At a given place on earth's surface the horizontal component of earths magnetic field is `2xx10^(-5)T` and resultant magnetic field is `4xx10^(-5)T`. The angles of dip at this place is

Options

A. `30^(@)`
B. `60^(@)`
C. `90^(@)`
D. `44^(@)`

Answer

B

Explanation

Since, `B_(H)=B cos delta`
`"Here", B=4xx10^(-5)T, B_(H)=2xx10^(-5)T`
`therefore cos delta=(B_(H))/(B)=(2xx10^(-5))/(4xx10^(-5))=(1)/(2)=cos 60^(@) Rightarrow delta=60^(@)`