Answer:
N= 238 turns
Explanation:
The induced Emf that goes through a solenoid can be calculated using the below formula;
Where ξ=induced Emf
L= self inductance
I= current
ξ= L|dⁱ/dt|
Making L which is the self inductance subject of formula we have
L=ξ/[|dⁱ|*|dt|]
The current here is changing at the rate of
.0260 A/s
L=NΦB/i
N=ξ/Φ|di|*|dt|
Magnitude of the induced Emf given= 12.6mV then if we convert to volt we have 12.6×10⁻³ V
The current I = 1.40A
Magnitude flux through the flux=/0.00285 Wb
Then if we substitute all this Value to equation above we have
N=(12.6×10⁻³ V×1.40A)/(0.00285 Wb×0.0260 A/s)
N=238turn
Therefore, there are 238turns in the solenoid
The number of turns that the solenoid has is;
238 turns
We are given;
Rate of change of current; di/dt = 0.026 A
Induced emf; E = 12.6 mv = 12.6 × 10⁻³ V
Average flux; Φ = 0.00285 Wb
Current; I = 1.4 A
We know that formula for the number of turns in the solenoid is;
N = EI/(Φ(di/dt)
Thus;
N = (12.6 × 10⁻³ × 1.4)/(0.00285 × 0.026)
N = 238.06
This is approximately N = 238 turns
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