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AMU Medical AMU Solved Paper-2003

  • question_answer
    At a certain place the horizontal component of the earths magnetic field is H and the angle of dip is 45°. The total intensity of the field at that place will be

    A)  H                                           

    B)  \[\sqrt{2}H\]

    C)  2 H                                        

    D)  \[{{\text{H}}^{\text{2}}}\]

    Correct Answer: B

    Solution :

                     The relation between angle of dip \[\theta \], and horizontal component \[(\mu )\] of earths magnetic field is given by                                 \[H=B\cos \theta \] Given,   \[\theta ={{45}^{o}}\], \[\therefore \]  \[H=B\,.\,\cos {{45}^{o}}\] \[\Rightarrow \]               \[H=\frac{B}{\sqrt{2}}\] \[\Rightarrow \]               \[I=\sqrt{2}\,H\]


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