• # question_answer If BH = $\frac{1}{\sqrt{3}}{{B}_{V}},$find the angle of dip. (where symbols have at their usual meanings) A)  $60{}^\circ$              B)  $30{}^\circ$ C)  $45{}^\circ$              D)  $90{}^\circ$

Magnetic dip or magnetic inclination is given by       $\tan \delta =\frac{{{B}_{V}}}{{{B}_{H}}}$ Given      ${{B}_{H}}=\frac{1}{\sqrt{2}}{{B}_{V}}$ $\frac{{{B}_{V}}}{{{B}_{H}}}=\sqrt{3}$ From Eqs. (i) and (ii), we get $\tan \delta =\sqrt{3}$ $\delta =60{}^\circ$