Final answer:
The number of turns required in the generator coil to produce a peak voltage of 6.7 KV in a 0.15 T magnetic field at a frequency of 60 Hz is approximately 106 turns.
Step-by-step explanation:
The question pertains to calculating the number of turns in a generator coil required to produce a certain peak voltage, given the frequency, the dimensions of the coil, and the magnetic field strength. To find this, we need to use the formula for the induced electromotive force (emf) in a generator, which is ε = NBA sin(ωt), where ε is the induced emf, N is the number of turns in the coil, B is the magnetic field strength, A is the area of the coil, and ω is the angular velocity.
Givent this formula and knowing that ε₀ (the peak emf) is proportional to N, we can rearrange the formula to solve for N:
N = ε₀ / (AB ω)
The area A of the rectangular coil is 70 cm by 1.2 m, which we convert to meters to get 0.7 m by 1.2 m. Therefore, A = 0.7 m * 1.2 m = 0.84 m².
The magnetic field strength B is given as 0.15 T. The angular velocity ω is related to the frequency f by the equation ω = 2πf. Given that the frequency is 60 Hz, we have ω = 2π * 60 Hz = 120π rad/s.
Now, we can plug these values into the rearranged formula:
N = 6700 V / (0.84 m² * 0.15 T * 120π rad/s)
After calculating, we find that the number of turns required is:
N = 6700 / (0.84 * 0.15 * 120π)
N ≈ 106 turns