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VIT Engineering VIT Engineering Solved Paper-2011

  • question_answer
    A solenoid has 2000 turns wound over a length of 0.30 m. The area of its cross-section is \[1.2\times {{10}^{-3}}{{m}^{2}}\]. Around its central section, a coil of 300 turn is wound. If an initial current of 2 A in the solenoid is reversed in 0.25 s, then the emf induced in the coil is

    A)  \[6\times {{10}^{-4}}V\]

    B)  \[4.8\times {{10}^{-3}}V\]

    C)  \[6\times {{10}^{-2}}V\]

    D)  \[48mV\]

    Correct Answer: D

    Solution :

    Induced emf \[e=M\frac{di}{dt}\] \[=\frac{{{\mu }_{0}}\,{{N}_{1}}{{N}_{2}}A}{l}.\frac{di}{dt}\] \[=\frac{4\pi \times {{10}^{-7}}\times 2000\times 300\times 1.2\times {{10}^{-3}}}{0.30}\] \[\times \frac{|2-(-2)|}{0.25}\] \[=48.2\times {{10}^{-3}}V=48mV\]


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