CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2009

  • question_answer
    The angle of dip at a place is\[{{37}^{o}}\]and the vertical component of the earths magnetic field is\[6\times {{10}^{-5}}T.\]The earths magnetic field at this place is\[(tan\text{ }{{37}^{o}}=3/4)\]

    A)  \[7\times {{10}^{-5}}T\]              

    B)         \[6\times {{10}^{-5}}T\]

    C)  \[5\times {{10}^{-5}}T\]              

    D)         \[{{10}^{-4}}T\]

    E)  \[4\times {{10}^{-5}}\,T\]

    Correct Answer: D

    Solution :

    Given: \[\tan {{37}^{o}}=\frac{3}{4}\] The vertical component of the earths magnetic field                 \[{{B}_{V}}=6\times {{10}^{-5}}T\]                 \[\sin {{37}^{o}}=\frac{3}{5}\] For vertical component                 \[{{B}_{V}}=B\sin \theta \] Or           \[B=\frac{{{B}_{V}}}{\sin \theta }\] Or           \[B=\frac{6\times {{10}^{-5}}}{3}\times 5\]          Or           \[B=10\times {{10}^{-5}}\] Or           \[B={{10}^{-4}}T\]


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