Question:
If the horizontal component of earth's magnetic field at a place is $2.8 \times 10^{-5}$ T and the magnetic field of the earth at this place is $5.6 \times 10^{-5}$ T, then the inclination at the place is:
If the horizontal component of earth's magnetic field at a place is $2.8 \times 10^{-5}$ T and the magnetic field of the earth at this place is $5.6 \times 10^{-5}$ T, then the inclination at the place is:
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Inclination (or dip) is the angle the total magnetic field makes with the horizontal component.
Updated On: Jun 10, 2026
- $45^\circ$
- $30^\circ$
- $60^\circ$
- $90^\circ$
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The Correct Option is C
Solution and Explanation
Step 1: Concept
The relationship between total magnetic field ($B$), horizontal component ($B_H$), and angle of inclination ($\delta$) is $B_H = B \cos \delta$.
Step 2: Analysis
$\cos \delta = B_H / B = (2.8 \times 10^{-5}) / (5.6 \times 10^{-5}) = 0.5$. $\cos \delta = 0.5 \implies \delta = 60^\circ$.
Step 3: Conclusion
The inclination is $60^\circ$.
Final Answer: (C)
The relationship between total magnetic field ($B$), horizontal component ($B_H$), and angle of inclination ($\delta$) is $B_H = B \cos \delta$.
Step 2: Analysis
$\cos \delta = B_H / B = (2.8 \times 10^{-5}) / (5.6 \times 10^{-5}) = 0.5$. $\cos \delta = 0.5 \implies \delta = 60^\circ$.
Step 3: Conclusion
The inclination is $60^\circ$.
Final Answer: (C)
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