Magnetic Induction at any point on the axis of a Solenoid

Let l represent the length of the solenoid and N the total number of turns in its winding. The number of turns per unit length is then N/L. a is the radius of the solenoid. A current i is flowing in the solenoid. The solenoid contains air in its core. Let us find the magnetic induction B at a point P on the axis of the solenoid.

Consider an elementary length dx of the solenoid, at a distance x from P. We can regard this element AB as a circular coil of radius a containing N dx / L turns.

Magnetic induction at P due to the element dx is

    dB = μ0 i a2 . Ndx . _____1______
        ____2       __L       (a² + x²)^3/2

Let us use the angle Φ instead of x as the independent variable. Then,
x = a cot Φ; dx = - a cosec² Φ dΦ
Substituting these values of x and dx in Eq. (1), we get

dB = μ0 i a² . N . a cosec2Φ dΦ    = - μ₀ iN sin Φ dΦ
        ___2_     L  [a² + a² cot² Φ]^3/2       2L

The magnetic induction at P due to the entire length of the solenoid is

    B = - μ₀ iN ∫ sin Φ dΦ
               2L

    = μ0 iN [cos α – cos β]
         2L

The direction of B is parallel to the axis of the solenoid.
    B = μ iN [cos α – cos β]
            2L
Where μ = μ₀ μ_r and μ = B/H.

Special cases

(i) At a point well inside a very long solenoid: α = 0, β = 180.
. :    B = μ₀ iN / L
(ii) At an axial point at one end of a long solenoid: α = 0, β = 90.
    B = μ₀ iN / 2L

Hence the magnetic induction at either end is one-half its magnitude at points well inside the solenoid.