Alternating-current generator
In the remainder of this chapter we apply the principles of Section 17–1 to analyze a number of the phenomena discussed in Chapter 16. We first look in more detail at the alternating-current generator. Such a generator consists basically of a coil of wire rotating in a uniform magnetic field. The same result can also be achieved by a fixed coil in a magnetic field whose direction rotates in the manner described in the last chapter. We will consider only the former case. Suppose we have a circular coil of wire which can be turned on an axis along one of its diameters. Let this coil be located in a uniform magnetic field perpendicular to the axis of rotation, as in Fig. 17–6. We also imagine that the two ends of the coil are brought to external connections through some kind of sliding contacts. Due to the rotation of the coil, the magnetic flux through it will be changing. The circuit of the coil will therefore have an emf in it. Let $S$ be the area of the coil and $\theta$ the angle between the magnetic field and the normal to the plane of the coil.1 The flux through the coil is then \begin{equation} \label{Eq:II:17:13} BS\cos\theta. \end{equation}  If the coil is rotating at the uniform angular velocity $\omega$, $\theta$ varies with time as $\theta=\omega t$. Each turn of the coil will have an emf equal to the rate of change of this flux. If the coil has $N$ turns of wire the total emf will be $N$ times larger, so  \begin{equation} \label{Eq:II:17:14} \emf=-N\,\ddt{}{t}(BS\cos\omega t)=NBS\omega\sin\omega t. \end{equation} 
  \begin{align} \emf&=-N\,\ddt{}{t}(BS\cos\omega t)\notag\\[1ex] \label{Eq:II:17:14} &=NBS\omega\sin\omega t. \end{align}   
 If we bring the wires from the generator to a point some distance from the rotating coil, where the magnetic field is zero, or at least is not varying with time, the curl of $\FLPE$ in this region will be zero and we can define an electric potential. In fact, if there is no current being drawn from the generator, the potential difference $V$ between the two wires will be equal to the emf in the rotating coil. That is, \begin{equation*} V=NBS\omega\sin\omega t=V_0\sin\omega t. \end{equation*}  The potential difference between the wires varies as $\sin\omega t$. Such a varying potential difference is called an alternating voltage.  Since there is an electric field between the wires, they must be electrically charged. It is clear that the emf of the generator has pushed some excess charges out to the wire until the electric field from them is strong enough to exactly counterbalance the induction force. Seen from outside the generator, the two wires appear as though they had been electrostatically charged to the potential difference $V$, and as though the charge was being changed with time to give an alternating potential difference. There is also another difference from an electrostatic situation. If we connect the generator to an external circuit that permits passage of a current, we find that the emf does not permit the wires to be discharged but continues to provide charge to the wires as current is drawn from them, attempting to keep the wires always at the same potential difference. If, in fact, the generator is connected in a circuit whose total resistance is $R$, the current through the circuit will be proportional to the emf of the generator and inversely proportional to $R$. Since the emf has a sinusoidal time variation, so also does the current. There is an alternating current \begin{equation*} I=\frac{\emf}{R}=\frac{V_0}{R}\sin\omega t. \end{equation*}  The schematic diagram of such a circuit is shown in Fig. 17–7. We can also see that the emf determines how much energy is supplied by the generator. Each charge in the wire is receiving energy at the rate $\FLPF\cdot\FLPv$, where $\FLPF$ is the force on the charge and $\FLPv$ is its velocity. Now let the number of moving charges per unit length of the wire be $n$; then the power being delivered into any element $ds$ of the wire is \begin{equation*} \FLPF\cdot\FLPv n\,ds. \end{equation*}  For a wire, $\FLPv$ is always along $d\FLPs$, so we can rewrite the power as \begin{equation*} nv\FLPF\cdot d\FLPs. \end{equation*}  The total power being delivered to the complete circuit is the integral of this expression around the complete loop: \begin{equation} \label{Eq:II:17:15} \text{Power}=\oint nv\FLPF\cdot d\FLPs. \end{equation}  Now remember that $qnv$ is the current $I$, and that the emf is defined as the integral of $F/q$ around the circuit. We get the result \begin{equation} \label{Eq:II:17:16} \text{Power from a generator}=\emf I. \end{equation}  When there is a current in the coil of the generator, there will also be mechanical forces on it. In fact, we know that the torque on the coil is proportional to its magnetic moment, to the magnetic field strength $B$, and to the sine of the angle between. The magnetic moment is the current in the coil times its area. Therefore the torque is \begin{equation} \label{Eq:II:17:17} \tau=NISB\sin\theta. \end{equation}  The rate at which mechanical work must be done to keep the coil rotating is the angular velocity $\omega$ times the torque: \begin{equation} \label{Eq:II:17:18} \ddt{W}{t}=\omega\tau=\omega NISB\sin\theta. \end{equation}  Comparing this equation with Eq. (17.14), we see that the rate of mechanical work required to rotate the coil against the magnetic forces is just equal to $\emf I$, the rate at which electrical energy is delivered by the emf of the generator. All of the mechanical energy used up in the generator appears as electrical energy in the circuit. As another example of the currents and forces due to an induced emf, let’s analyze what happens in the setup described in Section 17–1, and shown in Fig. 17–1. There are two parallel wires and a sliding crossbar located in a uniform magnetic field perpendicular to the plane of the parallel wires. Now let’s assume that the “bottom” of the U (the left side in the figure) is made of wires of high resistance, while the two side wires are made of a good conductor like copper—then we don’t need to worry about the change of the circuit resistance as the crossbar is moved. As before, the emf in the circuit is \begin{equation} \label{Eq:II:17:19} \emf=vBw. \end{equation}  The current in the circuit is proportional to this emf and inversely proportional to the resistance of the circuit: \begin{equation} \label{Eq:II:17:20} I=\frac{\emf}{R}=\frac{vBw}{R}. \end{equation}  Because of this current there will be a magnetic force on the crossbar that is proportional to its length, to the current in it, and to the magnetic field, such that \begin{equation} \label{Eq:II:17:21} F=BIw. \end{equation}  Taking $I$ from Eq. (17.20), we have for the force \begin{equation} \label{Eq:II:17:22} F=\frac{B^2w^2}{R}\,v. \end{equation}  We see that the force is proportional to the velocity of the crossbar. The direction of the force, as you can easily see, is opposite to its velocity. Such a “velocity-proportional” force, which is like the force of viscosity, is found whenever induced currents are produced by moving conductors in a magnetic field. The examples of eddy currents we gave in the last chapter also produced forces on the conductors proportional to the velocity of the conductor, even though such situations, in general, give a complicated distribution of currents which is difficult to analyze. It is often convenient in the design of mechanical systems to have damping forces which are proportional to the velocity. Eddy-current forces provide one of the most convenient ways of getting such a velocity-dependent force. An example of the application of such a force is found in the conventional domestic wattmeter. In the wattmeter there is a thin aluminum disc that rotates between the poles of a permanent magnet. This disc is driven by a small electric motor whose torque is proportional to the power being consumed in the electrical circuit of the house. Because of the eddy-current forces in the disc, there is a resistive force proportional to the velocity. In equilibrium, the velocity is therefore proportional to the rate of consumption of electrical energy. By means of a counter attached to the rotating disc, a record is kept of the number of revolutions it makes. This count is an indication of the total energy consumption, i.e., the number of watthours used. We may also point out that Eq. (17.22) shows that the force from induced currents—that is, any eddy-current force—is inversely proportional to the resistance. The force will be larger, the better the conductivity of the material. The reason, of course, is that an emf produces more current if the resistance is low, and the stronger currents represent greater mechanical forces. We can also see from our formulas how mechanical energy is converted into electrical energy. As before, the electrical energy supplied to the resistance of the circuit is the product $\emf I$. The rate at which work is done in moving the conducting crossbar is the force on the bar times its velocity. Using Eq. (17.22) for the force, the rate of doing work is \begin{equation*} \ddt{W}{t}=\frac{v^2B^2w^2}{R}. \end{equation*}  We see that this is indeed equal to the product $\emf I$ we would get from Eqs. (17.19) and (17.20). Again the mechanical work appears as electrical energy.