To solve the problem, we need to find the value of the Earth's horizontal magnetic field at a second location based on the oscillation frequency of a magnet at two different places.
### Step-by-Step Solution:
1. **Understanding the Given Data:**
- At the first location, the magnet makes 40 oscillations per minute.
- The magnetic field intensity at this location (B_H1) is \(0.1 \times 10^{-5} \, \text{T}\).
- At the second location, the time period (T2) for one complete vibration is given as 2.5 seconds.
2. **Convert Oscillations to Time Period:**
- The frequency (f1) of oscillation at the first location can be calculated as:
\[
f_1 = \frac{40 \, \text{oscillations}}{1 \, \text{minute}} = \frac{40}{60} \, \text{Hz} = \frac{2}{3} \, \text{Hz}
\]
- The time period (T1) is the reciprocal of frequency:
\[
T_1 = \frac{1}{f_1} = \frac{1}{\frac{2}{3}} = \frac{3}{2} \, \text{seconds}
\]
3. **Relate Time Periods to Magnetic Fields:**
- The time period of oscillation is inversely proportional to the magnetic field intensity:
\[
T \propto \frac{1}{B_H}
\]
- Therefore, we can write the ratio of the time periods and magnetic fields:
\[
\frac{T_1}{T_2} = \frac{B_{H2}}{B_{H1}}
\]
4. **Substituting Known Values:**
- We have:
\[
T_1 = \frac{3}{2} \, \text{s}, \quad T_2 = 2.5 \, \text{s}, \quad B_{H1} = 0.1 \times 10^{-5} \, \text{T}
\]
- Plugging these values into the ratio:
\[
\frac{\frac{3}{2}}{2.5} = \frac{B_{H2}}{0.1 \times 10^{-5}}
\]
5. **Calculating the Ratio:**
- Simplifying the left side:
\[
\frac{3/2}{2.5} = \frac{3/2}{5/2} = \frac{3}{5}
\]
- Thus, we have:
\[
\frac{3}{5} = \frac{B_{H2}}{0.1 \times 10^{-5}}
\]
6. **Finding \(B_{H2}\):**
- Rearranging gives:
\[
B_{H2} = 0.1 \times 10^{-5} \times \frac{3}{5}
\]
- Calculating \(B_{H2}\):
\[
B_{H2} = 0.1 \times 10^{-5} \times 0.6 = 0.06 \times 10^{-5} \, \text{T} = 0.36 \times 10^{-6} \, \text{T}
\]
### Final Answer:
The value of the Earth's horizontal magnetic field at the second location is:
\[
B_{H2} = 0.36 \times 10^{-6} \, \text{T}
\]
