A long solenoid with 15 turns per cm has a small loop of area 2.0 cm² placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, what is the induced e.m.f. in the loop while the current is changing? (Answer: 7.54 ｘ10⁻⁶ V). 



Data Given:

Number of turn of the solenoid = N =  15 turn / cm = 1500 turn / m

Numbers of turns per unit length = n = 1500 turns

Area of the solenoid  = A = 2.0 cm² = 0.0002 m²

Initial current in the solenoid = I₁ = 2 A

Final current in the solenoid = I₂ = 4 A

So change in current △I = I₂ - I₁ = 4 A - 2 A = 2 A

Change in time interval = △t = 0.1 s

Permeability of free space = µ₀ = 4 ㄫ ｘ 10⁻⁷ H m⁻¹

To Find:

Induced e.m.f. in the solenoid = Ɛ = ?


Solution:

According to Faraday's Law

Ɛ = `\frac {Δф}{Δt}`

where Δф = ΔB . A and ΔB = μ₀ n ΔI ,  So

Ɛ = `\frac {ΔB . A}{Δt}`

Ɛ = `\frac {μ₀ n ΔI A}{Δt}`

by putting the corresponding values

Ɛ = `\frac {4 ㄫ ｘ 10⁻⁷ H m⁻¹ ｘ 1500 ｘ 2 A ｘ 0.0002 m²}{0.1 s}`

Ɛ = `\frac {4 ｘ 3.1416 ｘ 10⁻⁷ H m⁻¹ ｘ 1500 ｘ 2 A ｘ 0.0002 m²}{0.1 s}`

Ɛ = `\frac {7.540 ｘ 10⁻⁷ H A m}{0.1 s}`

Ɛ = 75.40 ｘ 10⁻⁷ V

or

Ɛ =  7.54  ｘ 10⁻⁶ V ---------------Ans.





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