At a place, the horizontal component of Earth's magnetic field is `3 xx 10^(-5) T` and angle of dip is 45 . A train passes with a speed of `80 km h^(-1)` on a railway track running north-south. Calculate the emf induced between the rails if the distance between both the rails is 80 cm.

Answer

To solve the problem, we need to calculate the electromotive force (emf) induced between the rails of the railway track due to the motion of the train in the presence of Earth's magnetic field.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Horizontal component of Earth's magnetic field, \( B_H = 3 \times 10^{-5} \, T \)
- Angle of dip, \( \delta = 45^\circ \)
- Speed of the train, \( V = 80 \, km/h \)
- Distance between the rails, \( L = 80 \, cm = 0.8 \, m \)

2. **Convert Speed to SI Units:**
- Convert \( V \) from km/h to m/s:
\[ V = 80 \, km/h \times \frac{1000 \, m}{1 \, km} \times \frac{1 \, h}{3600 \, s} = \frac{80000}{3600} \approx 22.22 \, m/s \]

3. **Calculate the Vertical Component of the Magnetic Field:**
- Since the angle of dip is \( 45^\circ \), the vertical component \( B_V \) can be calculated using:
\[ B_V = B_H \cdot \tan(\delta) = B_H \cdot \tan(45^\circ) = B_H \cdot 1 = B_H = 3 \times 10^{-5} \, T \]

4. **Determine the Induced EMF:**
- The formula for the motional EMF induced in a conductor moving through a magnetic field is given by:
\[ \text{EMF} = B_V \cdot L \cdot V \]
- Substituting the values:
\[ \text{EMF} = (3 \times 10^{-5} \, T) \cdot (0.8 \, m) \cdot (22.22 \, m/s) \]

5. **Calculate the EMF:**
- Performing the multiplication:
\[ \text{EMF} = 3 \times 10^{-5} \cdot 0.8 \cdot 22.22 \approx 0.533 \times 10^{-3} \, V = 0.533 \, mV \]

### Final Answer:
The induced EMF between the rails is approximately \( 0.533 \, mV \).

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