Question:
The earth's magnetic field at a certain place has a total strength of 0.5 Gauss and the horizontal component of 0.3 Gauss. Then the angle of dip at that place is
The earth's magnetic field at a certain place has a total strength of 0.5 Gauss and the horizontal component of 0.3 Gauss. Then the angle of dip at that place is
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Think of the classic 3-4-5 right triangle. Since the horizontal base is 3 and the hypotenuse is 5, the vertical side must be 4, making $\tan\delta = \frac{\text{opposite}}{\text{adjacent}} = \frac{4}{3}$.
Updated On: Jun 3, 2026
- $\tan^{-1}\frac{3}{4}$
- $\tan^{-1}\frac{4}{3}$
- $\sin^{-1}\frac{3}{4}$
- $\sin^{-1}\frac{3}{5}$ \
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The Correct Option is B
Solution and Explanation
Step 1: Concept
Earth's total magnetic field $B$ can be resolved into a horizontal component $B_{H} = B \cos\delta$ and a vertical component $B_{V} = B \sin\delta$, where $\delta$ is the angle of dip.
Step 2: Meaning
The relation between the horizontal component, total field, and angle of dip gives $\cos\delta = \frac{B_{H}}{B}$. From this, the vertical component can be found using $B_{V} = \sqrt{B^2 - B_{H}^2}$.
Step 3: Analysis
Given $B = 0.5\text{ Gauss}$ and $B_{H} = 0.3\text{ Gauss}$, we get $\cos\delta = \frac{0.3}{0.5} = \frac{3}{5}$. In a right-angled triangle with an adjacent side of 3 and a hypotenuse of 5, the opposite side (vertical component $B_{V}$) is $\sqrt{5^2 - 3^2} = 4$. Therefore, $\tan\delta = \frac{B_{V}}{B_{H}} = \frac{4}{3}$.
Step 4: Conclusion
Solving for the angle of dip gives $\delta = \tan^{-1}\frac{4}{3}$.
Final Answer: (B)
Earth's total magnetic field $B$ can be resolved into a horizontal component $B_{H} = B \cos\delta$ and a vertical component $B_{V} = B \sin\delta$, where $\delta$ is the angle of dip.
Step 2: Meaning
The relation between the horizontal component, total field, and angle of dip gives $\cos\delta = \frac{B_{H}}{B}$. From this, the vertical component can be found using $B_{V} = \sqrt{B^2 - B_{H}^2}$.
Step 3: Analysis
Given $B = 0.5\text{ Gauss}$ and $B_{H} = 0.3\text{ Gauss}$, we get $\cos\delta = \frac{0.3}{0.5} = \frac{3}{5}$. In a right-angled triangle with an adjacent side of 3 and a hypotenuse of 5, the opposite side (vertical component $B_{V}$) is $\sqrt{5^2 - 3^2} = 4$. Therefore, $\tan\delta = \frac{B_{V}}{B_{H}} = \frac{4}{3}$.
Step 4: Conclusion
Solving for the angle of dip gives $\delta = \tan^{-1}\frac{4}{3}$.
Final Answer: (B)
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