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$$\nabla \cdot \vec{E}_{induced} = 0$$ $$\nabla \times \vec{E}_{induced} = -\frac{d\vec{B}}{dt}$$

$$\nabla \cdot \vec{E}_{induced} = 0$$ $$\nabla \times \vec{E}_{induced} = -\frac{\partial\vec{B}}{\partial t}$$

$$\nabla \cdot \vec{E}_{induced} = 0$$ $$\nabla \times \vec{E}_{induced} = \vec{B}$$

$$\nabla \cdot \vec{E}_{induced} = 0$$ $$\nabla \times \vec{E}_{induced} = -\frac{d\vec{B}}{dt}$$

Note: If a conductor is located within the electric field, the induced electric field will cause the electrons in the conductor to rearrange. This rearrangement will cause a reaction electric field \$\vec{E}_{reaction}\$ to be created which satisfies the equations

$$\nabla \cdot \vec{E}_{reaction} = \frac{\rho}{\epsilon_0}$$ $$\nabla \times \vec{E}_{reaction} = 0$$

The total electric field \$\vec{E}_{total}\$ is given by

$$\vec{E}_{total} = \vec{E}_{induced} + \vec{E}_{reaction}$$

Note: If a conductor is located within the electric field, the induced electric field will cause the electrons in the conductor to rearrange. This rearrangement will cause a reaction electric field \$\vec{E}_{reaction}\$ to be created which satisfies the equations

$$\nabla \cdot \vec{E}_{reaction} = \frac{\rho}{\epsilon_0}$$ $$\nabla \times \vec{E}_{reaction} = 0$$

The total electric field \$\vec{E}_{total}\$ is given by

$$\vec{E}_{total} = \vec{E}_{induced} + \vec{E}_{reaction}$$

The voltage drop along a curve \$\gamma\$ which begins at point \$P_1\$ and ends at \$P_2\$ is given by the integral

$$V_{\gamma}=\int_{\gamma} \vec{E}_{total} \cdot d \vec{\ell}$$

and corresponds to the work per charge associated with moving a test charge from \$P_1\$ to \$P_2\$ along \$\gamma\$. It also corresponds to the voltage used in Ohm's law.

$$V_{\gamma} = I_{\gamma}R_{\gamma}$$