Respuesta :
Answer:
0.074 V
Explanation:
Parameters given:
Number of turns, N = 121
Radius of coil, r = 2.85 cm = 0.0285 m
Time interval, dt = 0.179 s
Initial magnetic field strength, Bin = 55.1 mT = 0.0551 T
Final magnetic field strength, Bfin = 97.9 mT = 0.0979 T
Change in magnetic field strength,
dB = Bfin - Bin
= 0.0979 - 0.0551
dB = 0.0428 T
The magnitude of the average induced EMF in the coil is given as:
|Eavg| = |-N * A * dB/dt|
Where A is the area of the coil = pi * r² = 3.142 * 0.0285² = 0.00255 m²
Therefore:
|Eavg| = |-121 * 0.00255 * (0.0428/0.179)|
|Eavg| = |-0.074| V
|Eavg| = 0.074 V
Answer:
The magnitude of the average EMF = 73.83 mv
Explanation:
From faradays law of induction, EMF is given as;
EMF = NA(ΔB/Δt)
We are given that;
N = 121 turns
B1 = 55.1 mT
B2 = 97.9 mT
Thus, ΔB = 97.9 mT - 55.1 mT = 42.8 mT
t = 0.179 s
r = 2.85cm = 0.0285 m
Area = πr² = π x (0.0285)² = 0.0025518 m²
Plugging in the relevant values, we can calculate EMF as;
EMF = (121)(0.0025518)(42.8 mT/0.179) = 73.83 mv
