Explanation:
it is given that,
Frequency of AC generator, f = 60 Hz
Magnetic field, B = 0.12 T
Area of the coil, A = 0.03 m²
Peak out put, V = 270 V
Resistance of the coil, R = 04 ohms
1. Let N is the number of loops the coil must contain. The peak voltage of AC generator is given by :
[tex]V=NBA\omega[/tex]
[tex]N=\dfrac{V}{BA\omega}[/tex]
[tex]N=\dfrac{270}{0.12\times 0.03\times 2\pi \times 60}[/tex]
N = 199
2. Applying Ohm's law as :
[tex]I=\dfrac{V}{R}[/tex]
[tex]I=\dfrac{270}{0.4}[/tex]
I = 675 A
3. Power, [tex]P=V\times I[/tex]
[tex]P=270\times 675[/tex]
P = 182250 watts
Hence, this is the required solution.