Answer:
10.85 kV
Explanation:
The maximum value of the emf (ε) produced can be calculated using Faraday's law equation:
[tex]\epsilon = - \frac{N*d(\phi)}{dt} = - \frac{N*d(BAcos(\theta))}{dt}[/tex] (1)
Where:
N: is the turns of wire = 130 turns
Φ: is the magnetic flux = BAcos(θ)
B: is the magnetic field = 3.82 T
A: is the area of the coil = a*b = 0.746 m*0.249 m = 0.186 m²
θ: is the angle between the magnetic field lines and the normal to A = ωt = 2πft
f: is the frequency = 1120 rev/min = 18.7 rev/s
From equation (1) we have:
[tex]\epsilon_{max} = NBA2\pi f sin(2\pi ft) = NAB2\pi f = 130*0.186 m^{2}*3.82 T*2*\pi*18.7 rev/s = 10852.8 V = 10.85 kV[/tex]
Therefore, the maximum value of the emf produced is 10.85 kV.
I hope it helps you!
Answer:
The maximum value of the emf produced is 10819.098 V
Explanation:
Given data:
N = number of turn = 130
The area is:
[tex]A=74.6*24.9=1857.54cm^{2} =0.185754m^{2}[/tex]
B = magnetic field = 3.82 T
w = angular speed = 1120 rev/min = 117.286 rad/s
The maximum value of the emf produced is:
[tex]E=BANw=3.82*0.185754*130*117.286=10819.098V[/tex]
