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BVP Medical BVP Medical Solved Paper-2007

  • question_answer
    A small piece of metal wire is dragged across the gap between the poles of a magnet is 0.4 s. If the change in magnetic flux in the wire is \[{{E}_{n}}\] Wb, then emf induced in the wire is

    A)                 \[4{{E}_{n}}\]                    

    B)                 \[{{E}_{n}}/4\]

    C)                 \[2{{E}_{n}}\]                    

    D)                 \[{{E}_{n}}/2\]

    Correct Answer: D

    Solution :

                    From Faradays law of electromagnetic induction, when the magnetic flux through a circuit is changing an induced emf (e) is set up in the circuit whose magnitude is equal to the negative rate of change of magnetic flux \[f=\frac{v}{v+{{v}_{s}}}\,\,f\]                                 \[=\frac{320}{320+4}\times 240\] Given, \[=237Hz\] \[=f-f=\]              \[\Rightarrow \]                 \[n=243-237=6\] Note: The direction of induced emf is such as to oppose the change that produced it.


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