Discussion
A magnetometer reads a field of \( 20 \ \mathrm{\mu T} \) north and \( 5 \ \mathrm{\mu T} \) east in the horizontal plane. The magnetic heading offset is approximately
Answer: A
Correct answer: 14.0 degrees
The heading offset from magnetic north is the angle of the horizontal field vector:
\[ \theta = \arctan\left(\frac{B_E}{B_N}\right) = \arctan\left(\frac{5}{20}\right) = \arctan(0.25) = 14.0\ \mathrm{degrees} \]
The option \( 76.0 \) is the complement, from dividing the components the other way round - a common slip, avoided by remembering that the angle is measured from north, so the north component belongs in the denominator. Only the horizontal components are used; the vertical component carries no heading information, which is why the sensor must first be levelled using the attitude solution.
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