Faraday's Law

Faraday's Law - From integral to differential

Faraday's Law can be converted from integral to differential form using our definition of curl! Here is the integral form of Faraday's Law again $$ \oint_C \vec{E} \circ d\vec{l} = -\frac{d}{dt} \int_S \vec{B} \circ \hat{n}dA $$ And here is our definition of curl again $$ (\vec{\nabla} \times \vec{E}) \circ \hat{n} = \lim_{\Delta A \rightarrow 0} \frac{1}{\Delta A}\oint_C \vec{E} \circ d\vec{l} $$ If we substitute Faraday's law into this definition, we get the following: $$ (\vec{\nabla} \times \vec{E}) \circ \hat{n} =- \lim_{\Delta A \rightarrow 0} \frac{1}{\Delta A}\frac{d}{dt} \int_S \vec{B} \circ \hat{n}dA $$ In the limit that $ \Delta A \rightarrow 0 $ we can say that $ \vec{B} $ is constant across this surface, therefore $$ (\vec{\nabla} \times \vec{E}) \circ \hat{n} = -\lim_{\Delta A \rightarrow 0} \frac{1}{\Delta A} \frac{d}{dt} (\vec{B} \circ \hat{n} \Delta A) $$ $$ (\vec{\nabla} \times \vec{E}) \circ \hat{n} = -\frac{d}{dt} \vec{B} \circ \hat{n} $$ Now if $ \hat{n} $ is constant in time, the differential operator $ \frac{d}{dt} $ can be brought into the dot product. We use a partial derivative as we have kept the area $ \Delta A $ at a constant orientation in space and also because $ \vec{B} $ can depend not only on time, but on position too. $$ (\vec{\nabla} \times \vec{E}) \circ \hat{n} = -\frac{\partial \vec{B}}{\partial t} \circ \hat{n} $$ $$ \vec{\nabla} \times \vec{E} = -\frac{\partial \vec{B}}{\partial t} $$ This is Faraday's Law in differential form. It states the curl of the electric field at a point is equal to the negative rate of change of magnetic field at that point.