A copper disc of radius `10cm` is rotating in magnetic field `B=0.4 G` with `10 rev /s`. What will be the potential differenee across the peripheral points of disc?

Answer

To solve the problem of finding the potential difference across the peripheral points of a rotating copper disc in a magnetic field, we can follow these steps:

### Step 1: Understand the given parameters
- Radius of the disc, \( r = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Magnetic field strength, \( B = 0.4 \, \text{G} = 0.4 \times 10^{-4} \, \text{T} \)
- Frequency of rotation, \( f = 10 \, \text{rev/s} \)

### Step 2: Calculate the angular velocity
The angular velocity \( \omega \) can be calculated using the formula:
\[ \omega = 2\pi f \]
Substituting the value of \( f \):
\[ \omega = 2\pi \times 10 \, \text{rev/s} = 20\pi \, \text{rad/s} \]

### Step 3: Calculate the potential difference
The potential difference \( V \) across the peripheral points of the disc can be calculated using the formula:
\[ V = \frac{1}{2} B \omega r^2 \]
Substituting the values of \( B \), \( \omega \), and \( r \):
\[ V = \frac{1}{2} \times (0.4 \times 10^{-4} \, \text{T}) \times (20\pi \, \text{rad/s}) \times (0.1 \, \text{m})^2 \]

### Step 4: Simplify the expression
Calculating \( r^2 \):
\[ r^2 = (0.1)^2 = 0.01 \, \text{m}^2 \]
Now substituting this back into the equation:
\[ V = \frac{1}{2} \times (0.4 \times 10^{-4}) \times (20\pi) \times (0.01) \]
\[ V = (0.2 \times 10^{-4}) \times (20\pi \times 0.01) \]
\[ V = (0.2 \times 10^{-4}) \times (0.2\pi) \]
\[ V = 0.04\pi \times 10^{-4} \]

### Step 5: Calculate the numerical value
Using \( \pi \approx 3.14 \):
\[ V \approx 0.04 \times 3.14 \times 10^{-4} = 0.1256 \times 10^{-4} \, \text{V} = 1.256 \times 10^{-5} \, \text{V} \]

### Final Answer
The potential difference across the peripheral points of the disc is approximately:
\[ V \approx 1.256 \times 10^{-5} \, \text{V} \, \text{or} \, 12.56 \, \mu\text{V} \]