A long solenoid of length 'l' having carries a current I. Deduce the expression for the magnetic field in the interior of the solenoid.

Explanation

Consider a long solenoid having n turns per unit length as shown in Fig. The upper view of dots in the figure is like a uniform current sheet coming out of the plane of the paper . From the right hand rule, the field due to this is to the left at point Q (above) and to the right at point P (below). The lower row of crosses in the figure is like a uniform current sheet going into the plane of the paper. The field at any point above it (P as well as Q) is to the right. The two field reinforce each other at P and exactly cancel at Q.
Thus, a uniform magnetic field `vecB` is present along the axis of solenoid at any point inside the solenoid and the zero at any point outside the solenoid.
Consider a rectangular Amperian loop abcd. Along cd the magnetic field is zero as explained is at right angle to bc or da.
`:. oint vecB . vecdl = int_(a)^(b) vecB . vecdl + int_b^c vecB . vecdl + int_c^d vecB . vecdl + int_d^a vecB . vecdl`
`= int_a^b vecB . vecdl = int_a^b B dl = B int_a^b dl = B. Delta l " [where " ab = Delta l `(say)]
According to Ampere.s circuital law
`oint vecB cdot vecdl = mu_0` (current enclosed in length `Delta l` ) `= mu_0 (n Delta l I) " " [ :. ` Number of turns = `n Delta l` ]
Hence, we have
`B Delta l = mu_0 n Delta l I`
`implies B = mu_0 n l`
The direction of the field is given by the right hand rule.
If the solenoid has a total length l and total number of turns N, then
`B = (mu_0 N I)/(l)`.