In New Hampshire the average horizontal component of Earth's magnetic field in 1912 was 16 uT, and the average inclination or "dip" was 73 degrees. What was the corresponding mag-nitude of Earth's magnetic field?
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To calculate the magnitude of Earth's magnetic field using the horizontal component and inclination, you can use the trigonometric relationship between the three components: horizontal (H), inclination (I), and total field (F).
The relationship is given by the equation:
F = H / cos(I)
Where:
F is the total magnetic field (magnitude of Earth's magnetic field).
H is the horizontal component of Earth's magnetic field.
I is the inclination (or dip) angle.
In this case, the horizontal component (H) is given as 16 uT and the inclination (I) is given as 73 degrees.
First, convert the inclination angle from degrees to radians:
I (radians) = I (degrees) * pi / 180
I (radians) = 73 * pi / 180
I (radians) ≈ 1.274090
Next, substitute the values into the equation:
F = H / cos(I)
F = 16 uT / cos(1.274090)
Evaluate the cosine of the inclination angle using a scientific calculator:
cos(1.274090) ≈ 0.45223
Finally, calculate the magnitude of Earth's magnetic field:
F ≈ 16 uT / 0.45223
F ≈ 35.407 uT
Therefore, the corresponding magnitude of Earth's magnetic field in New Hampshire in 1912 was approximately 35.407 uT.