Magnetic field due to a long current carrying solenoid:
Consider a solenoid of length L having N turns. The diametre of the solenoid is assumed to be much smaller when compared to its length and the coil is wound very closely.
In order to calculate the magnetic field at any point inside the solenoid, we use Ampere’s circuital law. Consider a rectangular loop abed. Then from Ampere’s circuital law
= µ0 x (total current enclosed by Amperian loop)
The left hand side of the equation is
Since the elemental lengths along bc and da are perpendicular to the magnetic field which is along the axis of the solenoid, the integrals
Since the magnetic field outside the solenoid is zero, the integral
where the length of the loop ab is h. But the choice of length of the loop ab is arbitrary. We ean take very large loop such that it is equal to the length of the solenoid L. Therefore the integral is
let N I be the current passing through the solenoid of N turns, then
The number of turns per unit length is given
by \(\frac{NL}{L}\) = n, then
B = µ0 \(\frac{nLI}{L}\) = µ0nI
Since n is a constant for a given solenoid and p0 is also constant. For a fixed current I, the magnetic field inside the solenoid is also a constant
