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A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at $22^{\circ}$ with the horizontal. The horizontal component of the earth's magnetic field at the place is known to be $0.35$ G. Determine the magnitude of the earth's magnetic field at the place.
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Verified Answer
Given: Horizontal component of earth's magnetic field $\mathrm{H}$ $=0.35 \mathrm{G}$
angle of $\operatorname{dip} \delta=22^{\circ}$
To find: Earth's magnetic field B.
Formula: $\mathrm{H}=\mathrm{B} \cos \delta$
$$
\begin{aligned}
\mathrm{B} &=\frac{\mathrm{H}}{\cos \delta}=\frac{0.35}{\cos 22^{\circ}} \\
\Rightarrow \mathrm{B} &=\frac{0.35}{0.9272}=0.38 \mathrm{G}
\end{aligned}
$$
angle of $\operatorname{dip} \delta=22^{\circ}$
To find: Earth's magnetic field B.
Formula: $\mathrm{H}=\mathrm{B} \cos \delta$
$$
\begin{aligned}
\mathrm{B} &=\frac{\mathrm{H}}{\cos \delta}=\frac{0.35}{\cos 22^{\circ}} \\
\Rightarrow \mathrm{B} &=\frac{0.35}{0.9272}=0.38 \mathrm{G}
\end{aligned}
$$
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