The horizontal component of the earth's m | vedclass

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(c)By using ${B_H} = B\cos \phi $
$==>$ $\cos \phi = \frac{{{B_H}}}{B} = \frac{{0.22}}{{0.4}}$
$==>$ $	an \phi = \frac{{\sqrt {{{(0.4)}^2} -

Vedclass · Accepted Answer

${	an ^{ - 1\,}}(1.518)$

Vedclass · Answer

(c)By using ${B_H} = B\cos \phi $
$==>$ $\cos \phi = \frac{{{B_H}}}{B} = \frac{{0.22}}{{0.4}}$
$==>$ $	an \phi = \frac{{\sqrt {{{(0.4)}^2} - {{(0.22)}^2}} }}{{0.22}}$
$==>$ $\phi = {	an ^{ - 1}}(1.518)$

The horizontal component of the earth's magnetic field is $0.22$ $Gauss$ and total magnetic field is $0.4$ $Gauss$. The angle of dip. is

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