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MagnetismEasy

Magnetic field of a wire and solenoid

(a) An infinite straight wire carries current I = 8 A. Calculate the field B at distance r = 4 cm.\n(b) A solenoid with n = 1200 turns/m carries current I = 3 A. Calculate the internal field B.
Given data
I_wire = 8 Ar = 0.04 mn = 1200 turns/mI_sol = 3 Aμ₀ = 4π×10⁻⁷ T·m/A
Review the theory: Magnetismo
Steps
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  1. Calculate the magnetic field B of the wire at r = 4 cm (in μT = 10⁻⁶ T).
  2. Calculate the internal magnetic field B of the solenoid (in mT = 10⁻³ T).
Full worked solution
  1. Calculate the magnetic field B of the wire at r = 4 cm (in μT = 10⁻⁶ T).
    Bfilo=μ0I2πr=4π×10−7⋅82π⋅0.04B_{filo} = \dfrac{\mu_0 I}{2\pi r} = \dfrac{4\pi\times10^{-7} \cdot 8}{2\pi \cdot 0.04}
    B=μ0I2πr=4π×10−7×82π×0.04=2×10−7×80.04=16×10−70.04=40×10−6 T=40 μTB = \frac{\mu_0 I}{2\pi r} = \frac{4\pi\times10^{-7} \times 8}{2\pi \times 0.04} = \frac{2\times10^{-7} \times 8}{0.04} = \frac{16\times10^{-7}}{0.04} = \mathbf{40\times10^{-6}\,T} = \mathbf{40\,\mu T}. The magnetic field around a straight wire decreases with distance rr.
  2. Calculate the internal magnetic field B of the solenoid (in mT = 10⁻³ T).
    Bsol=μ0nI=4π×10−7⋅1200⋅3B_{sol} = \mu_0 n I = 4\pi\times10^{-7} \cdot 1200 \cdot 3
    B=μ0nI=4π×10−7×1200×3=4π×10−7×3600≈4.52×10−3 T=4.52 mTB = \mu_0 n I = 4\pi\times10^{-7} \times 1200 \times 3 = 4\pi\times10^{-7} \times 3600 \approx \mathbf{4.52\times10^{-3}\,T} = \mathbf{4.52\,mT}. The field inside a long solenoid is uniform and depends on the turn density nn and current II.
Result:B_wire = 40 μT at 4 cm. B_solenoid = 4.52 mT internal.
TheoryMagnetismoFormula sheetPhysics 2

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