Question

The magnetic field due to the earth has a horizontal component of 26 μT at a place where the dip is 60°. Find the vertical component and the magnitude of the field.

Answer 1

Given :
Horizontal component of Earth's magnetic field, B = 26 μT ​     
Angle of dip, δ = 60°

The horizontal component of Earth's magnetic field `(B_H)` is given by 

`B_H = Bcosδ`

Here,
B = Total magnetic field of Earth
On substituting the respective values, we get

`26 xx 10^-6 = B xx 1/2`

⇒ B = `52 xx 10^-6 = 52  "uT"`

The vertical component of Earth's magnetic field `(B_y)` is given by 

`B_y = Bsinδ`

⇒ `B_y = 52 xx 10^-6 xx (sqrt(3))/2`

⇒ `B_y = 44.98  "uT" = 45  "uT"`