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RAJASTHAN PMT Rajasthan - PMT Solved Paper-1995

  • question_answer
    The length of a coil is \[0.3\,m\]and the number of turns is 2000. The area of cross-section of the coil is\[1.2\times {{10}^{-3}}{{m}^{2}}\]. Another coil is 300 turns is wrapped over the above coil. A current of 2A is passed through the solenoid and its direction is changed in\[0.25\,sec\]. Then the e.m.f induced in volt will be:

    A)  \[7.6\times {{10}^{-2}}\]       

    B)                         \[4.8\times {{10}^{-3}}\]

    C)         \[3.2\times {{10}^{-4}}\]             

    D)                         \[3.2\times {{10}^{-2}}\]

    Correct Answer: A

    Solution :

                    Magnetic field \[B=\frac{{{\mu }_{0}}ni}{l}\] \[=\frac{4\pi \times {{10}^{-7}}\times 2000\times 2}{0.3}\] Induced e.nr.f. \[E=-A\frac{B}{t}\] Substituting the values \[E=7.6\times {{10}^{-2}}\text{ }volt\]


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