Question

The magnetic field due to the earth has a horizontal component of $26\mu$ T at a place where the dip is ${60}^0$. Find the vertical component and the magnitude of the field.

Answer 1

Here $\delta \left(dip\right)={60}^0$

Since $B_H=B\text{cos}{60}^0$

$\Rightarrow 26\times {10}^{-6}=B\times \frac{1}{2}$

$\Rightarrow B=52\times {10}^{-6}=52\mu T$

$B\gamma =B\text{sin}\delta =52\times {10}^{-6}\times \frac{\sqrt{3}}{2}$

= $44.98\mu T=45\mu T$.