At certain place, the horizontal component of earth's magnetic field is 3.0G and the angle dip at the place is `30^(@)`. The magnetic field of earth at that location

Options

A. 4.5G
B. 5.1G
C. 3.5G
D. 6.0G

Answer

To find the total magnetic field of the Earth at a given location where the horizontal component of the Earth's magnetic field (BH) is 3.0 Gauss and the angle of dip (θ) is 30 degrees, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Components of the Magnetic Field**:
The total magnetic field (B) can be resolved into two components: the horizontal component (BH) and the vertical component (BV). The angle of dip (θ) is the angle between the total magnetic field and the horizontal component.

2. **Use the Relationship Between Components**:
The relationship between the total magnetic field (B), the horizontal component (BH), and the angle of dip (θ) is given by:
\[ BH = B \cdot \cos(\theta) \]
Rearranging this formula gives us:
\[ B = \frac{BH}{\cos(\theta)} \]

3. **Substitute the Known Values**:
We know that:
- \( BH = 3.0 \, \text{Gauss} \)
- \( \theta = 30^\circ \)
- The cosine of 30 degrees is \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \).

Plugging these values into the equation gives:
\[ B = \frac{3.0}{\cos(30^\circ)} = \frac{3.0}{\frac{\sqrt{3}}{2}} = 3.0 \cdot \frac{2}{\sqrt{3}} = \frac{6.0}{\sqrt{3}} \]

4. **Calculate the Total Magnetic Field**:
To simplify \( \frac{6.0}{\sqrt{3}} \):
\[ B \approx \frac{6.0}{1.732} \approx 3.464 \, \text{Gauss} \]

5. **Final Result**:
The total magnetic field at that location is approximately \( 3.464 \, \text{Gauss} \), which can be rounded to \( 3.5 \, \text{Gauss} \).