In a region of uniform magnetic inductance `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 direction of `B` and which forms a dimater of the coil. If the coil rotates at `200` r.p.m the amplitude of the alternatic current induced in the coil is

Options

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

Explanation

To solve the problem, we need to calculate the amplitude of the alternating current induced in the circular coil when it is rotated in a uniform magnetic field. Here are the steps to arrive at the solution:

### Step 1: Identify the given values
- Magnetic field strength, \( B = 10^{-2} \, \text{T} \)
- 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 from rpm to radians per second
To convert revolutions per minute (rpm) to radians per second (rad/s), we use the formula:
\[ \omega = 2\pi \times \left(\frac{n}{60}\right) \]
Substituting the values:
\[ \omega = 2\pi \times \left(\frac{200}{60}\right) = \frac{400\pi}{60} = \frac{20\pi}{3} \, \text{rad/s} \]

### 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:
\[ A = \pi (0.3)^2 = \pi (0.09) = 0.09\pi \, \text{m}^2 \]

### Step 4: Calculate the maximum EMF induced in the coil
The maximum EMF \( E_{\text{max}} \) induced in the coil can be calculated using the formula:
\[ E_{\text{max}} = N \cdot B \cdot A \cdot \omega \]
Assuming the number of turns \( N = 1 \):
\[ E_{\text{max}} = 1 \cdot (10^{-2}) \cdot (0.09\pi) \cdot \left(\frac{20\pi}{3}\right) \]
Calculating this:
\[ E_{\text{max}} = 10^{-2} \cdot 0.09\pi \cdot \frac{20\pi}{3} = \frac{10^{-2} \cdot 0.09 \cdot 20 \cdot \pi^2}{3} \]
\[ E_{\text{max}} = \frac{0.18 \cdot 10^{-2} \cdot \pi^2}{3} = 0.06 \cdot 10^{-2} \cdot \pi^2 \, \text{V} \]

### Step 5: Calculate the amplitude of the induced current
The amplitude of the induced current \( I_{\text{max}} \) can be calculated using Ohm's law:
\[ I_{\text{max}} = \frac{E_{\text{max}}}{R} \]
Substituting the values:
\[ I_{\text{max}} = \frac{0.06 \cdot 10^{-2} \cdot \pi^2}{\pi^2} = 0.06 \cdot 10^{-2} = 6 \times 10^{-3} \, \text{A} \]
Converting to milliamperes:
\[ I_{\text{max}} = 6 \, \text{mA} \]

### Final Answer:
The amplitude of the alternating current induced in the coil is \( 6 \, \text{mA} \).

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