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A jet plane is travelling west at $450 \mathrm{~m} / \mathrm{s}$. If the horizontal component of earth's magnetic field is $4 \times 10^{-4} \mathrm{~T}$ and angle of dip is $30^{\circ}$, then the vertical component is
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Verified Answer
The correct answer is:
$2.3 \times 10^{-4} \mathrm{~T}$
Given, $B_{H}=4 \times 10^{-4} \mathrm{~T}$
Angle of $\operatorname{dip}, \delta=30^{\circ}$
We know that,
$$
\begin{aligned}
\tan \delta &=\frac{B_{V}}{B_{H}} \\
\Rightarrow \quad B_{V} &=B_{H} \tan \delta \\
&=4 \times 10^{-4} \tan 30^{\circ} \\
&=\frac{4}{\sqrt{3}} \times 10^{-4} \\
&=2.3 \times 10^{-4} \mathrm{~T}
\end{aligned}
$$
Angle of $\operatorname{dip}, \delta=30^{\circ}$
We know that,
$$
\begin{aligned}
\tan \delta &=\frac{B_{V}}{B_{H}} \\
\Rightarrow \quad B_{V} &=B_{H} \tan \delta \\
&=4 \times 10^{-4} \tan 30^{\circ} \\
&=\frac{4}{\sqrt{3}} \times 10^{-4} \\
&=2.3 \times 10^{-4} \mathrm{~T}
\end{aligned}
$$
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