A long cylindrical conductor of radius R carries current density J(r) = J₀(1 − r²/R²) in the +z direction. Derive the total current and magnetic field for r R. Express the interior result in terms of total current I, verify continuity at R, and find the radius inside where B is maximal. Assume magnetic permeability μ₀.

Question

A long cylindrical conductor of radius R carries current density J(r) = J₀(1 − r²/R²) in the +z direction. Derive the total current and magnetic field for r < R and r > R. Express the interior result in terms of total current I, verify continuity at R, and find the radius inside where B is maximal. Assume magnetic permeability μ₀.

Accepted Answer

I = (πJ₀R²)/2. Inside B = μ₀I(2r/R² − r³/R⁴)/(2π); outside B = μ₀I/(2πr). The interior maximum occurs at r = R√(2/3). Integrate current density over the cross-section: I = ∫₀ᴿJ₀(1−s²/R²)2πs ds = 2πJ₀(R²/2−R²/4) = πJ₀R²/2. For an Amperian circle at r < R, I_enc = 2πJ₀(r²/2−r⁴/(4R²)) = πJ₀(r²−r⁴/(2R²)). Ampère's law B(2πr) = μ₀I_enc gives B = μ₀J₀[r/2−r³/(4R²)] = μ₀I[2r/R²−r³/R⁴]/(2π). For r > R, B = μ₀I/(2πr). At r = R the interior expression becomes μ₀I/(2πR), exactly matching the exterior field, so no unphysical jump occurs. Differentiate the interior factor 2r/R²−r³/R⁴: its derivative is 2/R²−3r²/R⁴. Setting this to zero gives r = R√(2/3), where the field reaches its interior maximum.

Nonuniform current in a cylindrical conductor

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