Question

At a given place on the earth's surface, the horizontal component of earth's magnetic field is $3\times {10}^{-6}T$ and resultant magnetic field is $6\times {10}^{-6}T$. Angle of dip at this place is :

• A. ${30}^o$
• B. ${40}^o$
• C. ${50}^o$
• D. ${60}^o$

Answer 1

The correct option is D ${60}^o$

$B_H=B_e\cos \theta$
where $\theta$ is the dip angle.
$\cos \theta =\frac{B_H}{B_e}=\frac{3\times {10}^{-6}}{6\times {10}^{-6}}=\frac{1}{2}=\cos {60}^{\circ }$
$\Rightarrow \theta ={60}^{\circ }$