Inductance of a Solenoid
Consider a long solenoid of cross-sectional area A. What is the inductance per unit length near its middle?
To use the defining equation for inductance (Eq. 31-30), we must calculate the flux linkage setup by a given current in the solenoid windings. Consider a length I. near the middle of this solenoid. The flux linkage for this section of the solenoid is Ncf>B = (n/)(BA) .in which n is the number of turns per unit length of the solenoid and B is the
magnitude of the magnetic field within the solenoid.The magnitude B is given by Inductance-like capacitance-depends only on the geometry of the device. The dependence on the square of the number of turns per unit length is to be expected. If you, say, triple n, you not only triple the number of turns (N) but you also triple the flux (cf>B = BA = Jloina) through each turn, multiplying the flux linkage NSF>B and thus the inductance L by a factor of 9.
If the solenoid is very much longer than its radius, then Eq. 31-3 gives its inductance to a good approximation. This approximation neglects the spreading of the magnetic field lines near the ends of the solenoid, just as the parallel-plate capacitor formula (C = oA/d) neglects the fringing of the electric field lines near the
edges of the capacitor plates. From Eq. 31-32, and recalling that n is a number per unit length, we can see that an inductance can be written as a product of the permeability constant JLo and a quantity with the dimensions of a length. This means that JLo can be expressed in the unit henry per meter: B = Join,