The two rails of a railway track insulated from each other and the ground are connected to a millivoltmeter. What will be the reading of the millivoltmeter, when a train travels at a speed of 180 km/h along the track? Given that the vertical component of Earth's field is `2 xx 10^(-5) Wb//m^(2)` and the rails are separated by 1 m.

Answer

To solve the problem, we need to calculate the electromotive force (EMF) induced between the two rails of the railway track when a train travels at a certain speed. We will use the formula for EMF induced in a conductor moving through a magnetic field.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Speed of the train, \( V = 180 \, \text{km/h} \)
- Vertical component of Earth's magnetic field, \( B = 2 \times 10^{-5} \, \text{Wb/m}^2 \)
- Separation between the rails, \( L = 1 \, \text{m} \)

2. **Convert Speed from km/h to m/s:**
- To convert km/h to m/s, use the conversion factor \( \frac{5}{18} \):
\[ V = 180 \, \text{km/h} \times \frac{5}{18} = 50 \, \text{m/s} \]

3. **Use the Formula for Induced EMF:**
- The formula for the induced EMF (E) in a conductor moving through a magnetic field is given by:
\[ E = B \times L \times V \]
where:
- \( E \) is the induced EMF,
- \( B \) is the magnetic field strength,
- \( L \) is the length of the conductor (distance between the rails),
- \( V \) is the velocity of the conductor.

4. **Substitute the Values into the Formula:**
- Now, substituting the values we have:
\[ E = (2 \times 10^{-5} \, \text{Wb/m}^2) \times (1 \, \text{m}) \times (50 \, \text{m/s}) \]

5. **Calculate the Induced EMF:**
- Performing the multiplication:
\[ E = 2 \times 10^{-5} \times 1 \times 50 = 100 \times 10^{-5} \, \text{V} = 1 \times 10^{-3} \, \text{V} = 1 \, \text{mV} \]

6. **Conclusion:**
- The reading of the millivoltmeter will be \( 1 \, \text{mV} \).