NEET AIPMT SOLVED PAPER SCREENING 2008

  • question_answer
    A long solenoid has 500 rums. When a current of 2 A is passed through it, the resulting magnetic flux linked with each turn of the solenoid is \[4\times {{10}^{-3}}\]Wb. The self-inductance of the solenoid is

    A) 2.5 H      

    B)                        2.0 H

    C) 1.0 H

    D) 4.0 H

    Correct Answer: C

    Solution :

                                    Key Idea : Inductance of a coil is numerically equal to the emf induced in the coil when the current in the coil changes at the rate of \[1A{{s}^{-1}}.\] If \[I\] is the current flowing in the circuit, then flux linked with the circuit is observed to be proportional to I, ie, \[\phi \propto I\]or\[\phi =LI\]          ...(i) where L is called the self-inductance or coefficient of self-inductance or simply inductance of the coil. Net flux through solenoid, \[\phi =500\times 4\times {{10}^{-3}}=2Wb\] or \[2=L\times 2\][after putting values in Eq. (i)] or \[L=1H\]


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