At a certain place, the horizontal component of the earth's magnetic field is `B_(0)` and the angle of dip is `45^(circ)`. If the total intensity of the field at that place is `sqrt(alpha) B_(0)`, then find `alpha`

Answer

To solve the problem, we need to use the relationship between the total magnetic field (B), the horizontal component (B₀), and the vertical component (Bᵥ) of the Earth's magnetic field, as well as the angle of dip (δ).

1. **Understanding the Components of the Magnetic Field**:
- The horizontal component of the Earth's magnetic field is given as \( B_0 \).
- The angle of dip (δ) is given as \( 45^\circ \).
- The total intensity of the magnetic field (B) can be expressed in terms of its horizontal (B₀) and vertical (Bᵥ) components using the formula:
\[ B = \sqrt{B_0^2 + B_v^2} \]

2. **Finding the Vertical Component**:
- The vertical component of the magnetic field (Bᵥ) can be related to the horizontal component and the angle of dip using the formula:
\[ B_v = B \sin(\delta) \]
- Since δ = \( 45^\circ \), we have:
\[ \sin(45^\circ) = \frac{1}{\sqrt{2}} \]
- Therefore, the vertical component becomes:
\[ B_v = B \cdot \frac{1}{\sqrt{2}} \]

3. **Expressing Total Magnetic Field**:
- We can also express the total magnetic field in terms of the horizontal component:
\[ B = \sqrt{B_0^2 + \left(B \cdot \frac{1}{\sqrt{2}}\right)^2} \]
- This simplifies to:
\[ B = \sqrt{B_0^2 + \frac{B^2}{2}} \]

4. **Rearranging the Equation**:
- Squaring both sides gives:
\[ B^2 = B_0^2 + \frac{B^2}{2} \]
- Rearranging this, we get:
\[ B^2 - \frac{B^2}{2} = B_0^2 \]
- This simplifies to:
\[ \frac{B^2}{2} = B_0^2 \]
- Multiplying both sides by 2 results in:
\[ B^2 = 2B_0^2 \]

5. **Finding the Value of α**:
- We know from the problem statement that the total intensity of the field is given as:
\[ B = \sqrt{\alpha} B_0 \]
- Squaring both sides gives:
\[ B^2 = \alpha B_0^2 \]
- From our earlier result, we have \( B^2 = 2B_0^2 \). Therefore, we can equate the two expressions:
\[ \alpha B_0^2 = 2B_0^2 \]
- Dividing both sides by \( B_0^2 \) (assuming \( B_0 \neq 0 \)):
\[ \alpha = 2 \]

Thus, the value of \( \alpha \) is **2**.