Solenoid Class Notes | Lecture Note

University:
Santa Fe College
Course:
ETS 2527C | Electromechanical Components and Mechanisms
Academic year:

2024
Views:

46

Pages:

6
Author:

Jabin F.

# Solenoid -> I=free current Condition-I Airfilled (μ=μ₀) M=0 Bin = μ₀nIẑ B = μ₀(H+M) B = μH Hin = nIẑ Cond" - II [Iron filled (μ=μr, M≠0)] B = μmIẑ μr = μ/μ₀ B = μH => H depends on External Current Hin = nIẑ M = (μnI/μ₀ - nI)ẑ Jb = ∇ x m = 0 Kb = m x ẑ = (μnI/μ₀ - nI)(ẑ x ϕ) Kb = nI(μ/μ₀ - 1)ϕ if μ given => B = μH if μr given => B = μ₀μrH if χm given => B = μ₀(1 + χm)H If M is given then B = μ₀(H + M) ∮ H.dl = Ifenc. Inside: Ifenc. = 0 Hin = 0 Outside: Ifenc. = 0, Hout = 0 Bin = μ₀(Hin + Min) = μ₀(0 + M) = μ₀M Bout = μ₀(Hout + Mout) = μ₀(0 + 0) = 0 Additional Question if [M = Mẑ] Kb=?, Jb=? Jb = ∇ x m = 0 Kb = (m x ẑ) = (Mẑ x ẑ) = 0 Kb = m x ẑ = M(ẑ x ẑ) = Mϕ Solenoid Bin = μ₀nIẑ = μ₀Kẑ R = Rb + Rf Bin = μ₀Mz Ques. # A long wire of radius a has made of iron with susceptibility (χm) Carries a uniformly distributed current I. Find magnetic fied (B) inside and out side. # If = I Jf = I/πa² Inside ∮ H.dl = Ifenc. Ifenc = ∫(I/πa²)dsL = ∫∫(I/πa²)ρdρdϕ = Is²/a² => Hin 2πρ = Is²/a² => Hin = Is/(2πρa²)ϕ Bin = μHin = μ₀μrHIn = μ₀(1 + χm)Hin Bin = μ₀(1 + χm)Is/(2πρa²)ϕ Min = (B/μ₀) - Hin Min = ((1 + χm)Is/(2πρa²))ϕ - (Is/(2πρa²))ϕ Min = (χmIs/(2πρa²))ϕ Min = (Iχm / (2πρa²))ϕ # outside Ifenc = I Ifenc. = ∫∫(I/πa²)ρdρdϕ = I Hout 2πρ = I => Hout = I/(2πρ)ϕ Bout = μ₀Hout = μ₀I/(2πρ)ϕ Mout = 0