A 2 m long metallic rod rotates with an angular frequency `200" rod "s^(-1)` about on axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant megnetic field of 0.5 T parallel to axis exises everywhere. The emf developed between the centre and the ring is

Options

A. 100V
B. 200V
C. 300V
D. 400V

Explanation

To solve the problem of finding the EMF developed between the center and the ring of a rotating metallic rod, we can follow these steps:

### Step 1: Identify the given values
- Length of the rod (L) = 2 m
- Angular frequency (ω) = 200 rad/s
- Magnetic field strength (B) = 0.5 T

### Step 2: Determine the linear velocity at the end of the rod
The linear velocity (v) of the end of the rod can be calculated using the formula:
\[ v = L \cdot \omega \]
Substituting the values:
\[ v = 2 \, \text{m} \cdot 200 \, \text{rad/s} = 400 \, \text{m/s} \]

### Step 3: Calculate the average linear velocity
Since we need the average linear velocity between the center and the end of the rod, we can use:
\[ v_{\text{avg}} = \frac{v_{\text{end}} + v_{\text{center}}}{2} \]
Here, the linear velocity at the center is 0, so:
\[ v_{\text{avg}} = \frac{400 \, \text{m/s} + 0}{2} = 200 \, \text{m/s} \]

### Step 4: Use the formula for EMF
The EMF (E) induced in the rod can be calculated using the formula:
\[ E = B \cdot L \cdot v_{\text{avg}} \]
Substituting the values:
\[ E = 0.5 \, \text{T} \cdot 2 \, \text{m} \cdot 200 \, \text{m/s} \]

### Step 5: Calculate the EMF
Now, calculating the EMF:
\[ E = 0.5 \cdot 2 \cdot 200 = 200 \, \text{V} \]

### Final Answer
The EMF developed between the center and the ring is **200 V**.

---