A solenoid of length 8 cm and cross-sectional area 0.5 cm² turns. Find the self-inductance of the solenoid when the core is air. If the current in the solenoid increases through 1.5 A in 0.2 sec, find the magnitude of induced emf in it. (Answer: 1.6 ｘ10⁻³ V,  2.12 ｘ10⁻⁴ H)



Data Given:

Length of the solenoid = l = 8 cm = 0.08 m

Area of the circuit = A = 0.5 cm² = 0.00005 m² 

Numbers of turns = N = 520 turns

increased current = ΔI = 1.5 A

Change in time interval = △t = 0.2 s

µ₀ = 4π ｘ10⁻⁷ Wb A⁻¹ m⁻¹



To Find:

Self-inductance = L = ?

Induced e.m.f. = Ɛ = ?




Solution:


The formula for self-induction is given by:

L = µ₀ n² l A

n = Nl

L = µ₀ (Nl)² l A

L = µ₀ N2l2 l A

or

L = µ₀ N2l A


putting values

L = 4 ｘ3.1416 ｘ10⁻⁷ Wb A⁻¹ m⁻¹ ｘ(520)20.08m ｘ0.00005 m²

L = 12.5664 ｘ10⁻⁷ Wb A⁻¹ m⁻¹ ｘ2704000.08m ｘ0.00005 m²

L = 12.5664 ｘ10⁻⁷ Wb A⁻¹ m⁻¹ ｘ3380000 ｘ 0.00005 m²

L = 2123.7216ｘ10⁻⁷ H

or

L = 2.12ｘ10⁻⁴ H ------------Ans.1

------------------------------------

To find the Induced e.m.f. by Using another formula for self-inductance 

L = Ɛ ΔtΔI

or 

Ɛ =  L ΔIΔt

by putting the corresponding values

Ɛ =  2.12ｘ10⁻⁴ H ｘ 1.5A0.2s

Ɛ =  2.12ｘ10⁻⁴ H ｘ 7.5 A s⁻¹

Ɛ =  15.9ｘ10⁻⁴ V

or

Ɛ =  1.6ｘ10⁻³ V-------------------Ans.(2)



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