A coil of area `80` square cm and 50 turns is rotating with `2000` revolution per minut about an axis perpendicular to a magnetic field field of `0.05` Telsa. The maximum value of the e.m.f. developed in it is

Options

A. `200pi "volt"`
B. `(10pi)/(3)"volts"`
C. `(4pi)/(3)"volts"`
D. `(2)/(3)"volt"`

Explanation

To find the maximum value of the electromotive force (e.m.f.) developed in a rotating coil, we can use the formula for induced e.m.f. in a coil rotating in a magnetic field. The formula is given by:

\[ \text{e.m.f.} = n \cdot B \cdot A \cdot \omega \]

Where:
- \( n \) = number of turns in the coil
- \( B \) = magnetic field strength (in Tesla)
- \( A \) = area of the coil (in square meters)
- \( \omega \) = angular velocity (in radians per second)

### Step 1: Convert the area from square centimeters to square meters
The area \( A \) is given as \( 80 \) square cm. To convert it to square meters:

\[ A = 80 \, \text{cm}^2 = 80 \times 10^{-4} \, \text{m}^2 = 0.008 \, \text{m}^2 \]

### Step 2: Convert the rotational speed from revolutions per minute to radians per second
The coil is rotating at \( 2000 \) revolutions per minute (rpm). To convert this to radians per second, we use the conversion factor \( 2\pi \) radians per revolution and \( 60 \) seconds per minute:

\[ \omega = 2000 \, \text{rpm} \times \frac{2\pi \, \text{radians}}{1 \, \text{revolution}} \times \frac{1 \, \text{minute}}{60 \, \text{seconds}} = \frac{2000 \times 2\pi}{60} \, \text{radians/second} \]

Calculating this gives:

\[ \omega \approx \frac{2000 \times 6.2832}{60} \approx 209.44 \, \text{radians/second} \]

### Step 3: Substitute the values into the e.m.f. formula
Now we have all the values needed to calculate the maximum e.m.f. developed in the coil:

- \( n = 50 \) turns
- \( B = 0.05 \, \text{T} \)
- \( A = 0.008 \, \text{m}^2 \)
- \( \omega \approx 209.44 \, \text{radians/second} \)

Substituting these values into the e.m.f. formula:

\[ \text{e.m.f.} = n \cdot B \cdot A \cdot \omega \]

\[ \text{e.m.f.} = 50 \cdot 0.05 \cdot 0.008 \cdot 209.44 \]

Calculating this step-by-step:

1. Calculate \( 50 \cdot 0.05 = 2.5 \)
2. Calculate \( 2.5 \cdot 0.008 = 0.02 \)
3. Finally, calculate \( 0.02 \cdot 209.44 \approx 4.1888 \)

Thus, the maximum value of the e.m.f. developed in the coil is approximately:

\[ \text{e.m.f.} \approx 4.19 \, \text{V} \]

### Final Answer
The maximum value of the e.m.f. developed in the coil is approximately \( 4.19 \, \text{V} \).

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