At a certain place, the angle of dip is ` 60^(@)` and the horizontal component of the earth's magnetic field `(B_(H))` is `0.8xx10^(-4)` T. The earth's overall magnetic field is

Options

A. `1.5xx10^(-4) T`
B. `1.6xx10^(-3) T`
C. `1.5xx10^(-3) T`
D. `1.6xx10^(-4) T`

Answer

D

Explanation

`B_(H)=0.8xx10^(-4)T,theta=60` given
`B_(H)=B_(e)"cos"theta`
`B_(e)=(B_(H))/("cos"theta)=(0.8xx10^(-4))/("cos"theta60)=(0.8xx10^(-4))/(1/2)`
`=1.6xx10^(-4)T`