Uttarakhand PMT Uttarakhand PMT Solved Paper-2010

  • question_answer
    A solenoid 30 cm long is made by winding 2000 loops of wire on an iron rod whose cross-section is 1.5 cm2. If the relative permeability of the iron is 6000, what is the self-inductance of the solenoid?

    A)  1.5 H          

    B)  2.5 H

    C)  3.5 H          

    D)  0.5 H

    Correct Answer: A

    Solution :

     Self-inductance of the salenoid \[L=\frac{{{\mu }_{r}}{{\mu }_{0}}{{N}^{2}}A}{l}\] \[=\frac{600\times 4\pi \times {{10}^{-7}}\times {{(2000)}^{2}}\times (1.5\times {{10}^{-4}})}{0.3}\] \[=1.5H\]


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