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CMC Medical CMC-Medical VELLORE Solved Paper-2014

  • question_answer
    A toroidal solenoid with an air case has an average radius of 15\[cm\], area of cross-section 12\[c{{m}^{2}}\] and 1200 turns. Ignoring the field variation across the cross-section of the toroid, the self-inductance of the toroid is

    A)  4.6 mH                                

    B)  6.9 mH

    C)  2.3 mH                                

    D)  9.2 mH

    Correct Answer: C

    Solution :

                    Magnetic field \[B={{\mu }_{0}}ni\] \[n=\frac{N}{2\pi r}\] \[B=\frac{{{\mu }_{0}}Ni}{2\pi r}\] Flux linked with the solenoid is \[\phi =NBA\] \[\phi =\frac{{{\mu }_{0}}{{N}^{2}}iA}{2\pi r}\] Self-inductance of the toroid \[L=\frac{\phi }{i}=\frac{{{\mu }_{0}}{{N}^{2}}A}{2\pi r}\]    \[=\frac{4\pi \times {{10}^{-7}}\times {{(1200)}^{2}}\times 12\times {{10}^{-4}}}{2\pi \times 15\times {{10}^{-2}}}\] \[L=2.3\times {{10}^{-3}}H=2.3\,mH\]


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