In the magnetic meridian of a certain place, the horizontal component of the earth's magnetic field is `0*26G` and dip angle is `60^@`. What is the magnetic field of earth at this location?

Explanation

It is given that `H_(E)=0.26G`, From Fig. 5.11, we have
` cos 60^(@)=(H_(E))/(B_(E))`
`B_(E)=(H_(E))/(cos60^(@))`
`=(0.26)/((1//2))=0.52G`