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. `6x10^(-5)T`
B. `6sqrt2x10^(-5)T`
C. `3.6sqrt3x10^(-5)T`
D. `sqrt2x10^(-5)T`

Answer

C

Explanation

`B_(H)=0.36xx10^(-4)T`
`Tan delta=(B_(V))/(B_(H))`