The value of the horizontal component of the earth's magnetic field and and angle of dip are `1.8xx10^(-5) Weder//m^(2)` and `30^(@)` respectively at some place. The total intensity of earth's magnetic field at that place will be

Options

A. `2.08xx10^(-5) "Weber"//m^(2)`
B. `3.67xx10^(-5) "Weber"//m^(2)`
C. `3.18xx10^(-5) "Weber"//m^(2)`
D. `5.0xx10^(-5) "Weber"//m^(2)`

Answer

A

Explanation

Horizontal component `B_(H)=B cos varphi`
Total intensity of earth magnetic field `B=B_(H)/(cos varphi)`
`=(1.8xx10^(5))/(cos 30^(@))=(1.8xx10^(-5))/(sqrt(3)//2)=2.08xx10^(-5) Wb//m^(2)`