If a magnet is suspended at an angle 30^o to | vedclass

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Question

(a)Let the real dip be  $\phi $, then $	an \phi = \frac{{{B_V}}}{{{B_H}}}$
For apparent dip,
$	an {\phi }' = \frac{{{B_V}}}{{{B_H}\c

Vedclass · Accepted Answer

${	an ^{ - 1}}(\sqrt 3 /2)$

Vedclass · Answer

(a)Let the real dip be  $\phi $, then $	an \phi = \frac{{{B_V}}}{{{B_H}}}$
For apparent dip,
$	an {\phi }' = \frac{{{B_V}}}{{{B_H}\cos \beta }} = \frac{{{B_V}}}{{{B_H}\cos {{30}^o}}} = \frac{{2{B_V}}}{{\sqrt 3 {B_H}}}$
or $	an {45^o} = \frac{2}{{\sqrt 3 }}.	an \phi $ or $\phi = {	an ^{ - 1}}\left( {\frac{{\sqrt 3 }}{2}} ight)$

If a magnet is suspended at an angle $30^o$ to the magnetic meridian, it makes an angle of $45^o$ with the horizontal. The real dip is

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