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The rotational of the electric field: $$\nabla\times\mathbf E = -\frac{\partial\mathbf B}{\partial t}$$ Using Stokes' Theorem on this equation, we get the integral form of this equation: $$\varepsilon = \oint_{\gamma}\mathbf E\cdot\mathbf{dl} = -\frac{d}{dt}\iint_S\mathbf B\cdot\mathbf{dS} = -\frac{d\Phi}{dt}$$ Which means, the electric field in a ...