At a certain place, the angle of dip is `30^(@)` and the horizontal component of earth's magnetic field is `50muT`. The total magnetic field (in `muT`) of the earth at this place, is

Options

A. `100sqrt3muT`
B. `100muT`
C. `(100)/(sqrt3)muT`
D. `200muT`

Answer

To solve the problem, we need to find the total magnetic field (B) of the Earth at a place where the angle of dip (θ) is 30 degrees and the horizontal component of the Earth's magnetic field (Bh) is 50 µT.

### 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 that the total magnetic field makes with the horizontal plane.

2. **Use the Relationship Between Components**:
- The horizontal component (Bh) is related to the total magnetic field (B) and the angle of dip (θ) by the formula:
\[ Bh = B \cdot \cos(θ) \]
- Rearranging this formula gives us:
\[ B = \frac{Bh}{\cos(θ)} \]

3. **Substitute the Known Values**:
- We know that Bh = 50 µT and θ = 30 degrees. We can substitute these values into the equation.
- The cosine of 30 degrees is:
\[ \cos(30^\circ) = \frac{\sqrt{3}}{2} \]
- Therefore, substituting the values:
\[ B = \frac{50 \, \mu T}{\cos(30^\circ)} = \frac{50 \, \mu T}{\frac{\sqrt{3}}{2}} = 50 \, \mu T \cdot \frac{2}{\sqrt{3}} = \frac{100 \, \mu T}{\sqrt{3}} \]

4. **Calculate the Total Magnetic Field**:
- To express the answer in a more standard form, we can rationalize the denominator:
\[ B = \frac{100 \, \mu T}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{100\sqrt{3} \, \mu T}{3} \]
- This gives us the total magnetic field in terms of µT.

5. **Final Answer**:
- The total magnetic field (B) at this place is:
\[ B \approx 57.74 \, \mu T \quad (\text{using } \sqrt{3} \approx 1.732) \]

### Conclusion:
The total magnetic field of the Earth at this place is approximately \( \frac{100\sqrt{3}}{3} \, \mu T \) or about 57.74 µT.