At a certain place, the angle of dip is $60^{\circ}$ and the horizontal component of the earth's magnetic…

1. $1.5 \times 10^{-4} \mathrm{~T}$
2. $1.6 \times 10^{-3} \mathrm{~T}$
3. $1.5 \times 10^{-3} \mathrm{~T}$
4. $1.6 \times 10^{-4} \mathrm{~T}$

Solution

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

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

Asked in: AP EAMCET 2014

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