Question
The magnetic field due to the earth has a horizontal component of $26\mu\text{T}$ at a place where the dip is 60°. Find the vertical component and the magnitude of the field.
✓
Answer
$\delta(\text{dip}) =60^\circ$$\text{B}_\text{H}=\text{B}\cos60^\circ$
$\Rightarrow\text{B}=52\times10^{-6}=52\mu\text{T}$
$\text{B}_\text{v}=\text{B}\sin\delta=52\times10^{-6}\frac{\sqrt{3}}{2}$
$=44.98\mu\text{T} \approx45\mu\text{T}$
$\Rightarrow\text{B}=52\times10^{-6}=52\mu\text{T}$
$\text{B}_\text{v}=\text{B}\sin\delta=52\times10^{-6}\frac{\sqrt{3}}{2}$
$=44.98\mu\text{T} \approx45\mu\text{T}$
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