At a place the earth's horizontal component of magnetic field is `0.36xx10^(-4) "Weber"//m^(2)`. If the angle of dip at that place is `60^(@)`, then the vertical component of earth's field at that place in `"Weber"//m^(2)` will be approxmately

Options

A. `0.12xx10^(-4)`
B. `0.24xx10^(-4)`
C. `0.40xx10^(-4)`
D. `0.62xx10^(-4)`

Answer

D

Explanation

From the relation `B_(H)=B cos varphi` and `B_(V)=B sin varphi B_(V)/B_(H)=tan varphi` or `B_(V)=B_(H) tan varphi`
`=0.36xx10^(-4)xx tan 60^(@)=0.623 xx10^(-4) Wb//m^(2)`