Name: Magnetic Energy Density Stored in a Solenoid
Uploaded: 2019-07-19
Duration: 6 min 18 s
Description: The fourth state of matter is plasma and it is superheated matter. Where can we store this kind of energy? In a solenoid! Tune in to learn more about energy...

Question

Derive an expression for self inductance of a solenoid.

Accepted Answer

We are considering a solenoid with nn turns with length $l$ . The area of cross-section is $A$ . The solenoid carriers current $I$ and $B$ is the magnetic field inside the solenoid.

The magnetic field $B$ is given as,

$B = \frac{μ_{\circ} nI}{l}$

Where, $μ_{\circ}$ is the permeability of free space, $n$ is the number of turns and $I$ is the current in the solenoid and $l$ is the length.

The magnetic flux is the product of the magnetic field and area of the cross-section.

Here the magnetic flux per turn is given as,

$ϕ = B \times A$

Substituting the values in the above expression,

$ϕ = \frac{μ_{\circ} nI}{l} \times A$

Hence, there is $n$ number of turns, the total magnetic flux is given as,

$ϕ = \frac{μ_{\circ} nI}{l} \times A \times n$

$ϕ = \frac{μ_{\circ} n^{2} IA}{l} . . . . . (1)$

If $L$ is the coefficient of self-inductance of the solenoid, then

$ϕ = LI . . . . . (2)$

Comparing the two equations we get,

$LI = \frac{μ_{\circ} n^{2} IA}{l}$

$L = \frac{μ_{\circ} n^{2} A}{l}$

So the expression for the coefficient of self-inductance is $\frac{μ_{\circ} n^{2} A}{l}$ .

Derive an expression for self inductance of a solenoid.

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