The horizontal compound of the earth's magnetic field at a given place is `0.4xx10^(-4) Wb//m^(2)` and angle of dip `=30^(@)`. What is the total intensity of the earth's magnetic field at that place?

Options

A. `0.23xx10^(-4) Wb//m^(2)`
B. `0.46xx10^(-4) Wb//m^(2)`
C. `0.58xx10^(-4) Wb//m^(2)`
D. `0.35xx10^(-4) Wb//m^(2)`

Answer

B

Explanation

`:' B_(H)=B cos delta =B cos 30^(@)`
where B is the total intensity of the earth's magnetic field.
`:. B=B_(H)/(cos 30^(@))`
But `cos 30^(@)=sqrt(3)/2=1.732/2=0.866`
`:. B=(0.4xx10^(-4))/0.866=0.46xx10^(-4)" Wb/m"^(2)`