In the magnetic meridian of a certain plane,the horizontal component of earth's magnetic field is 0.36 Gauss and the dip angle is 60°. The magnetic field of the earth at location is:
In the magnetic meridian of a certain plane,the horizontal component of earth's magnetic field is 0.36 Gauss and the dip angle is 60°. The magnetic field of the earth at location is:
0.72 Gauss
0.18 Gauss
0.42 Gauss
0.56 Gauss
0.81 Gauss
The Correct Option is A
Approach Solution - 1
Given:
- Horizontal component of Earth's magnetic field (\( B_H \)) = 0.36 Gauss
- Dip angle (\( \theta \)) = \( 60^\circ \)
Step 1: Understand the Relationship Between Components
The Earth's total magnetic field (\( B \)) can be resolved into horizontal (\( B_H \)) and vertical (\( B_V \)) components:
\[ B_H = B \cos \theta \]
\[ B_V = B \sin \theta \]
Step 2: Calculate the Total Magnetic Field (\( B \))
Using the horizontal component formula:
\[ B = \frac{B_H}{\cos \theta} \]
Substitute the given values:
\[ B = \frac{0.36 \, \text{Gauss}}{\cos 60^\circ} \]
\[ \cos 60^\circ = 0.5 \]
\[ B = \frac{0.36}{0.5} = 0.72 \, \text{Gauss} \]
Conclusion:
The total magnetic field of the Earth at this location is 0.72 Gauss.
Answer: \(\boxed{A}\)
Approach Solution -2
Step 1: Recall the relationship between the components of Earth's magnetic field.
The total magnetic field of the Earth (\( B \)) can be resolved into two components:
- The horizontal component (\( B_H \)), and
- The vertical component (\( B_V \)).
The relationship between these components is given by:
\[ B_H = B \cos \theta, \]
where:
- \( B \) is the total magnetic field of the Earth,
- \( B_H \) is the horizontal component, and
- \( \theta \) is the dip angle.
We are tasked with finding \( B \), given \( B_H = 0.36 \, \text{Gauss} \) and \( \theta = 60^\circ \).
Step 2: Solve for \( B \).
Rearranging the formula \( B_H = B \cos \theta \) to solve for \( B \):
\[ B = \frac{B_H}{\cos \theta}. \]
Substitute the given values \( B_H = 0.36 \, \text{Gauss} \) and \( \cos 60^\circ = 0.5 \):
\[ B = \frac{0.36}{0.5} = 0.72 \, \text{Gauss}. \]
Final Answer: The magnetic field of the Earth at this location is \( \mathbf{0.72 \, \text{Gauss}} \), which corresponds to option \( \mathbf{(A)} \).
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