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The horizontal component of Earth's magnetic field at a place is $3 \times 10^{-5} \mathrm{~T}$. If the dip at that place is $45^{\circ}$, the resultant magnetic field at that place is
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The correct answer is:
{~T}$$3 \sqrt{2} \times 10^{-5} \mathrm{~T}$
Given, $\begin{aligned} B_H & =3 \times 10^{-5} \mathrm{~T} \\ \delta & =45\end{aligned}$
From the diagram
$\cos \delta=\frac{B_H}{B}$
$\begin{aligned} \Rightarrow \quad B & =\frac{B_H}{\cos \delta}=\frac{3 \times 10^{-5}}{\cos 45^{\circ}} \\ & =3 \sqrt{2} \times 10^{-5} \mathrm{~T}\end{aligned}$
From the diagram
$\cos \delta=\frac{B_H}{B}$
$\begin{aligned} \Rightarrow \quad B & =\frac{B_H}{\cos \delta}=\frac{3 \times 10^{-5}}{\cos 45^{\circ}} \\ & =3 \sqrt{2} \times 10^{-5} \mathrm{~T}\end{aligned}$
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