=5milli\ henry\]


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