Two parallel rails of a railway track insulated from each other and with the ground are connected to a millivoltmeter. The distance between the rails is one metre. A train is traveling with a velocity of `72 km ph` along the track. The reading of the millivoltmetre (in `m V`) is : (Vertical component of the earth's magnetic induction is `2 xx 10^(-5) T`)

Options

A. 1.44
B. 0.72
C. 0.4
D. 0.2

Answer

To solve the problem, we need to calculate the potential difference (voltage) induced across the rails of the railway track due to the motion of the train in the presence of a magnetic field. We will use the formula for electromagnetic induction:

\[ V = B \times v \times L \]

Where:
- \( V \) is the induced voltage (potential difference),
- \( B \) is the magnetic field strength,
- \( v \) is the velocity of the train,
- \( L \) is the length between the rails.

### Step-by-Step Solution:

1. **Convert the velocity of the train from km/h to m/s:**
\[ v = 72 \text{ km/h} = 72 \times \frac{5}{18} = 20 \text{ m/s} \]

2. **Identify the given values:**
- The vertical component of the Earth's magnetic induction, \( B = 2 \times 10^{-5} \text{ T} \)
- The distance between the rails (length), \( L = 1 \text{ m} \)
- The velocity of the train, \( v = 20 \text{ m/s} \)

3. **Substitute the values into the formula:**
\[ V = B \times v \times L \]
\[ V = (2 \times 10^{-5} \text{ T}) \times (20 \text{ m/s}) \times (1 \text{ m}) \]

4. **Calculate the induced voltage:**
\[ V = 2 \times 10^{-5} \times 20 = 4 \times 10^{-4} \text{ V} \]

5. **Convert the voltage from volts to millivolts:**
\[ V = 4 \times 10^{-4} \text{ V} = 0.4 \text{ mV} \]

### Final Answer:
The reading of the millivoltmeter is **0.4 mV**.

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