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NEET Sample Paper NEET Sample Test Paper-21

  • question_answer
    Two coils A and B having turns 300 and 600 respectively are placed near each other. On passing a current of 3.0 ampere in A, the flux linked with A is \[1.2\times {{10}^{-4}}\] Wb and with B it is \[9.0\times {{10}^{-5}}\]Wb. The mutual inductance of the system is:

    A) \[4\times {{10}^{-5}}H\]                

    B) \[3\times {{10}^{-5}}H\]    

    C)   \[2\times {{10}^{-5}}H\]    

    D)   \[1.8\times {{10}^{-2}}H\]    

    Correct Answer: D

    Solution :

    \[\left| {{e}_{s}} \right|={{N}_{s}}\frac{d{{\Phi }_{s}}}{dt}\]and \[\left| {{e}_{s}} \right|=M\frac{d{{I}_{s}}}{dt}\,\therefore \,{{N}_{s}}\frac{d{{\Phi }_{s}}}{dt}=M\frac{d{{I}_{s}}}{dt}\] \[M={{N}_{s}}\frac{d\Phi }{d{{I}_{p}}}=\frac{600\times 9.00\times {{10}^{-5}}}{3.0}=1800\times {{10}^{-5}}H.\] Hence, the correction option is(d).


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