Question

At a given place on earth's surface the horizontal component of earths magnetic field is $2\times {10}^{-5}$T and resultant magnetic field is $4\times {10}^{-5}$T. The angle of dip at this place is?

• A. ${30}^o$
• B. ${60}^o$
• C. ${90}^o$
• D. ${45}^o$

Answer 1

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

Since $B_H=B\cos \delta$
Here, $B=4\times {10}^{-5}$T, $B_H=2\times {10}^{-5}$T

$\therefore \cos \delta =\frac{B_H}{B}=\frac{2\times {10}^{-5}}{4\times {10}^{-5}}=\frac{1}{2}=\cos {60}^o\Rightarrow \delta ={60}^o$.