This section list formulas for inductances in specific situations. Beware that some of the equations are in Imperial units. The permeability of free space, μ0, is constant and is defined to be exactly equal to 4π×10-7 H m-1. == 1. Basic inductance formula for a cylindrical coil == L = μ 0 μ r N 2 A l {\displaystyle L={\frac {\mu _{0}\mu _{r}N^{2}A}{l}}} L = inductance / H μr = relative permeability of core material N = number of turns A = area of cross-section of the coil / m2 l = length of coil / m == 2. The self-inductance of a straight, round wire in free space == L s e l f = μ 0 b 2 π [ ln ( b a + 1 + b 2 a 2 ) − 1 + a 2 b 2 + a b + μ r 4 ] {\displaystyle L_{self}={\frac {\mu _{0}b}{2\pi }}\left[\ln \left({\frac {b}{a}}+{\sqrt {1+{\frac {b^{2}}{a^{2}}}}}\right)-{\sqrt {1+{\frac {a^{2}}{b^{2}}}}}+{\frac {a}{b}}+{\frac {\mu _{r}}{4}}\right]} Lself = self inductance / H b = wire length /m a = wire radius /m μ r {\displaystyle \mu _{r}} = relative permeability of wireIf you make the assumption that b >> a and that the wire is nonmagnetic ( μ r = 1 {\displaystyle \mu _{r}=1} ), then this equation can be approximated to L s e l f = μ 0 b 2 π [ ln ( 2 b a ) − 3 / 4 ] {\displaystyle L_{self}={\frac {\mu _{0}b}{2\pi }}\left[\ln \left({\frac {2b}{a}}\right)-3/4\right]} (for low frequencies) L s e l f = μ 0 b 2 π [ ln ( 2 b a ) − 1 ] {\displaystyle L_{self}={\frac {\mu _{0}b}{2\pi }}\left[\ln \left({\frac {2b}{a}}\right)-1\right]} (for high frequencies due to the skin effect)L = inductance / H b = wire length / m a = wire radius / mThe inductance of a straight wire is usually so small that it is neglected in most practical problems. If the problem deals with very high frequencies (f > 20 GHz), the calculation may become necessary. For the rest of this book, we will assume that this self-inductance is negligible. == 3. Inductance of a short air core cylindrical coil in terms of geometric parameters: == L = r 2 N 2 9 r + 10 l {\displaystyle L={\frac {r^{2}N^{2}}{9r+10l}}} L = inductance in μH r = outer radius of coil in inches l = length of coil in inches N = number of turns == 4. Multilayer air core coil == L = 0.8 r 2 N 2 6 r + 9 l + 10 d {\displaystyle L={\frac {0.8r^{2}N^{2}}{6r+9l+10d}}} L = inductance in μH r = mean radius of coil in inches l = physical length of coil winding in inches N = number of turns d = depth of coil in inches (i.e., outer radius minus inner radius) == 5. Flat spiral air core coil == L = r 2 N 2 ( 2 r + 2.8 d ) × 10 5 {\displaystyle L={\frac {r^{2}N^{2}}{(2r+2.8d)\times 10^{5}}}} L = inductance / H r = mean radius of coil / m N = number of turns d = depth of coil / m (i.e. outer radius minus inner radius)Hence a spiral coil with 8 turns at a mean radius of 25 mm and a depth of 10 mm would have an inductance of 5.13µH. == 6. Winding around a toroidal core (circular cross-section) == L = μ 0 μ r N 2 r 2 D {\displaystyle L=\mu _{0}\mu _{r}{\frac {N^{2}r^{2}}{D}}} L = inductance / H μr = relative permeability of core material N = number of turns r = radius of coil winding / m D = overall diameter of toroid / m