A metallic rod of 1 m length is rotated with a frequency of 50 `rev//s`, with on end hinged at the centre and the other end at the circumference of a circular metallic ring of radius 1 m, about an axis passing through the centre and perpendicular at to the plane of the ring. A constant uniform magnetic field of 1 T parallel to the axis is persent eveywhere. what is the e.m.f. between the centre and the metallic ring?

Options

A. 157V
B. 117V
C. 127V
D. 137V

Answer

To solve the problem of finding the electromotive force (e.m.f.) generated between the center and the metallic ring due to the rotation of the rod 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 (f) = 50 rev/s
- Magnetic field (B) = 1 T

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

3. **Use the formula for e.m.f. (E):**
- The formula for the induced e.m.f. in a rotating rod in a magnetic field is:
\[ E = \frac{1}{2} B \omega L^2 \]

4. **Substitute the values into the e.m.f. formula:**
- Now substituting B, ω, and L into the formula:
\[ E = \frac{1}{2} \times 1 \, \text{T} \times (100\pi \, \text{rad/s}) \times (1 \, \text{m})^2 \]
- This simplifies to:
\[ E = \frac{1}{2} \times 1 \times 100\pi \times 1 \]
\[ E = 50\pi \, \text{V} \]

5. **Calculate the numerical value:**
- Approximating π as 3.14:
\[ E \approx 50 \times 3.14 = 157 \, \text{V} \]

### Final Answer:
The e.m.f. between the center and the metallic ring is approximately **157 V**.

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