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 $2\times {10}^{-5}$ T and resultant magnetic field is $4\times {10}^{-5}$T. The angle of dip at this place is

• A. ${30}^0$
• B. ${60}^0$
• C. ${90}^0$
• D. ${45}^0$

Answer 1

The correct option is B ${60}^0$

Since $B_H=Bcos\delta$
Here, $B=4\times {10}^{-5}T,B_H=2\times {10}^{5}T$
$\therefore cos\delta =\frac{B_H}{B}=\frac{2\times {10}^{-5}}{4\times {10}^{-5}}=\frac{1}{2}=cos{60}^0\Rightarrow \delta ={60}^0$