Question

At a certain place, the angle of dip is ${60}^o$ and the horizontal component of earth's magnetic field $\left(B_H\right)$ is $0.8\times {10}^{-4}T$. The earth's overall magnetic field is:

• A. $1.5\times {10}^{-4}T$
• B. $1.6\times {10}^{-3}T$
• C. $1.5\times {10}^{-3}T$
• D. $1.6\times {10}^{-4}T$

Answer 1

The correct option is D $1.6\times {10}^{-4}T$

Given : $B_H=0.8\times {10}^{-4}T$ $\varphi ={60}^o$

Let the earth's overall magnetic field be $B$
$\therefore$ $B_H=Bcos\varphi$

$0.8\times {10}^{-4}=Bcos{60}^o$

or, $0.8\times {10}^{-4}=B\times 0.5$ $⟹B=1.6\times {10}^{-4}T$