At a plane the value of horizontal component of the eart's magnetic field `H` is `3xx10^(-5)weber//m^(2)`. A metallic rod `AB` of length `2m` placed in east-west direction, having the end `A` towards east, falls vertically downward with a constant velocity of `50 m//s`. which end of the rod becomes positively charged and what is the value of induced potential difference between the two ends?

Options

A. End A , `3 xx 10^(-3)` mV
B. End A , 3mV
C. End B , `3 xx 10^(-3)` mV
D. End B ,3 mV

Answer

To solve the problem, we need to determine which end of the metallic rod becomes positively charged and calculate the induced potential difference between the two ends of the rod.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Horizontal component of the Earth's magnetic field, \( H = 3 \times 10^{-5} \, \text{Wb/m}^2 \)
- Length of the rod, \( L = 2 \, \text{m} \)
- Velocity of the rod, \( V = 50 \, \text{m/s} \)

2. **Use the Formula for Induced EMF:**
The induced electromotive force (emf) can be calculated using the formula:
\[ E = B \cdot L \cdot V \]
where \( B \) is the magnetic field, \( L \) is the length of the rod, and \( V \) is the velocity of the rod.

3. **Substitute the Values into the Formula:**
\[ E = (3 \times 10^{-5} \, \text{Wb/m}^2) \cdot (2 \, \text{m}) \cdot (50 \, \text{m/s}) \]

4. **Calculate the Induced EMF:**
\[ E = 3 \times 10^{-5} \cdot 2 \cdot 50 = 3 \times 10^{-3} \, \text{V} \]
This can be converted to millivolts:
\[ E = 3 \, \text{mV} \]

5. **Determine the Direction of Induced Current:**
To find out which end of the rod becomes positively charged, we can use Fleming's Right-Hand Rule. According to this rule:
- Thumb: Direction of motion (downwards)
- First Finger: Direction of magnetic field (from North to South, which is towards the geographic North)
- Second Finger: Direction of induced current

In this case, if the rod is falling downwards and the magnetic field is horizontal (from West to East), the induced current will flow from end A (East) to end B (West). Therefore, end A becomes positively charged.

6. **Final Answer:**
The end A of the rod becomes positively charged, and the induced potential difference between the two ends is \( 3 \, \text{mV} \).

### Summary:
- **Induced Potential Difference:** \( 3 \, \text{mV} \)
- **Positively Charged End:** End A