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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
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Verified Answer
The correct answer is:
$1.6 \times 10^{-4} \mathrm{~T}$
Given,
$\begin{aligned}
B_H & =0.8 \times 10^{-4} \mathrm{~T} \\
\theta & =60^{\circ} \\
B_e & =?
\end{aligned}$
We know that,
$\begin{aligned}
B_H & =B_e \cos \theta \\
0.8 \times 10^{-4} & =B_e \cos 60^{\circ}
\end{aligned}$
$\begin{aligned} B_e & =\frac{0.8 \times 10^{-4}}{\frac{1}{2}} \\ & =1.6 \times 10^{-4} \mathrm{~T}\end{aligned}$
$\begin{aligned}
B_H & =0.8 \times 10^{-4} \mathrm{~T} \\
\theta & =60^{\circ} \\
B_e & =?
\end{aligned}$
We know that,
$\begin{aligned}
B_H & =B_e \cos \theta \\
0.8 \times 10^{-4} & =B_e \cos 60^{\circ}
\end{aligned}$
$\begin{aligned} B_e & =\frac{0.8 \times 10^{-4}}{\frac{1}{2}} \\ & =1.6 \times 10^{-4} \mathrm{~T}\end{aligned}$
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