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  3. CEE Kerala Engineering

CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2008

  • question_answer
    In a certain place, die horizontal component of magnetic field is\[\frac{1}{\sqrt{3}}\]times to vertical component. The angle of dip at this place is

    A)  zero                     

    B)         \[\pi /3\]            

    C)         \[\pi /2\]               

    D)         \[\pi /6\]

    E)  \[\pi /4\]

    Correct Answer: B

    Solution :

    Given that \[{{B}_{H}}=\frac{1}{\sqrt{3}}{{B}_{V}}\] Also,      \[\tan \theta =\frac{{{B}_{V}}}{{{B}_{H}}}=\sqrt{3}=\tan {{60}^{o}}\] \[\therefore \]  \[\theta =60{}^\circ =\frac{\pi }{3}\]


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