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MagnetismHard

Solenoid — self-inductance and magnetic energy

A solenoid has N = 800 turns, length l = 40 cm, cross-section A = 12 cm² and carries current I = 5 A.\nCalculate: (a) self-inductance L, (b) magnetic energy U, (c) internal field B, (d) energy density u.
Given data
N = 800l = 0.40 mA = 12×10⁻⁴ m²I = 5 Aμ₀ = 4π×10⁻⁷ T·m/A
Review the theory: Magnetismo
Steps
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  1. Calculate the self-inductance L (in mH = 10⁻³ H).
  2. Calculate the magnetic energy U (in mJ = 10⁻³ J).
  3. Calculate the internal magnetic field B (in mT).
  4. Calculate the magnetic energy density u (in J/m³).
Full worked solution
  1. Calculate the self-inductance L (in mH = 10⁻³ H).
    L=μ0N2lA=4π×10−7⋅80020.40⋅12×10−4L = \mu_0 \dfrac{N^2}{l} A = 4\pi\times10^{-7} \cdot \dfrac{800^2}{0.40} \cdot 12\times10^{-4}
    L=μ0N2Al=4π×10−7×8002×12×10−40.40=4π×10−7×640000×12×10−40.40=4π×10−7×1920≈2.413×10−3 H=2.41 mHL = \mu_0 \frac{N^2 A}{l} = 4\pi\times10^{-7} \times \frac{800^2 \times 12\times10^{-4}}{0.40} = 4\pi\times10^{-7} \times \frac{640000 \times 12\times10^{-4}}{0.40} = 4\pi\times10^{-7} \times 1920 \approx \mathbf{2.413\times10^{-3}\,H} = \mathbf{2.41\,mH}.
  2. Calculate the magnetic energy U (in mJ = 10⁻³ J).
    U=12LI2=12⋅2.412×10−3⋅52U = \dfrac{1}{2}LI^2 = \dfrac{1}{2} \cdot 2.412\times10^{-3} \cdot 5^2
    U=12LI2=12×2.413×10−3×52=12×2.413×10−3×25=30.16×10−3 J=30.16 mJU = \frac{1}{2}L I^2 = \frac{1}{2} \times 2.413\times10^{-3} \times 5^2 = \frac{1}{2} \times 2.413\times10^{-3} \times 25 = \mathbf{30.16\times10^{-3}\,J} = \mathbf{30.16\,mJ}.
  3. Calculate the internal magnetic field B (in mT).
    B=μ0nI=μ0NlI=4π×10−7⋅8000.40⋅5B = \mu_0 n I = \mu_0 \dfrac{N}{l} I = 4\pi\times10^{-7} \cdot \dfrac{800}{0.40} \cdot 5
    n=N/l=800/0.40=2000 turns/mn = N/l = 800/0.40 = 2000\,\mathrm{turns/m}. B=μ0nI=4π×10−7×2000×5=4π×10−7×10000=4π×10−3≈12.57×10−3 T=12.57 mTB = \mu_0 n I = 4\pi\times10^{-7} \times 2000 \times 5 = 4\pi\times10^{-7} \times 10000 = 4\pi\times10^{-3} \approx \mathbf{12.57\times10^{-3}\,T} = \mathbf{12.57\,mT}.
  4. Calculate the magnetic energy density u (in J/m³).
    u=B22μ0=(12.57×10−3)22⋅4π×10−7u = \dfrac{B^2}{2\mu_0} = \dfrac{(12.57\times10^{-3})^2}{2 \cdot 4\pi\times10^{-7}}
    u=B22μ0=(12.57×10−3)22×4π×10−7=1.580×10−42.513×10−6≈62.9 J/m3u = \frac{B^2}{2\mu_0} = \frac{(12.57\times10^{-3})^2}{2 \times 4\pi\times10^{-7}} = \frac{1.580\times10^{-4}}{2.513\times10^{-6}} \approx \mathbf{62.9\,J/m^3}. Verification: U=u⋅Al=62.9×(12×10−4×0.40)≈30.2 mJU = u \cdot A l = 62.9 \times (12\times10^{-4} \times 0.40) \approx 30.2\,\mathrm{mJ} ✓.
Result:L = 2.41 mH — U = 30.15 mJ — B = 12.57 mT — u = 62.9 J/m³.
TheoryMagnetismoFormula sheetPhysics 2

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