Electrodynamics/Faraday's Law

Faraday's Law
In 1831, Faraday discovered the following three phenomena:
 * 1. If a coil of wire is pulled across a magnetic field, a current is induced.
 * 2. When a magnet is moved, a current is also induced.
 * 3. When the strength of the magnetic field is changed, a current is induced.

It is easy to understand the first phenomenon. We know that a coil of wire has free electrons inside. When the current is moved, these charges are in motion, so a magnetic force will cause them to move, creating a current.

Of course, we would immediately recognize that phenomenon 1 would imply phenomenon 2 by the relativity principle. However, its explanation in classical mechanics is less straightfoward. Clearly, since the wire isn't moving, there is no magnetic force ($$ \mathbf{F}_B=q\mathbf{v}\times\mathbf{B}$$). Thus, it must be the electric field creating the current. Somehow, a changing magnetic field must induce an electric field.

It turns out that the induced emf in all 3 cases is given by $$\xi=- \frac{\partial\Phi_B}{\partial t}$$, where $$\Phi_B$$ is the magnetic flux. Most people would call this Faraday's Law, but this equation really encompasses two physical laws: the Lorentz Force Law, and the induction of electric fields. It is a miracle that the same equation would describe two different phenomena.

The direction of the induced current is given by Lenz's Law, which states that the induced current will try to oppose the change in flux. This explains the negative sign in Faraday's equation. Lenz's Law is necessary for the Law of Conservation of Energy to hold.

Consider a metal plate falling through a space between two magnets. As it is falling, the magnetic flux would change, giving rise to a induced current, also known as an eddy current. Lenz's Law tells us that this current will oppose the change in flux, and so the magnetic force would act upwards on the plate. Thus, its velocity will slow. This application is used to damp pendulums, and is important enough to be considered for motor design.

How Generators Work
In an electric generator, a coil of wires is rotated in a magnetic field. As the coil rotates, the flux passing through a loop of the wire changes, and so a voltage is induced.

Suppose the coil has $$N$$ turns. Then the induced voltage is given by:


 * $$\xi= - N \frac{\partial\Phi}{\partial t} $$

By Lenz's Law, the resulting current flows in such a way that it will oppose the change in flux. Thus, if there were no external forces, the coil's rotation rate would slow down and eventually stop. This makes sense, for the current we induced carries energy, and because energy is conserved, the coil's rotation energy must decrease. Thus, a generator needs an external torque to turn the coil of wires.

Real Generators
Real generators generally spin a large magnet inside a Coil, in Australia we use three phase generators, meaning that there are in fact three coils, at 120 degree spacings. These generators produce voltages in the thousands. They are powered by sources such as coal, oil, and hydroelectric. In some countries Nuclear Fission is used to produce steam which turns turbines.

More on Faraday's Law
Let us try to relate the emf with the electric field. As the electric field moves a charge down the wire, it does work $$\mathbf{E} \cdot d\mathbf{l}$$ per unit charge. The total work per unit charge it does in moving the charge through the wire must be the emf, so
 * $$\xi=\oint_{\partial S} \mathbf{E} \cdot d\mathbf{l}$$

The law of fluxes would then be:
 * $$\oint_{\partial S} \mathbf{E} \cdot d\mathbf{l} = -{d\Phi \over dt}$$

You might object, saying that the integral of an electric field along a closed path is always 0, or that the electric field is conservative, but that is only true for electrostatic fields. Induced electric fields are never conservative, and there is no way of defining a potential energy for induced electric fields. $$E=-\nabla \phi$$ becomes false.

Faraday's Law is considered one of Maxwell's Equations.

We can write it in its differential form by noting that $$ \int_{\gamma} \mathbf{F} \cdot d\mathbf{s}=\int_{A} (\nabla \times \mathbf{F}) \cdot d\mathbf{A}$$. Thus,


 * $$ \int_{\gamma} \mathbf{E} \cdot ds=\int_{A} (\nabla \times \mathbf{E}) \cdot d\mathbf{A}=\int_{A} -\frac{\partial \mathbf{B}}{\partial t} d\mathbf{A}$$

so
 * $$\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}} {\partial t}$$