.A long solenoid of length L having N turns carries a current I.Deduce the expression for the magnetic field in the interior of the solenoid.

To derive the expression for the magnetic field inside a long solenoid, we can use Ampere's law. Ampere's law states that the line integral of the magnetic field (BB) around a closed path is equal to the permeability of the medium (μμ) times the total current passing through the surface enclosed by the path.

For a long solenoid, the magnetic field inside is essentially uniform and parallel to the axis of the solenoid. Therefore, we can choose a rectangular path parallel to the axis of the solenoid to apply Ampere's law.

Let's denote:

• BB as the magnetic field inside the solenoid,
• μμ as the permeability of the medium,
• II as the current passing through the solenoid,
• NN as the number of turns per unit length of the solenoid.

Considering a rectangular path with sides parallel to the solenoid axis, the line integral of BB along the sides parallel to the axis is zero since the magnetic field is parallel to the sides.

The line integral of BB along the sides perpendicular to the axis (top and bottom) is:

∮B⋅dl=B⋅2l∮B⋅dl=B⋅2l

where ll is the length of the solenoid.

By Ampere's law, this is equal to μμ times the total current passing through the surface enclosed by the path. Since the current is uniform throughout the solenoid, the total current passing through the surface is I⋅AI⋅A, where AA is the cross-sectional area of the solenoid.

Therefore, we have:

B⋅2l=μIAB⋅2l=μIA

From this equation, we can solve for BB, the magnetic field inside the solenoid:

B=μI2lAB=2lμIA

The cross-sectional area AA of the solenoid is equal to the area of each turn multiplied by the number of turns per unit length (NN):

A=πr2⋅NA=πr2⋅N

Substituting this expression for AA into the equation for BB, we get:

B=μI2lπr2⋅NB=2lμIπr2⋅N

Therefore, the expression for the magnetic field inside the solenoid is:

B=μI2⋅NB=2μI⋅N