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  = `\frac {△I_p}{△t}` = 20 A s⁻¹


To Find:


Change in flux = Δф = ?


Induced emf  = Ɛ = ?


Solution:

Using the following relation

Ɛ =  L `\frac {△I}{Δt}`  --------(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 `\frac {Δф}{Δt}` -------------(2)

by comparing equations (1) and (2)

N `\frac {Δф}{Δt}` = L `\frac {△I}{Δt}`

or

Δф = `\frac {L}{N}` ｘ ΔI 

by putting values

Δф = `\frac {2.4 ｘ10⁻³ H}{250}` ｘ 2 A

Δф = 0.0192 ｘ10⁻³ Web

or

Δф = 1.9 ｘ 10⁻⁵ Web ------------Ans.2




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