A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at `22^@` with the horizontal. The horizontal component of the earth's magnetic field at the place is known to be `0*35G`. Determine the strength of the earth's magnetic field at the place.

Explanation

Horizontal component of earth's magnetic field, `B_(a)=0.35G`
Angle mde by the needle with the horizontal plan e =Angle of dip=`delta=22^(@)`
Earth's magnetic field strenght=B
We can relate B and `B_(H)` as:
`B_(H)=B cos theta`
`thereforeB=(B_(R))/(cos delta)`
`=(0.35)/(cos22^(@))=0.377G` ltbr. Hence, the strengh of earth's magnetic field at the given location is 0.377G.