A circular copper disc of radius 25 cm is rotating about its own axis with a constant angular velocity of `130 rad//s` . If a magnetic field of `5xx10^(-3)` Tesla is applied at right angles to the disc, then the induced potential difference between the centre and the rim of the disc will approximately be __

Options

A. `20xx10^(-3) V`
B. `20xx10^(-6)V`
C. `20xx10^(-9)`
D. Zero

Answer

To solve the problem of finding the induced potential difference between the center and the rim of a rotating copper disc in a magnetic field, we can follow these steps:

### Step 1: Identify the given values
- Radius of the disc, \( R = 25 \, \text{cm} = 0.25 \, \text{m} \)
- Angular velocity, \( \omega = 130 \, \text{rad/s} \)
- Magnetic field strength, \( B = 5 \times 10^{-3} \, \text{T} \)

### Step 2: Understand the induced EMF in a rotating disc
The induced EMF (\( \text{d}E \)) in a small element of the disc can be calculated using the formula:
\[ \text{d}E = B \cdot v \cdot \text{d}L \]
where:
- \( v \) is the linear velocity of the small element,
- \( \text{d}L \) is the length of the small element.

### Step 3: Calculate the linear velocity
The linear velocity \( v \) of a point at a distance \( x \) from the center of the disc is given by:
\[ v = \omega \cdot x \]

### Step 4: Set up the expression for induced EMF
Taking a small strip of thickness \( \text{d}x \) at a distance \( x \) from the center, the induced EMF can be expressed as:
\[ \text{d}E = B \cdot (\omega \cdot x) \cdot \text{d}x \]

### Step 5: Integrate to find total EMF
To find the total EMF (\( E \)) between the center (0) and the rim (R), we integrate:
\[ E = \int_0^R B \cdot (\omega \cdot x) \, \text{d}x \]
This simplifies to:
\[ E = B \cdot \omega \int_0^R x \, \text{d}x \]

### Step 6: Solve the integral
The integral \( \int_0^R x \, \text{d}x \) evaluates to:
\[ \int_0^R x \, \text{d}x = \left[ \frac{x^2}{2} \right]_0^R = \frac{R^2}{2} \]
Thus, we have:
\[ E = B \cdot \omega \cdot \frac{R^2}{2} \]

### Step 7: Substitute the known values
Now substituting the values:
\[ E = (5 \times 10^{-3} \, \text{T}) \cdot (130 \, \text{rad/s}) \cdot \frac{(0.25 \, \text{m})^2}{2} \]
Calculating \( (0.25)^2 = 0.0625 \):
\[ E = (5 \times 10^{-3}) \cdot (130) \cdot \frac{0.0625}{2} \]
\[ E = (5 \times 10^{-3}) \cdot (130) \cdot (0.03125) \]
\[ E = 5 \times 130 \times 0.03125 \times 10^{-3} \]
\[ E = 20.3125 \times 10^{-3} \, \text{V} \]

### Step 8: Round the result
Thus, the induced potential difference is approximately:
\[ E \approx 20 \times 10^{-3} \, \text{V} = 20 \, \text{mV} \]

### Final Answer
The induced potential difference between the center and the rim of the disc is approximately \( 20 \, \text{mV} \).

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