Use Stokes' Theorem to derive the integral form of Faraday's law, [int_{C} mathbf{E} cdot d mathbf{s}=-frac{partial}{partial t}

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Use Stokes' Theorem to derive the integral form of Faraday's law,

\[\int_{C} \mathbf{E} \cdot d \mathbf{s}=-\frac{\partial}{\partial t} \iint_{D} \mathbf{B} \cdot d \mathbf{S}\]

from the differential form of Maxwell's equations.

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