Problem A solenoid has N=400N=400N=400 turns wound on a cylindrical former of area A=10 cm2A=10\ \text{cm}^2A=10 cm2 and length ℓ=40 cm\ell=40\ \text{cm}ℓ=40 cm (air core). Calculate the inductance L=μ0N2AℓL=\dfrac{\mu_0 N^2 A}{\ell}L=ℓμ0N2A. Solution Data in SI units: μ0=4π⋅10−7 T⋅m/A\mu_0=4\pi\cdot 10^{-7}\ \text{T}\cdot\text{m/A}μ0=4π⋅10−7 T⋅m/A, A=10 cm2=1,0⋅10−3 m2A=10\ \text{cm}^2=1{,}0\cdot 10^{-3}\ \text{m}^2A=10 cm2=1,0⋅10−3 m2, ℓ=0,40 m\ell=0{,}40\ \text{m}ℓ=0,40 m, N2=4002=1,6⋅105N^2=400^2=1{,}6\cdot 10^{5}N2=4002=1,6⋅105. Calculation: L=μ0N2Aℓ=(4π⋅10−7)(1,6⋅105)(1,0⋅10−3)0,40L=\frac{\mu_0 N^2 A}{\ell}=\frac{(4\pi\cdot 10^{-7})(1{,}6\cdot 10^{5})(1{,}0\cdot 10^{-3})}{0{,}40}L=ℓμ0N2A=0,40(4π⋅10−7)(1,6⋅105)(1,0⋅10−3) L=2,01⋅10−40,40≈5,0⋅10−4 HL=\frac{2{,}01\cdot 10^{-4}}{0{,}40}\approx 5{,}0\cdot 10^{-4}\ \text{H}L=0,402,01⋅10−4≈5,0⋅10−4 H L≈5,0⋅10−4 H\ev{L\approx 5{,}0\cdot 10^{-4}\ \text{H}}L≈5,0⋅10−4 H Links Topics: Induzione elettromagnetica Concepts: Induttanza e autoinduzione Objects: Solenoide