The angle pf dip and the horizontal component of earth's magnetic field at a place are `30^(@)S` and 0.36 Oe . Determine the magnitude and direction of the vertical component of earth 's magnetic field at thet place .

Explanation

Here , `theta=30^(@)S` . So in equilibrium , the south pole of the magnetic needle leans downwards [ Fig] .
We know that , `tantheta=(V)/(H)`
`therefore ` V = vertical component of the earth's magnetic field
`=Htantheta`
`=0.36tan30^(@)[becauseH=0.36"Oe"]`
`=0.36xx(1)/(sqrt(3))=0.208"Oe"`
Since the north pole of the magnetic needle lies above the horizontal line , the , direction of the vertical component (V)will be vertically upwards .