A metal rod of length 2 m is rotating with an angualr velocity of 100`"rads"^(-1)` in plane perpendicular to a uniform magnetic field of 0.3 T. The potential difference between the ends of the rod is

Options

A. 30 V
B. 40 V
C. 60 V
D. 600 V

Explanation

To find the potential difference (induced EMF) between the ends of a rotating metal rod in a magnetic field, we can use the formula for motional EMF:

\[ \text{EMF} = \frac{1}{2} B \omega L^2 \]

where:
- \( B \) is the magnetic field strength,
- \( \omega \) is the angular velocity,
- \( L \) is the length of the rod.

### Step-by-step Solution:

1. **Identify the given values:**
- Length of the rod, \( L = 2 \, \text{m} \)
- Angular velocity, \( \omega = 100 \, \text{rad/s} \)
- Magnetic field strength, \( B = 0.3 \, \text{T} \)

2. **Substitute the values into the formula:**
\[ \text{EMF} = \frac{1}{2} B \omega L^2 \]
\[ \text{EMF} = \frac{1}{2} \times 0.3 \, \text{T} \times 100 \, \text{rad/s} \times (2 \, \text{m})^2 \]

3. **Calculate \( L^2 \):**
\[ L^2 = (2 \, \text{m})^2 = 4 \, \text{m}^2 \]

4. **Plug \( L^2 \) back into the equation:**
\[ \text{EMF} = \frac{1}{2} \times 0.3 \times 100 \times 4 \]

5. **Perform the multiplication:**
\[ \text{EMF} = \frac{1}{2} \times 0.3 \times 400 \]
\[ \text{EMF} = \frac{1}{2} \times 120 = 60 \, \text{V} \]

6. **Final result:**
The potential difference (induced EMF) between the ends of the rod is \( 60 \, \text{V} \).