A solenoid has 250 turns and its self-inductance is 2.4 mH. What is the flux through each turn when the current is 2 A? What is the induced emf when the current changes at 20 A s⁻¹? (Answer: 1.5 𝐱 10⁻⁵ Wb, 48 mV).
Data Given:
Number of turn solenoid = N = 250
Self Induction = L = 2.4 mH = 2.4 x10⁻³ H
Current in each coil = I = 2 A
Current changing rate in primary coil = △Ip△t = 20 A s⁻¹
To Find:
Change in flux = Δф = ?
Induced emf = Ɛ = ?
Solution:
Using the following relation
Ɛ = L △IΔt --------(1)
by putting the corresponding values
Ɛ = 2.4 x10⁻³ H x 20 A s⁻¹
Ɛ = 48 x10⁻³ H
Ɛ = 48 mV ---------------Ans.1
We have another relation for induced emf
Ɛ = N ΔфΔt -------------(2)
by comparing equations (1) and (2)
N ΔфΔt = L △IΔt
or
Δф = LN x ΔI
by putting values
Δф = 2.4x10⁻³H250 x 2 A
Δф = 0.0192 x10⁻³ Web
or
Δф = 1.9 x 10⁻⁵ Web ------------Ans.2
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