Question

In a certain region the average horizontal component of Earth's magnetic field is 17 μT, and the average inclination or "dip" is 79°. What is the corresponding magnitude of Earth's magnetic field in microteslas?

Answer 1

Answer:

the Earth magnetic field in that region is about 89.09 $\mu$T

Explanation:

If the magnetic field has an inclination of $79^o$, and its horizontal component is 17 microTesla, then we can calculate the magnitude of the full magnetic field via the cosine function. Notice that the decomposition of a vector in horizontal and perpendicular components originates a right angle triangle where the vector is the hypotenuse of the triangle.

$cos(\theta)=\frac{adjacent}{hypotenuse}\\cos(79^o)=\frac{17}{B} \\B=\frac{17}{cos(79^o)}\,\mu T \\B=89.09\,\,\mu T$

Answer 2

The magnitude of Earth's magnetic field is 89 μT

Trigonometric ratio is used to show the relationship between the sides and the angle of a right angle triangle.

The Earth's magnetic field is in the hypotenuse side.

Applying trigonometric ratios:

cos(θ) = adjacent / hypotenuse

cos(79) = 17 μT / hypotenuse

hypotenuse = 89 μT

The magnitude of Earth's magnetic field is 89 μT

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