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JEE Main & Advanced JEE Main Paper (Held On 11-Jan-2019 Morning)

  • question_answer
    There are two long co-axial solenoids of same length \[l\]. The inner and outer coils have radii \[{{r}_{1}}\] and \[{{r}_{2}}\]and number of turns per unit length \[{{n}_{1}}\] and \[{{n}_{2}}\] respectively. The ratio of mutual inductance to the self-inductance of the inner- coil is [JEE Main Online Paper (Held On 11-Jan-2019 Morning]

    A) \[\frac{{{n}_{2}}}{{{n}_{1}}}\]                                              

    B) \[\frac{{{n}_{2}}}{{{n}_{1}}}.\frac{r_{2}^{2}}{r_{1}^{2}}\]

    C) \[\frac{{{n}_{2}}}{{{n}_{1}}}.\frac{{{r}_{1}}}{{{r}_{2}}}\]  

    D)                  \[\frac{{{n}_{1}}}{{{n}_{2}}}\]

    Correct Answer: A

    Solution :

    The mutual inductance of the inner coil \[M={{\mu }_{0}}{{n}_{1}}{{n}_{2}}\pi r_{1}^{2}\]                                      ...(i) Self-inductance (L) of the inner coil is \[L={{\mu }_{0}}n_{1}^{2}\pi r_{1}^{2}\]                                         ...(ii) Using (i) and (ii),\[\frac{M}{L}=\frac{{{n}_{2}}}{{{n}_{1}}}\]


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