<Dl> <Dd> Z i n t = k ρ 2 π R J 0 (k R) J 1 (k R) (\ displaystyle \ mathbf (Z) _ (int) = (\ frac (k \ rho) (2 \ pi R)) (\ frac (J_ (0) (kR)) (J_ (1) (kR)))). </Dd> </Dl> <Dd> Z i n t = k ρ 2 π R J 0 (k R) J 1 (k R) (\ displaystyle \ mathbf (Z) _ (int) = (\ frac (k \ rho) (2 \ pi R)) (\ frac (J_ (0) (kR)) (J_ (1) (kR)))). </Dd> <P> The internal impedance is complex and may be interpreted as a resistance in series with an inductance . The inductance accounts for energy stored in the magnetic field inside the wire . It has a maximum value of μ 8 π (\ displaystyle (\ frac (\ mu) (8 \ pi))) H / m at zero frequency and goes to zero as the frequency increases . The zero frequency internal inductance is independent of the radius of the round wire . </P> <P> The effective resistance due to a current confined near the surface of a large conductor (much thicker than δ) can be solved as if the current flowed uniformly through a layer of thickness δ based on the DC resistivity of that material . The effective cross-sectional area is approximately equal to δ times the conductor's circumference . Thus a long cylindrical conductor such as a wire, having a diameter D large compared to δ, has a resistance approximately that of a hollow tube with wall thickness δ carrying direct current . The AC resistance of a wire of length L and resistivity ρ (\ displaystyle \ rho) is: </P>

Why there is no skin effect in dc