Inductor

Similar to how to Capacitors can store energy through an Electric Field, an inductor is a device designed to store energy through a Magnetic Field. A capacitor stores Charge, while an inductor ‘stores’ Magnetic Flux. Inductors are only really useful for non-constant Currents.

An inductor is defined as a coil of wire used for Inductance. The most common inductor is a Solenoid, a helical loop of wire.

Inductance §

Inductance of a Solenoid

L = \dfrac{\mu_{0}N^2A}{l}

>[!terms]- >* $L$ = [Inductance](Inductance.md) (in $\text{H}$) >* $\mu_{0}$ = Permeability of free space $\approx 1.26 \times 10^{-6}$ (in $\text{N/A}^2$) >* $N$ = Number of turns in the solenoid ($\text{unitless}$) >* $A$ = Area of the circle enclosed by the solenoid (in $\text{m}^2$) >* $l$ = Length of solenoid (in $\text{m}$)

Notice how, just like the capacitance of a capacitor can be given purely geometrically (without voltage or charge), the inductance of an inductor can be given purely geometrically (without current or flux)!

Induced EMF §

Induced EMF in a Solenoid

$$\mathcal{E}=-L\frac{\text{d}I}{\text{d}t}$$

Note

• $\mathcal{E}$ = Static-Induced EMF, as the conductor does not move, but the magnetic fields that are generated from the electric fields fluctuate. (in $\text{V}$)
• $L$ = Inductance (in $\text{H}$)
• $\frac{\text{d}I}{\text{d}t}$ = Change in Current (in $\text{A/s}$)

Derivations §

Inductance §

We can derive the inductance of a solenoid easily. Let’s start with the magnetic field of a Solenoid: we know that Magnetic Flux is simply the Dot Product of the magnetic field and area:

In a single loop of the solenoid , $\vec{A}$ is the circle enclosed by the loop and is always parallel to $\vec{B}$:

$${\Phi }_{loop}=\vec{B}\cdot \vec{A}=BA$$

$${\Phi }_{loop}=\frac{{\mu }_{0}NIA}{l}$$

Note that this one a single loop, even though we have an $N$ factor, it increases the magnetic field through the loop, through superposition.

Hence the total magnetic flux of the solenoid is:

$${\Phi }_{net}=N{\Phi }_{loop}=\frac{{\mu }_0{N}^{2}IA}{l}$$

And now we just divide the current:

$$L=\frac{{\Phi }_B}{I}=\frac{{\mu }_0{N}^{2}A}{l}$$

Induced EMF §

From Faraday’s Law of Induction:

$${\Phi }_B=LI$$

If the inductance remains constant, but the current changes, then:

$$\text{d}{\Phi }_B=L\text{d}I$$

$$\frac{\text{d}{\Phi }_B}{\text{d}t}=L\frac{\text{d}I}{\text{d}t}$$

$$\mathcal{E}=-L\frac{\text{d}I}{\text{d}t}$$

Applications §

• Spark plugs