A coil of area `80cm^(2)` and number of turns `50` is rotating about an axis perpendicular to magneitc field of `0.05T` at `200` rotations per minute. The maximum value of e.m.f. induced in it will be

Options

A. `200pi` volt
B. `(10pi)/(3)` volt
C. `(4pi)/(30)` volt
D. `(2)/(3)` volt

Answer

To find the maximum value of the induced electromotive force (e.m.f.) in the rotating coil, we can use the formula for the induced e.m.f. in a rotating coil:

\[ \text{e.m.f.} = N \cdot B \cdot A \cdot \omega \cdot \sin(\theta) \]

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)
- \( \theta \) = angle between the magnetic field and the normal to the plane of the coil

Since we are looking for the maximum value of e.m.f., we take \( \sin(\theta) = 1 \) (which occurs when the plane of the coil is perpendicular to the magnetic field). Therefore, the formula simplifies to:

\[ \text{e.m.f.}_{\text{max}} = N \cdot B \cdot A \cdot \omega \]

### Step 1: Convert the area from cm² to m²
Given:
- Area \( A = 80 \, \text{cm}^2 \)

To convert cm² to m²:
\[ A = 80 \, \text{cm}^2 = 80 \times 10^{-4} \, \text{m}^2 = 0.008 \, \text{m}^2 \]

### Step 2: Convert rotations per minute to radians per second
Given:
- Rotations per minute = 200

To convert to radians per second:
\[ \text{Rotations per second} = \frac{200}{60} = \frac{10}{3} \, \text{rotations/second} \]
\[ \omega = \frac{10}{3} \times 2\pi \, \text{radians/second} = \frac{20\pi}{3} \, \text{radians/second} \]

### Step 3: Substitute the values into the e.m.f. formula
Given:
- Number of turns \( N = 50 \)
- Magnetic field strength \( B = 0.05 \, \text{T} \)

Now substituting the values into the formula:
\[ \text{e.m.f.}_{\text{max}} = N \cdot B \cdot A \cdot \omega \]
\[ \text{e.m.f.}_{\text{max}} = 50 \cdot 0.05 \cdot 0.008 \cdot \frac{20\pi}{3} \]

### Step 4: Calculate the e.m.f.
Calculating step by step:
\[ = 50 \cdot 0.05 = 2.5 \]
\[ = 2.5 \cdot 0.008 = 0.02 \]
\[ = 0.02 \cdot \frac{20\pi}{3} = \frac{0.4\pi}{3} \, \text{V} \]

### Final Result
Thus, the maximum value of the induced e.m.f. is:
\[ \text{e.m.f.}_{\text{max}} = \frac{0.4\pi}{3} \, \text{V} \]