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Math Keeps Me Busy
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A time varying magnetic field \$\vec{B}\$ induces an electric field \$\vec{E}_{induced}\$ which satisfies the equations

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

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 EMF induced in a curve \$\gamma\$ which begins at point \$P_1\$ and ends at point \$P_2\$ is given by

$$\mathscr{E}_{induced}=\int_{\gamma} \vec{E}_{induced} \cdot d\vec{\ell}$$

In your diagram, there are two paths between any two points. One path going clockwise around the circle, and the other path going counter-clockwise. Therefore, there is not one EMF induced between two points in your diagram, but two, one for each path.

So what is the emf between AC?

$$\mathscr{E}_{induced}=\int_{\gamma AC} \vec{E}_{induced} \cdot d\vec{\ell}$$

where \$\gamma AC\$ is either the clockwise or counter-clockwise path from A to C.

is it different from the emf between AB?

Yes, the emf "between" A and B is

$$\mathscr{E}_{induced}=\int_{\gamma AB} \vec{E}_{induced} \cdot d\vec{\ell}$$

where \$\gamma AB\$ is either the clockwise or counter-clockwise path from A to B.

Math Keeps Me Busy
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