A copper disc of radius 0.1 m is rotated about its centre with 20 revolution per second in a uniform magnetic field of 0.1 T with its plane perpendicular to the field. The emf induced across the radius of the disc is-

Options

A. `(pi)/(20)`volt
B. `(pi)/(10)` volt
C. `20 pi`millivolt
D. `100 pi` millivolt

Explanation

To solve the problem of finding the induced EMF across the radius of a rotating copper disc in a magnetic field, we can follow these steps:

### Step 1: Identify the given values
- Radius of the disc, \( R = 0.1 \, \text{m} \)
- Frequency of rotation, \( f = 20 \, \text{rev/s} \)
- Magnetic field strength, \( B = 0.1 \, \text{T} \)

### Step 2: Convert frequency to angular velocity
The angular velocity \( \omega \) can be calculated using the formula:
\[ \omega = 2\pi f \]
Substituting the given frequency:
\[ \omega = 2\pi \times 20 = 40\pi \, \text{rad/s} \]

### Step 3: Set up the expression for induced EMF
The induced EMF (\( E \)) in a rotating disc can be calculated using the formula:
\[ E = \int_0^R B \cdot V \cdot dx \]
Where \( V \) is the linear velocity at a distance \( x \) from the center, given by:
\[ V = x \cdot \omega \]
Thus, we can rewrite the expression for EMF as:
\[ E = \int_0^R B \cdot (x \cdot \omega) \cdot dx \]

### Step 4: Substitute and integrate
Substituting \( V \) into the EMF equation:
\[ E = B \cdot \omega \cdot \int_0^R x \, dx \]
The integral of \( x \) from 0 to \( R \) is:
\[ \int_0^R x \, dx = \frac{R^2}{2} \]
Thus, we have:
\[ E = B \cdot \omega \cdot \frac{R^2}{2} \]

### Step 5: Substitute the known values
Now substituting the values of \( B \), \( \omega \), and \( R \):
\[ E = 0.1 \cdot (40\pi) \cdot \frac{(0.1)^2}{2} \]
Calculating this step-by-step:
\[ E = 0.1 \cdot 40\pi \cdot \frac{0.01}{2} \]
\[ E = 0.1 \cdot 40\pi \cdot 0.005 \]
\[ E = 0.1 \cdot 0.2\pi \]
\[ E = 0.02\pi \, \text{V} \]

### Step 6: Convert to millivolts
To express this in millivolts:
\[ E = 20\pi \, \text{mV} \]

### Step 7: Calculate the numerical value
Using \( \pi \approx 3.14 \):
\[ E \approx 20 \times 3.14 = 62.8 \, \text{mV} \]

### Final Answer
The induced EMF across the radius of the disc is approximately \( 62.8 \, \text{mV} \).

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