A train is moving in the north-south direction with a speed of `108 kmh^(-1)`. Find the e.m.f. generated between two wheels, if the length of the axle is 2m. Assume that the vertical component of earth's field is `8.0 xx 10^(-5) Wbm^(-2)`

Answer

To find the electromotive force (e.m.f.) generated between the two wheels of a train moving in the north-south direction, we can use the formula for induced e.m.f. in a conductor moving through a magnetic field. The formula is given by:

\[ E = B \cdot L \cdot v \]

Where:
- \(E\) is the induced e.m.f. (in volts),
- \(B\) is the magnetic field strength (in webers per square meter, Wb/m²),
- \(L\) is the length of the conductor (in meters),
- \(v\) is the speed of the conductor (in meters per second).

### Step-by-Step Solution:

1. **Convert the speed from km/h to m/s**:
- The speed of the train is given as \(108 \, \text{km/h}\).
- To convert km/h to m/s, we use the conversion factor: \(1 \, \text{km/h} = \frac{5}{18} \, \text{m/s}\).
- Thus,
\[ v = 108 \times \frac{5}{18} = 30 \, \text{m/s} \]

2. **Identify the given values**:
- Length of the axle \(L = 2 \, \text{m}\)
- Vertical component of Earth's magnetic field \(B = 8.0 \times 10^{-5} \, \text{Wb/m}^2\)

3. **Substitute the values into the e.m.f. formula**:
- Now we can substitute the values into the formula \(E = B \cdot L \cdot v\):
\[ E = (8.0 \times 10^{-5}) \cdot (2) \cdot (30) \]

4. **Calculate the e.m.f.**:
- Performing the multiplication:
\[ E = 8.0 \times 10^{-5} \cdot 2 \cdot 30 = 8.0 \times 60 \times 10^{-5} = 480 \times 10^{-5} = 4.8 \times 10^{-3} \, \text{V} \]

5. **Convert the e.m.f. to millivolts**:
- Since \(1 \, \text{V} = 1000 \, \text{mV}\):
\[ E = 4.8 \times 10^{-3} \, \text{V} = 4.8 \, \text{mV} \]

### Final Answer:
The e.m.f. generated between the two wheels of the train is \(4.8 \, \text{mV}\).