CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2011

  • question_answer
    A metal conductor of length 1 m rotates vertically about one of its ends at angular velocity\[5\text{ }rad{{s}^{-1}}\]. If the horizontal component of earths magnetic field is\[0.2\times {{10}^{-4}}T,\]then the emf developed between the ends of the conductor is

    A)  \[5\mu V\]                        

    B)         \[5mV\]                             

    C)  \[50\mu V\]                     

    D)  \[50mV\]

    E)  \[0.5\text{ }mV\]

    Correct Answer: C

    Solution :

    \[e=\frac{1}{2}B{{\lambda }^{2}}\omega =5\mu V\]


You need to login to perform this action.
You will be redirected in 3 sec spinner