# How to derive Inductance Formula of a Solenoid

Whenever current flows into a conductor the conductor generates magnetic flux and thereby magnetic field. Here it is shown how to derive Inductance formula of a Solenoid. Solenoid is a structure made by twisting a wire conductor in spiral form. Inductors used in circuits are solenoids.

Flowing picture shows a magnetic flux or magnetic field generated when current flows in a solenoid(inductor) with air as core.

Let the current flowing in the solenoid be I.

### 1.Determine Magnetic Flux Density(B)

Consider the section of the above solenoid as shown below. Consider the rectangular loop with length l and width w.

In the above diagram the field B is perpendicular to ds along the side 2,3 and 4 in the loop. Therefore we have,

$$\oint \vec{B} \cdot d\vec{s} = B l$$   ------------->(1)

From Ampere law we have,

$$\oint \vec{B} \cdot d\vec{s} = \mu_0 N I$$     ----------------->(2)

where N is the number of turns in the length l.

From (1) and (2) we have,

$$B l = \mu_0 N I$$     ----------------->(3)

or,

$$B = \frac{\mu_0 N I}{l}$$     ----------------->(4)

### 2. Determine Magnetic Flux ($$\phi$$)

By definition the magnetic flux $$\phi$$ is given by,

$$\phi = B A$$   --------------->(5)

where A is cross sectional area

Using (3) in (4),

$$\phi = \frac{\mu_0 N I A}{l}$$   --------------->(6)

### 3. Find Inductance(L)

After determining the magnetic flux we need to derive relation between inductance and magnetic flux. This is found using two equation for induced emf. The induced emf is given by the Faraday law.

$$\mathcal{E} = -N \frac{d\phi}{dt}$$   --------------->(7)

But also  we know that the induced emf depends upon the rate of change of current in the circuit. So,

$$\mathcal{E} = - L \frac{dI}{dt}$$   --------------->(8)

where L is a constant called inductance of the loop.

Therefore from (5) and (6) we have,

$$L \frac{dI}{dt} = - N \frac{d\phi}{dt}$$   --------------->(9)

Therefore,

$$L = \frac{N \phi}{I}$$   --------------->(10)

Using value of magnetic flux $$\phi$$ from (4) we get,

$$L = \frac{N^2 \mu_0 I A}{I l}$$   --------------->(11)

Thus the inductance is,

$L = \frac{\mu_0 N^2 A}{l} ------>(12)$