J & K CET Engineering J and K - CET Engineering Solved Paper-2003

  • question_answer
    At a place the angle of dip is \[{{30}^{o}}\]. If the horizontal component of earth's magnetic field is H, then the total field intensity is

    A)  \[\frac{H}{2}\]

    B)  \[\frac{2H}{\sqrt{3}}\]

    C)  \[H\sqrt{2}\]

    D)  \[H\sqrt{3}\]

    Correct Answer: B

    Solution :

    Angle of dip is defined as the angle at a particular point on the earth's surface between the direction of the earth's magnetic field and horizontal. Therefore, \[H=B\,cos\theta \] where B is total field intensity, H the horizontal component of earth's magnetic field and \[\theta \] the dip angle. Given,   \[\theta ={{30}^{o}},\]      \[\because \]  \[\cos \,{{30}^{o}}=\frac{\sqrt{3}}{2}\] \[\therefore \] \[B=\frac{H}{\cos \,{{30}^{o}}}=\frac{2H}{\sqrt{3}}\]


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