Question

The horizontal component of the earth’s magnetic field is $3.6\times 10^{-5}$ tesla where the dip angle is $60^{\circ }$. The magnitude of the earth’s magnetic field is

• A. $2.8\times 10^{-4}$ tesla
• B. $2.1\times 10^{-4}$ tesla
• C. $7.2\times 10^{-5}$ tesla
• D. $3.6\times 10^{-5}$ tesla

Answer 1

Answer: (C)

Horizontal component of earth's field,
$H=B\cos \theta$,
since, $\theta =60^{\circ }$

$3.6\times 10^{-5}=B\times \frac{1}{2}$
$\Rightarrow B=7.2\times 10^{-5}$ Tesla