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 $\left(B_H\right)$ is given by
B_H = Bcosδ
Here,
B = Total magnetic field of Earth
On substituting the respective values, we get
$$\begin{gathered} 26\times {10}^{-6}=B\times \frac{1}{2} \\ \Rightarrow B=52\times {10}^{-6}=52\mathrm{\mu T} \end{gathered}$$
The vertical component of Earth's magnetic field $\left(B_y\right)$ is given by
$$\begin{gathered} B_y\mathit{=}B\sin \delta \\ \Rightarrow B_y=52\times {10}^{-6}\times \frac{\sqrt{3}}{2} \\ \Rightarrow B_y=44.98\mathrm{\mu T}=45\mathrm{\mu T} \end{gathered}$$