# Rigorous derivation of Faraday's induction

When deriving Faraday's law of induction, one needs to calculate the time derivative of the flux $\Phi = \int_{S(t)} \vec{B} \cdot \vec{n} \;dS$, where $S(t)$ is the surface over which we define this flux. Now, Jackson and Wikipedia (see first equation) state that it is easy to prove that $$\frac{d}{dt}\int_{S(t)} \vec{B} \cdot \vec{n} \;dS = \int_{S(t)} \frac{\partial \vec{B}}{\partial t} \cdot \vec{n}\;dS + \oint_{\partial S(t)} \vec{B} \cdot \vec{v} \times d\vec{l},$$ where I assume $\vec{v}(t,\vec{x})$ is the vector field describing the velocity of the surface $S(t)$.

I fail to see how the second term is obvious and how I could get it from straightforward differentiation, no drawings.

My first idea would be to use that this is in fact a material derivative, and so the second term should come from $\vec{v} \cdot \vec{\nabla} \vec{B}$ plus some kind of Stoke's theorem. Is that possible?

So, let's take some closed loop enclosing a surface $S$ within the fluid and see how the magnetic flux through that loop changes as the loop gets carried along by the fluid. Over a time interval $dt$, $S$ moves a distance $\vec{v} \cdot dt$ and ends up as $S'$. The change in the flux is

$$d\Phi=\int_{S'} \vec{B}(t+dt) \cdot d\vec{a} \; - \int_{S} \vec{B}(t) \cdot d\vec{a} \;$$

Now, since $\vec{\nabla} \cdot \vec{B} = 0$ always, we can integrate $\vec B(t+dt)$ over the closed surface formed by the three surfaces $-S$, $S'$ and $R$ (the surface between $S$ and $S'$) taken together, then use the divergence theorem (this might be the "kind of Stoke's theorem" you were thinking of) to obtain

$$\int_{V} \vec{\nabla} \cdot \vec{B}d^3\vec{r} \; = \int_{S'} \vec{B}(t+dt) \cdot d\vec{a} \; + \int_{R} \vec{B}(t+dt) \cdot d\vec{a} \; - \int_{S} \vec{B}(t+dt) \cdot d\vec{a} = 0\;$$

($V$ is the volume enclosed by the three surfaces). Then, our preceding equation becomes

$$d\Phi=\int_{S} \vec{B}(t+dt) \cdot d\vec{a} \; - \int_{S} \vec{B}(t) \cdot d\vec{a} \; - \int_{R} \vec{B}(t+dt) \cdot d\vec{a} \; = dt\int_{S} \frac{\partial \vec{B}}{\partial t} \cdot d\vec{a} \; - \int_{R} \vec{B}(t+dt) \cdot d\vec{a} \;$$

The differential area $d\vec{a}$ is a parallelogram bounded by a line element $d\vec{l}$ around the loop $\partial S$ and the distance vector $\vec{v} \cdot dt$, so

$$d\vec{a}=d\vec{l} \times \vec{v} \cdot dt$$

Meaning that integrating over $d\vec{a}$ is equivalent to adding up these area elements as we move around the loop. That is, our second term becomes

$$\int_{R} \vec{B}(t+dt) \cdot d\vec{a} \; = \oint_{\partial S} \vec{B} (t+dt) \cdot (d\vec{l} \times \vec{v} \cdot dt) = dt \oint_{\partial S} \vec{B} (t) \cdot (d\vec{l} \times \vec{v})$$

Where I have taken a first order approximation in the last step. Finally, after some arrangement, we recover the expression you were asking for:

$$\frac{d}{dt}\Phi=\int_{S} \frac{\partial \vec{B}}{\partial t} \cdot d\vec{a}\; + \oint_{\partial S} \vec{B} (t) \cdot (\vec{v} \times d\vec{l})$$

Interestingly enough, and since you asked, Alfvén's theorem can be interpreted in terms of material curves, and the ideal induction equation can be written as a material derivative. See this notes for more details (section 2.1: Physical interpretation of MHD). Also, notice the resemblance between Alfvén's and Kevin's circulation theorem in inviscid fluid dynamics.

• A nice answer; same approach was taken by Purcell in his book. – user36790 Nov 7 '16 at 16:00