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

• A

  $1.5 \times 10^{-4} \mathrm{~T}$
• B

  $1.6 \times 10^{-3} \mathrm{~T}$
• C

  $1.5 \times 10^{-3} \mathrm{~T}$
• Correct answer
  D

  $1.6 \times 10^{-4} \mathrm{~T}$

Solution

Given,
$\begin{aligned}<br />B_H & =0.8 \times 10^{-4} \mathrm{~T} \\<br />\theta & =60^{\circ} \\<br />B_e & =?<br />\end{aligned}$

We know that,
$\begin{aligned}<br />B_H & =B_e \cos \theta \\<br />0.8 \times 10^{-4} & =B_e \cos 60^{\circ}<br />\end{aligned}$
$\begin{aligned} B_e & =\frac{0.8 \times 10^{-4}}{\frac{1}{2}} \\ & =1.6 \times 10^{-4} \mathrm{~T}\end{aligned}$

Hence, the correct answer is option: 'D'.