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RAJASTHAN PMT Rajasthan - PMT Solved Paper-2010

  • question_answer
    The horizontal component of .the earths magnetic field at a place is \[3\times {{10}^{-4}}T\] and the dip is \[{{\tan }^{-1}}\left( \frac{4}{3} \right).\] A metal rod of length 0.25 m placed in the north-south position and is moved at a constant speed of 10 cm/s towards the east. The emf induced in the rod will be

    A)  zero                                     

    B)  \[1\,\mu V\]

    C)  \[5\,\mu V\]                    

    D)         \[10\,\mu V\]

    Correct Answer: D

    Solution :

    Rod is moving towards east, so induced emf across its end will be\[e={{\beta }_{V}}vl=({{B}_{H}}\tan \phi )vl\] \[\therefore \]  \[e=3\times {{10}^{-4}}\times \frac{4}{3}\times (10\times {{10}^{-2}})\times 0.25\]                    \[={{10}^{-5}}V=10\mu V\]


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