Question

The magnetic field due to the earth has a horizontal components of $26\mu T$ at a place where the dip is ${60}^{\circ }$. Find the vertical components and the magnitude of the field.

Answer 1

$\delta \left(dip\right)={60}^o$

$B_H=B\cos {60}^o$

$\Rightarrow B=52\times {10}^{-5}=52\mu T$

$B_v=B\sin \delta =52\times {10}^{-5}\frac{\sqrt{3}}{2}=44.98\mu T\approx 45\mu T$