An a.c. generator consists of a coil of 100 turns and cross sectional area of `3 m^(2)`, rotating at a constant angular speed of `60 rad//sec` in a uniform magnetic field of 0.04 T. The resistance of the coil is `500 Omega`. Calculate (i) maximum current drawn from the generator and (ii) max. power dissipation in the coil.

Options

A. `518.4W`
B. 1036W
C. `259.2W`
D. Zero

Explanation

To solve the problem step by step, we will calculate (i) the maximum current drawn from the generator and (ii) the maximum power dissipation in the coil.

### Given Data:
- Number of turns (N) = 100
- Cross-sectional area (A) = 3 m²
- Angular speed (ω) = 60 rad/s
- Magnetic field (B) = 0.04 T
- Resistance of the coil (R) = 500 Ω

### Step 1: Calculate the Maximum EMF (E₀)
The maximum electromotive force (EMF) generated in the coil can be calculated using the formula:
\[ E_0 = N \cdot B \cdot A \cdot \omega \]

Substituting the values:
\[ E_0 = 100 \cdot 0.04 \cdot 3 \cdot 60 \]

Calculating:
\[ E_0 = 100 \cdot 0.04 = 4 \]
\[ E_0 = 4 \cdot 3 = 12 \]
\[ E_0 = 12 \cdot 60 = 720 \, \text{V} \]

### Step 2: Calculate the Maximum Current (I₀)
The maximum current can be calculated using Ohm's law:
\[ I_0 = \frac{E_0}{R} \]

Substituting the values:
\[ I_0 = \frac{720}{500} \]

Calculating:
\[ I_0 = 1.44 \, \text{A} \]

### Step 3: Calculate the Maximum Power Dissipation (P₀)
The maximum power dissipated in the coil can be calculated using the formula:
\[ P_0 = \frac{E_0^2}{R} \]

Substituting the values:
\[ P_0 = \frac{720^2}{500} \]

Calculating:
\[ P_0 = \frac{518400}{500} \]
\[ P_0 = 1036.8 \, \text{W} \]

However, the maximum power dissipation can also be expressed in terms of the maximum current:
\[ P_0 = I_0^2 \cdot R \]

Substituting the values:
\[ P_0 = (1.44)^2 \cdot 500 \]

Calculating:
\[ P_0 = 2.0736 \cdot 500 \]
\[ P_0 = 1036.8 \, \text{W} \]

### Final Answers:
(i) Maximum current drawn from the generator: **1.44 A**
(ii) Maximum power dissipation in the coil: **1036.8 W**