Faraday's Law of Induction

Faraday’s Law of Electromagnetic Induction relates Motional EMF to a changing Magnetic Flux. Actually, I’ll be talking about a combination of both Faraday’s Law (which gives the magnitude), and Lenz’s Law (which gives the sign).

Definition §

The induced emf in a loop of wire is equal to the negative of the change in magnetic flux over time. It is also proportional to the number of loops.

Formula - 1 §

Faraday's Law of Induction with Magnetic Flux

$$\mathcal{E}=-\frac{\text{d}{\Phi }_B}{\text{d}t}$$

Formula - 2 §

Faraday's Law of Induction with Induced Electric Field

$$\oint {\vec{E}}_{ind}\cdot d\vec{l}=-\frac{\text{d}{\Phi }_B}{\text{d}t}$$

Formula - 3 §

Loops on Faraday's Law

$${\mathcal{E}}_{net}=\sum {\mathcal{E}}_{loop}=n{\mathcal{E}}_{loop}$$

Terms §

• $\mathcal{E}$ = Induced EMF (in $\text{V}$)
• ${\Phi }_B$ = Magnetic Flux
• $t$ = Time period (in $\text{s}$)
• ${\vec{E}}_{ind}$ = Induced Electric Field
• $\vec{l}$ = Length vector for wire loop
• $n$ = Number of loops of wire (assuming same emf)\

Derivation (Formula 1) - From Motional EMF §

The standard equation can be derived by observing the Motional EMF of a moving rod in a magnetic field, attached in a U-shaped circuit:

Now, here’s where it gets tricky. Magnetic Flux is given as the Dot Product of the Magnetic Field and the Area vector. For now, let’s assume the rod is infinitely thin. How exactly do we calculate an area?

Simply put, we can’t. It’s not possible to calculate the Magnetic Flux of an infinitely thin line of current, because we have no concept of area.

However, Faraday’s law of induction doesn’t care about the magnetic flux. It cares about the magnetic flux with respect to time. And this we can calculate by looking at the area swept by the rod:

We know that the magnetic field is parallel to the area vector, so the dot product is the product of the magnitudes.

#ask bit confused about the whole area sweeping thing

$${\Phi }_B=BA$$

Area can be given in terms of displacement of the rod and the length of the rod, $x$ and $l$

$$A=lx$$

$${\Phi }_B=Blx$$

$$\text{d}{\Phi }_B=Bl\text{d}x$$

$$\frac{\text{d}{\Phi }_B}{\text{d}t}=Bl\frac{\text{d}x}{\text{d}t}$$

And we know $v=\frac{\text{d}x}{\text{d}t}$:

$$\frac{\text{d}{\Phi }_B}{\text{d}t}=Blv$$

Now recall the formula for a moving rod in a magnetic field

Since the motion is perpendicular to the magnetic field:

$$\mathcal{E}=Bvl=\frac{\text{d}{\Phi }_B}{\text{d}t}$$

The sign comes from Lenz’s Law

#ask HOW?!

#todo derivation with Solenoid to express $n$

Derivation (Formula 2) - With Induced Electric Field, EMF & Work §

We know that in the case of Static-Induced EMF, an Induced Electric Field is generated. We can use the definition of EMF to calculate the induced EMF from this electric field.

Let’s start with an external uniform Magnetic Field. As we start increasing the intensity of the magnetic field, we have a change in Magnetic Flux, specifically whenever $\frac{\text{d}\vec{B}}{\text{d}t}\ne 0$, which creates an Induced Electric Field.

We cannot use Electrical Potential, as this induced electric field is non-conservative. However, we can still use the general formula of work done:

Assume we place a test Point Charge of charge $+q$. We have two forces acting on this charge given by the Lorentz Force:

However, the magnetic field is always perpendicular to the motion, so the work (i.e. the Dot Product) is zero. Only the induced electric field applies work onto the charge. Therefore the work required to move the charge a distance of $\text{d}\vec{x}$ is given by: Electric Field#todo

$$\text{d}W=\vec{F}\cdot \text{d}\vec{x}=q\vec{E}\cdot \text{d}\vec{x}$$

$$W=q\oint \vec{E}\cdot d\vec{x}$$

And recall that EMF is just work over time:

$$\mathcal{E}=\oint \vec{E}\cdot d\vec{x}$$

And since formula 1 already tells us what $\mathcal{E}$ is…

$$\oint {\vec{E}}_{ind}\cdot d\vec{l}=-\frac{\text{d}{\Phi }_B}{\text{d}t}$$