Obtain relations between |
(i) earth's magnetic field and its horizontal and vertical components, |
(ii) angle of dip and the horizontal and vertical components of the earth's magnetic field. |
Answer:
We know that, \[{{B}_{H}}={{B}_{E}}\cos \delta \] ?(i) and \[{{B}_{V}}={{B}_{E}}\sin \delta \] ?(ii) Here, \[{{B}_{E}}\]= Earth?s magnetic field, \[{{B}_{H}}\] = horizontal component of magnetic field, \[{{B}_{V}}\]= vertical component of magnetic field and \[\delta \] = Angle of dip. (i) Squaring Eqs, (i) and (ii) and then adding, we get \[B_{H}^{2}+B_{V}^{2}=B_{E}^{2}\] or \[{{B}_{E}}=\sqrt{B_{H}^{2}+B_{V}^{2}}\] (ii) Dividing Eq. (ii) by Eq. (i), we get \[\frac{{{B}_{V}}}{{{B}_{H}}}=\tan \delta \] or \[{{B}_{V}}={{B}_{H}}.\tan \delta \]
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