Exercise 9. Suppose \textbf{E} is a time-dependent electric field and \textbf{H} is a time dependent magnetic field and let \textbf{S} be a surface with boundary \textbf{C} = \partial \textbf{S}. We have:
$\oint_C \textbf{E} \cdot d\textbf{s}$ = Voltage around C
$\iint_S \textbf{H} \cdot d\textbf{S}$ = Magnetic flux across S
Faraday's law states:
$\oint_C \textbf{E} \cdot d\textbf{s} = -\frac{\partial}{\partial t} \iint_S \textbf{H} \cdot d\textbf{S}$
That is, the voltage around C is equal to the negative rate of change of magnetic flux. (This is magnetic induction.) Show that if \textbf{E} is perpendicular to C, then the magnetic flux must be constant in time.
