At a place , the horizontal component of earth's magnetic field is B and angle of dip is `60^(@)` . What is the value of horizontal component of earth's magnetic field at equator ?

Answer

1

Explanation

Given , horizontal component of earth's magnetic field , H=B.
Angle of dip , `theta=60^(@)` .
`H=Vcos` [ V=vertical component of earth's magnetic field ]
`H=Vcos60^(@)`
`(V)/(2)`
`thereforeV=2H`
At equator , `theta=0^(@)`
`thereforeH_(eq)=Vcos0^(@)" or, " H_(eq)=2H`