1. Given data

• Number of turns: $N = 100$
• Area of each turn: $A = 14 \times 10^{-2} \, \text{m}^2 = 0.14 \, \text{m}^2$
• Magnetic field: $B = 3.0 \, \text{T}$
• Rotation speed: $360 \, \text{rev/min}$

We need the maximum emf induced in the rotating coil.

2. Formula for maximum emf in an AC generator

For a coil rotating in a uniform magnetic field,

$E_{\max} = N B A \omega$

where $\omega$ is the angular speed.

3. Convert rotational speed to angular speed

Given:

$360 \, \text{rev/min} = \frac{360}{60} = 6 \, \text{rev/s}$

Now,

$\omega = 2\pi f = 2\pi \times 6 = 12\pi \text{ rad/s}$

Using $\pi = \frac{22}{7}$,

$\omega = 12 \times \frac{22}{7} = \frac{264}{7} \text{ rad/s}$

4. Substitute into the formula

$E_{\max} = N B A \omega$

$E_{\max} = 100 \times 3.0 \times 0.14 \times \frac{264}{7}$

First simplify:

$3.0 \times 0.14 = 0.42$

So,

$E_{\max} = 100 \times 0.42 \times \frac{264}{7}$

$100 \times 0.42 = 42$

Thus,

$E_{\max} = 42 \times \frac{264}{7}$

$\frac{42}{7} = 6$

$E_{\max} = 6 \times 264 = 1584 \, \text{V}$

5. Final answer

$\boxed{1584}$

6. Comparison with stored correct answer

Stored correct answer = $1584$

My derived answer matches it exactly.