The number of turns in the coil of an ac genrator is `5000` and the area of the coil is `0.25m^(2)`. The coil is rotate at the rate of `100 "cycles"//"sec"` in a magnetic field of `0.2W//m^(2)`. The peak value of the emf generated is nearly

Options

A. 786 kV
B. 440 kV
C. 220 kV
D. 157 kV

Explanation

To find the peak value of the emf generated by the AC generator, we can use the formula for the emf (E) generated in an AC generator:

\[ E = N \cdot B \cdot A \cdot \omega \]

Where:
- \( E \) = peak emf
- \( N \) = number of turns in the coil
- \( B \) = magnetic field strength
- \( A \) = area of the coil
- \( \omega \) = angular velocity in radians per second

### Step 1: Identify the given values
- Number of turns, \( N = 5000 \)
- Area of the coil, \( A = 0.25 \, m^2 \)
- Frequency, \( f = 100 \, cycles/sec \)
- Magnetic field strength, \( B = 0.2 \, Wb/m^2 \)

### Step 2: Calculate the angular velocity (\( \omega \))
The angular velocity is related to the frequency by the formula:

\[ \omega = 2 \pi f \]

Substituting the value of frequency:

\[ \omega = 2 \cdot \pi \cdot 100 \]

Calculating this gives:

\[ \omega \approx 2 \cdot 3.14 \cdot 100 \approx 628.32 \, rad/sec \]

### Step 3: Substitute the values into the emf formula
Now, we can substitute the values into the emf formula:

\[ E = N \cdot B \cdot A \cdot \omega \]

Substituting the known values:

\[ E = 5000 \cdot 0.2 \cdot 0.25 \cdot 628.32 \]

### Step 4: Perform the calculations
Calculating step by step:

1. Calculate \( 5000 \cdot 0.2 = 1000 \)
2. Calculate \( 1000 \cdot 0.25 = 250 \)
3. Finally, calculate \( 250 \cdot 628.32 \)

\[ E \approx 250 \cdot 628.32 \approx 157080 \, V \]

### Step 5: Convert to kilovolts
To convert volts to kilovolts, we divide by 1000:

\[ E \approx \frac{157080}{1000} \approx 157.08 \, kV \]

### Final Answer
The peak value of the emf generated is approximately **157 kV**.

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