## Using Faraday's Law of Electromagnetic Induction to Prove One of Maxwell's Laws

Faraday's Law of Electromagnetic Induction states that

$\oint_C \mathbf{E} d \mathbf{r} =- \frac{1}{c} \frac{\partial}{\partial t} \int \int \mathbf{H} \cdot \mathbf{n} dS$

From Stoke's Theorem
$\oint_C \mathbf{F} \cdot d \mathbf{r} = \int \int_S (\mathbf{\nabla} \times \mathbf{F}) \cdot \mathbf{n} dS$
for a vector field
$\mathbf{F}$
with continuous derivatives.
If
$\mathbf{H}$
varies smoothly with time then we can write
$\frac{1}{c} \frac{\partial}{\partial t} \int \int \mathbf{H} \cdot \mathbf{n} dS = \frac{1}{c} \int \int \frac{\partial \mathbf{H}}{\partial t} \cdot \mathbf{n} dS$

Hence
$\int \int_S (\mathbf{\nabla} \times \mathbf{E}) \cdot \mathbf{n} dS= \int \int_S - \frac{1}{c} \frac{\partial \mathbf{H}}{\partial t} \cdot \mathbf{n} dS$

Equating the integrands gives
$\mathbf{\nabla} \times \mathbf{E} \cdot \mathbf{n}= - \frac{1}{c} \frac{\partial \mathbf{H}}{\partial t} \cdot \mathbf{n}$