A copper disc of radius `0.1 m` rotates about its centre with `10` revolutuion per second in a uniform magnetic field of `0.1` tesla with its plane perpendicular to the field. The emf induced across the radius of the disc is

Options

A. `(pi)/(10)V`
B. `(2pi)/(10)V`
C. `pixx10^(-2)V`
D. `2pixx10^(-2)V`

Explanation

To find the induced emf across the radius of a rotating copper disc in a magnetic field, we can use the formula for induced emf in a rotating conductor:

\[ \text{emf} = \frac{1}{2} B \omega r^2 \]

Where:
- \( B \) is the magnetic field strength (in tesla),
- \( \omega \) is the angular velocity (in radians per second),
- \( r \) is the radius of the disc (in meters).

### Step-by-Step Solution:

1. **Identify the given values:**
- Radius of the disc, \( r = 0.1 \, \text{m} \)
- Frequency of rotation, \( f = 10 \, \text{revolutions per second} \)
- Magnetic field strength, \( B = 0.1 \, \text{T} \)

2. **Convert frequency to angular velocity:**
The angular velocity \( \omega \) in radians per second can be calculated using the formula:
\[ \omega = 2\pi f \]
Substituting the given frequency:
\[ \omega = 2\pi \times 10 = 20\pi \, \text{radians/second} \]

3. **Substitute the values into the emf formula:**
Now we can substitute \( B \), \( \omega \), and \( r \) into the emf formula:
\[ \text{emf} = \frac{1}{2} B \omega r^2 \]
\[ \text{emf} = \frac{1}{2} \times 0.1 \times (20\pi) \times (0.1)^2 \]

4. **Calculate \( r^2 \):**
\[ r^2 = (0.1)^2 = 0.01 \]

5. **Calculate the emf:**
\[ \text{emf} = \frac{1}{2} \times 0.1 \times (20\pi) \times 0.01 \]
\[ = \frac{1}{2} \times 0.1 \times 20 \times \pi \times 0.01 \]
\[ = \frac{1}{2} \times 0.1 \times 20 \times 0.01 \times \pi \]
\[ = 0.001 \times 10 \times \pi \]
\[ = 0.01\pi \, \text{volts} \]

6. **Final Result:**
The induced emf across the radius of the disc is:
\[ \text{emf} \approx 0.0314 \, \text{volts} \quad (\text{using } \pi \approx 3.14) \]