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VMMC VMMC Medical Solved Paper-2011

  • question_answer
    A square loop of wire of each side 50 cm is kept, so that its plane makes an angle 9 with a uniform magnetic field of induction 1 T. The magnetic field is withdrawn in 0.1 s. It is found that the induced emf across the loop is 125 mV. The angle 0 is

    A) \[90{}^\circ \]

    B) \[60{}^\circ \]

    C) \[45{}^\circ \]

    D)        \[30{}^\circ \]

    Correct Answer: B

    Solution :

                            \[e=nAB\sin ({{90}^{o}}-\theta )\] \[=nAB\cos \theta \]                 \[\therefore \]  \[125\times {{10}^{-3}}=1\times 0.5\times 0.5\times 1\times \cos \theta \]                                 \[\cos \theta =\frac{1}{2}\] \[\theta ={{60}^{o}}\]


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