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?

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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the Earth magnetic field in that region is about 89.09 T

Explanation:

If the magnetic field has an inclination of , 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.

at positions a, g kinetic energy is maximum.

at position d, potential energy is maximum.

what is the actual question though?