In the magnetic meridian of a certain place, the horizontal component of earth's magnetic fied is `0.26 G` and the dip angle is `60^@`. Find a. Vertical component of earth's magnetic field b. the net magnetic field at this place
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Given, H=0.26G and `theta=60^(@)` (i) `tan theta=(V)/(H)` `therefore " " V=H tan theta=(0.26)tan60^(@)=0.45G` (ii) `H=B_(e) cos theta` `therefore " " B_(e)=(H)/(cos theta)=(0.26)/(cos 60^(@))=0.52G`
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