Question:
The magnitude of horizontal and vertical components of the total magnetic field at a particular location are 40500 nT and 36450 nT, respectively. The magnetic inclination at the same location in degrees is ________ (rounded off to one decimal place).
The magnitude of horizontal and vertical components of the total magnetic field at a particular location are 40500 nT and 36450 nT, respectively. The magnetic inclination at the same location in degrees is ________ (rounded off to one decimal place).
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Magnetic inclination is always calculated using \(\tan I = \frac{Z}{H}\). If \(Z > H\), inclination will be greater than 45° (field dipping steeply); if \(Z < H\), inclination will be less than 45° (field more horizontal).
Updated On: Aug 22, 2025
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Solution and Explanation
Step 1: Recall formula for magnetic inclination
Magnetic inclination (\(I\)) is the angle between the total magnetic field and the horizontal plane. It is given by: \[ \tan I = \frac{Z}{H} \] where: - \(Z = \) Vertical component of magnetic field = \(36450 \, \text{nT}\)
- \(H = \) Horizontal component of magnetic field = \(40500 \, \text{nT}\) Step 2: Substitute values
\[ \tan I = \frac{36450}{40500} \] \[ \tan I = 0.9 \] Step 3: Take inverse tangent
\[ I = \tan^{-1}(0.9) \] Using calculator: \[ I = 42.0^\circ \, (\text{rounded to one decimal place}) \] Step 4: Final Answer
\[ \boxed{42.0^\circ} \]
Magnetic inclination (\(I\)) is the angle between the total magnetic field and the horizontal plane. It is given by: \[ \tan I = \frac{Z}{H} \] where: - \(Z = \) Vertical component of magnetic field = \(36450 \, \text{nT}\)
- \(H = \) Horizontal component of magnetic field = \(40500 \, \text{nT}\) Step 2: Substitute values
\[ \tan I = \frac{36450}{40500} \] \[ \tan I = 0.9 \] Step 3: Take inverse tangent
\[ I = \tan^{-1}(0.9) \] Using calculator: \[ I = 42.0^\circ \, (\text{rounded to one decimal place}) \] Step 4: Final Answer
\[ \boxed{42.0^\circ} \]
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