The coil of an a.c. generator has 100 turns, each of cross sectional area `2 m^(2)`. It is rotating at a constant angular speed of 30 radians`//`s, in a uniform magnetic field of `2 xx 10^(-2)`T. What is the maximum power dissipated in the circuit, if the resistance of the circuit including that of the coil is `600 Omega`?

Options

A. 6 W
B. 9 W
C. 12 W
D. 24 W

Explanation

To solve the problem step by step, we will calculate the maximum power dissipated in the circuit using the given parameters.

### Step 1: Identify the given values
- Number of turns (N) = 100
- Cross-sectional area (A) = 2 m²
- Angular speed (ω) = 30 rad/s
- Magnetic field (B) = 2 × 10^(-2) T
- Resistance (R) = 600 Ω

### Step 2: Calculate the maximum induced EMF (E₀)
The formula for the maximum induced EMF in a rotating coil is given by:
\[ E_0 = N \cdot B \cdot A \cdot \omega \]
Substituting the values:
\[ E_0 = 100 \cdot (2 \times 10^{-2}) \cdot 2 \cdot 30 \]

### Step 3: Perform the calculation for E₀
Calculating step by step:
1. Calculate \( B \cdot A \):
\[ B \cdot A = (2 \times 10^{-2}) \cdot 2 = 4 \times 10^{-2} \]
2. Now multiply by N and ω:
\[ E_0 = 100 \cdot (4 \times 10^{-2}) \cdot 30 \]
\[ E_0 = 100 \cdot 1.2 = 120 \text{ V} \]

### Step 4: Calculate the maximum power (P_max)
The formula for maximum power dissipated in the circuit is:
\[ P_{max} = \frac{E_0^2}{2R} \]
Substituting the values:
\[ P_{max} = \frac{(120)^2}{2 \cdot 600} \]

### Step 5: Perform the calculation for P_max
Calculating step by step:
1. Calculate \( E_0^2 \):
\[ E_0^2 = 120^2 = 14400 \]
2. Now calculate \( 2R \):
\[ 2R = 2 \cdot 600 = 1200 \]
3. Finally, calculate \( P_{max} \):
\[ P_{max} = \frac{14400}{1200} = 12 \text{ W} \]

### Final Answer
The maximum power dissipated in the circuit is \( 12 \text{ W} \).

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