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Electromagnetic Induction. Maxwells Equations

Problem 3.355

Using Maxwell's equations, show that (a) a time-dependent magnetic field cannot exist without an electric field; (b) a uniform electric field cannot exist in the presence of a timedependent magnetic field; (c) inside an empty cavity a uniform electric (or magnetic) field can be time-dependent.

Reveal Answer
3.355. (a) If B(t),\mathbf{B}(t), then ∇×E=−∂B/∂t≠0\boldsymbol{\nabla} \times \mathbf{E}=-\partial \mathbf{B} / \partial t \neq 0. The spatial derivatives of the field E,\mathbf{E}, however, may not be equal to zero (∇×E≠0)(\mathbf{\nabla} \times \mathbf{E} \neq 0) only in the presence of an electric field. (b) If B(t)\mathbf{B}(t), then ∇×E=−∂B/∂t≠0\boldsymbol{\nabla} \times \mathbf{E}=-\partial \mathbf{B} / \partial t \neq 0. But in the uniform field ∇×E=0\nabla \times \mathbf{E}=0 (c) It is assumed that E=af⁡(t),\mathbf{E}=\operatorname{af}(t), where a is a vector which is independent of the coordinates, f(t)f(t) is an arbitrary funetion of time. Then −∂B/∂t=∇×E=0,-\partial \mathbf{B} / \partial t=\boldsymbol{\nabla} \times \mathbf{E}=0, that is the field B\mathbf{B} does not vary with time. Generally speaking, this contradicts the equation ∇×H=\boldsymbol{\nabla} \times \mathbf{H}= =∂D/∂t=\partial \mathbf{D} / \partial t for in this case its left-hand side does not depend on time whereas its right-hand side does. The only exception is the case when f(t)f(t) is a linear function. In this case the uniform field EE can be time-dependent.
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