question_answer

Derive an expression for self-inductance of a long air-cored solenoid of length l, cross-sectional area A and having number of turns N.

Answer:

Consider a long air-cored solenoid having n number of turns per unit length. When an electric current I flows through it then, the magnetic field produced in it is given by \[B={{\mu }_{0}}n\,I\] where, \[{{\mu }_{0}}\] is the permeability of free space. If A is the area of cross-section of the solenoid, then magnetic field linked with each turn of the solenoid, \[\phi '=B.A={{\mu }_{0}}n\,IA\] Total number of turns in the solenoid N = nl Here, n = number of turns per unit length. \[\therefore \] Total magnetic flux linked with the solenoid \[\phi =\phi '\times N=\mu {}_{0}n\,\,I\,\,A\times n\,l\] \[\phi ={{\mu }_{0}}{{n}^{2}}I\,Al\]                           ?(i) But total magnetic flux linked with a solenoid is given by \[\phi =LI\]                                            ?(ii)S From Eqs. (i) and (ii), we get \[L={{\mu }_{0}}{{n}^{2}}Al\] or \[L={{\mu }_{0}}\frac{{{N}^{2}}}{{{l}^{2}}}Al\] or \[L=\frac{{{\mu }_{0}}{{N}^{2}}A}{l}\] This is the required expression.