Electromagnetism 179therefore, henry0rLNA2(5.12)hence henryLNS2(5.13) 5.19 Factors Aff ecting Inductance Consider a coil of N turns wound on to a non-magnetic core, of uniform csa A metre 2 and mean length metre. The coil carries a current of I amp, which produces a ﬂ ux of weber. From equation (5.11), we know that the inductance will be LNIBALNBAIhenry, but webertherefore, henry……………[]1 Also, magnetic ﬁ eld strength, HNI ; so IHN and substituting this expression for I into equation [1] LNBAHNBANH/22……………[]Now, equation [2] contains the term BH , which equals o r We also know that 0rreluctanceA , S Worked Example 5.23 Q A 600 turn coil is wound on to a non-magnetic core of eff ective length 45 mm and csa 4 cm 2 .(a) Calculate the inductance, (b) The number of turns is increased to 900. Calculate the inductance value now produced. (c) The core of the 900 turn coil is now replaced by an iron core having a relative permeability of 75, and of the same dimensions as the original. Calculate the inductancein this case. Notes:1 Equation (5.12) compares with CANdεεor()1 farad for a capacitor. 2 If the number of turns is doubled, then the inductance is quadrupled, i.e. L N2 .3 The terms A and in equation (5.12) refer to the dimensions of the core, and NOT the coil.