A conducting rod of 1m length rotating with a frequency of 50 rev/sec. about its one of end inside the uniform magnetic field of 6.28 mT. The valu of induced emf between end of rod is :-

Options

A. 1 V
B. 2V
C. 0.5 V
D. 0.25 V

Answer

To find the induced electromotive force (emf) between the ends of a conducting rod rotating in a magnetic field, we can follow these steps:

### Step-by-Step Solution

1. **Identify the Given Values:**
- Length of the rod (L) = 1 m
- Frequency of rotation (f) = 50 rev/sec
- Magnetic field strength (B) = 6.28 mT = 6.28 × 10^(-3) T

2. **Convert Frequency to Angular Velocity:**
- The angular velocity (ω) can be calculated using the formula:
\[ \omega = 2\pi f \]
- Substituting the given frequency:
\[ \omega = 2\pi \times 50 = 100\pi \, \text{rad/sec} \]

3. **Use the Formula for Induced EMF:**
- The induced emf (ε) in a rotating rod in a magnetic field is given by the formula:
\[ \epsilon = \frac{1}{2} B \omega L^2 \]
- Here, B is the magnetic field, ω is the angular velocity, and L is the length of the rod.

4. **Substitute the Values into the Formula:**
- Now we can substitute the values we have:
\[ \epsilon = \frac{1}{2} \times (6.28 \times 10^{-3}) \times (100\pi) \times (1^2) \]

5. **Calculate the Induced EMF:**
- First, calculate \(100\pi\):
\[ 100\pi \approx 314.16 \]
- Now substitute this back into the equation:
\[ \epsilon = \frac{1}{2} \times (6.28 \times 10^{-3}) \times 314.16 \]
- Calculate:
\[ \epsilon = \frac{1}{2} \times 1.973 \approx 0.9865 \, \text{V} \]
- Rounding off gives:
\[ \epsilon \approx 0.99 \, \text{V} \]

6. **Final Answer:**
- The value of the induced emf between the ends of the rod is approximately **0.99 V**.