In a region of uniform magnetic induction `B=10^(2)` tesla, a circular coil of radius `30 cm` and resistance `pi^(2)` ohm is rotated about an axis which is perpendicular to the directon of `B` and which form a diameter of the coil. If the coil rotates at `200rpm` the amplitude of the alternating current induced in the coil is

Options

A. `4pi^(2)mA`
B. `30mA`
C. `6mA`
D. `200mA`

Explanation

To solve the problem, we need to calculate the amplitude of the alternating current induced in a rotating circular coil placed in a uniform magnetic field. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the given values
- Magnetic induction \( B = 10^2 \, \text{T} = 100 \, \text{T} \)
- Radius of the coil \( r = 30 \, \text{cm} = 0.3 \, \text{m} \)
- Resistance of the coil \( R = \pi^2 \, \Omega \)
- Rotational speed \( N = 200 \, \text{rpm} \)

### Step 2: Convert rotational speed to radians per second
To convert the rotational speed from revolutions per minute (rpm) to radians per second (rad/s), we use the conversion factor:
\[ \omega = 2\pi \times \frac{N}{60} \]
Substituting the values:
\[ \omega = 2\pi \times \frac{200}{60} = \frac{400\pi}{60} = \frac{20\pi}{3} \, \text{rad/s} \]

### Step 3: Calculate the area of the coil
The area \( A \) of the circular coil is given by:
\[ A = \pi r^2 \]
Substituting the radius:
\[ A = \pi (0.3)^2 = \pi \times 0.09 = 0.09\pi \, \text{m}^2 \]

### Step 4: Calculate the amplitude of the induced EMF
The amplitude of the induced EMF \( \mathcal{E}_0 \) in the coil can be calculated using the formula:
\[ \mathcal{E}_0 = \omega B A \]
Substituting the values:
\[ \mathcal{E}_0 = \left(\frac{20\pi}{3}\right) \times 100 \times (0.09\pi) \]
Calculating this:
\[ \mathcal{E}_0 = \frac{20 \times 100 \times 0.09 \pi^2}{3} = \frac{180 \pi^2}{3} = 60 \pi^2 \, \text{V} \]

### Step 5: Calculate the amplitude of the alternating current
Using Ohm's law, the amplitude of the alternating current \( I_0 \) can be calculated as:
\[ I_0 = \frac{\mathcal{E}_0}{R} \]
Substituting the values:
\[ I_0 = \frac{60 \pi^2}{\pi^2} = 60 \, \text{A} \]

### Final Answer
The amplitude of the alternating current induced in the coil is \( 60 \, \text{A} \).

---