Question

At a given place on the earths surface, the horizontal component of magnetic field is $3.2\times {10}^{-5}T$ and the resultant magnetic field is $6.4\times {10}^{-5}T$. The angle of dip at that place is:

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

Answer 1

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

$B_H=3.2\times {10}^{-5}$ T
$B=6.4\times {10}^{-5}$ T
$B\cos \Theta =B_H$
$\cos \Theta =\frac{B_H}{B}$
$=\frac{3.2\times {10}^{-5}}{6.4\times {10}^{-5}}$
$=\frac{1}{2}$
$\Theta ={60}^{\circ }$