The angle of dip at a place is 40.6° and | vedclass

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Question

(d)From the relation ${B_V} = I\sin \phi $
$I = \frac{V}{{\sin \phi }} = \frac{{6 	imes {{10}^{ - 5}}}}{{\sin {{40.6}^o}}} = \frac{{6 	imes {{10}^{

Vedclass · Accepted Answer

$9.2 	imes {10^{ - 5}}\,tesla$

Vedclass · Answer

(d)From the relation ${B_V} = I\sin \phi $
$I = \frac{V}{{\sin \phi }} = \frac{{6 	imes {{10}^{ - 5}}}}{{\sin {{40.6}^o}}} = \frac{{6 	imes {{10}^{ - 5}}}}{{0.65}} = 9.2 	imes {10^{ - 5}}$ $tesla$

The angle of dip at a place is $40.6°$ and the intensity of the vertical component of the earth's magnetic field $V = 6 \times {10^{ - 5}}$ $Tesla$. The total intensity of the earth's magnetic field $ (I) $ at this place is

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