Faraday's first dynamo: An alternate analysis

In a recent paper,¹ Smith gave a brief history concerning the role of the dynamos in the understanding of electromagnetic induction by Michael Faraday in the early 1830s. These dynamos comprise a thin conducting disk rotating about its axis in the presence of an axial external magnetic field. In the first of Faraday's dynamos, the magnetic field was asymmetric, being concentrated over a circular area located at a fixed distance from the disk axis. In the symmetrical dynamo (also referred to as a unipolar or homopolar generator), the axial magnetic field is uniform over the disk area. Because of its relative simplicity, the symmetric case is frequently discussed in textbooks on electromagnetism^2–5 and papers.^6–8,10,11 The asymmetric dynamo has received less attention. Its analytic solution was given in detail in Ref. 1, starting from expressions for the magnetic scalar potential obtained by Smythe.¹² The magnetic scalar potential is a tool generally omitted or presented as of limited interest² in modern textbooks on electromagnetism.

In our approach to this problem, we start from the analysis of space charges induced on a conductor moving in a static magnetic field, and use tools that (for the steady-state) are mainly from electrostatics, as typically presented in physics and engineering textbooks on electromagnetism. These books usually emphasize that in electrostatics, isolated conductors do not support volume charge densities (ρ), and that extra charges are distributed only on their surfaces (σ). However, conductors can carry volume charges when they move in a region where there is an external magnetic field. These space charges and the fields associated with them have been studied in a number of conducting systems rotating in static and uniform magnetic fields, such as a rigid sphere,^8,13,14 a sphere with a conducting fluid inside,¹⁴ a circular loop,⁹ and the symmetrical Faraday's dynamo.^8,13,14

From the space charges on the cylinder of Faraday's first dynamo, we evaluate the electrostatic potential $V(r→)$ through the standard tools for boundary value problems associated with Laplace's equation in cylindrical coordinates. The current density $J→(r→)$ in the disk is then evaluated through Ohm's law, and our result is compared with the one obtained from the scalar magnetic potential given by Smythe. We also calculate the conduction current circulating on the cylinder as well as the dynamo's electromotive force.

We use the same notation for the dimensions and coordinates used in Refs. 1 and 12 (the only exception is for the disk thickness for which we prefer h in place of b or d). The conducting cylinder has radius A, electrical conductivity σ, permeability μ₀, permittivity ε, and rotates with constant angular velocity $ω→=ωẑ$⁠. The applied magnetic field $B→=Bẑ$ is uniform over the circular area of radius a with center located a distance c from the cylinder axis (see Fig. 1). We define three regions: Region I is the cylinder material where B = 0, given by ρ < A (and excluding Region II); Region II is the cylinder where B ≠ 0, given by s < a; and Region III is the vacuum outside the cylinder, given by ρ > A. These regions, as well as the coordinate systems used in our analysis, are shown in Fig. 1. Comparing with the regions defined in Ref. 1, Region II is the same in the two descriptions, Region I here is the same as Regions I and III there, and what we call Region III here is not mentioned in Smith's description.

In the analysis, we will change from one system of coordinates (s, θ) to the other $(ρ,ϕ)$ through the following relations:

and

with similar relations for the coordinates $(s′,θ′)$⁠.

We consider the magnetic field generated by the induced conduction currents on the cylinder and by the convection currents (due to the rotation of the space charges on the cylinder) to be negligible compared with the applied field $B→$⁠. Our analysis is for a steady-state situation, in which the space charges, potential, vector fields, and currents are all independent of time (the approach to the steady state inside moving conductors is discussed in Ref. 16). In addition, all quantities are assumed to be uniform throughout the thickness of the cylinder (coordinate z) and the components of the vector fields along the z-axis are neglected. We disregard the inertia of the conduction electrons. Note that many of these assumptions are common to those of Smythe for this problem. The exception is that Smythe considers the magnetic field produced by the eddy currents on the disk, whereas we disregard this field in our approach.

We consider non-relativistic speeds, for which (ωA)² ≪ c², with c being the speed of light, and we restrict ourselves to a first-order theory (in ωA/c). In the inertial rest frame of the axis of the cylinder, the electric field is $E→(r→)$⁠, the magnetic field is $B→(r→)$⁠, and the conduction current density is $J→(r→)$⁠. A point inside the cylinder moves at a velocity $v→(r→)=ωρϕ̂$ in relation to the axis frame.

We start by examining the dielectric response of the medium. The constitutive equation for the polarization $P→(r→)$ inside the moving cylinder, in the first-order theory, is (see Ref. 15)

The constitutive equation for the conduction current density $J(r→)$ (Ohm's law) has the form

Equations (5) and (6) take account of the fact that inside the moving material, the force responsible for distorting the electronic orbitals of the atoms (polarization) and for the drift of the charge carriers (current) is $q(E→+v→×B→)$ instead of $qE→$⁠.

In a steady-state configuration $∇→·J→(r→)=0$⁠, and it follows that the space charge density of bound charges is zero, since $∇→·P→(r→)=0$⁠. From Gauss law, we obtain the volume free-charge density $ρF(r→)$ given by

showing that the $v→(r→)×B→(r→)$ field leads to free-charge accumulation at points in the conducting material where its divergence is non-zero. Thus, we have a uniform negative charge density in Region II, given by

An equation for the surface charge density $σF(r→)$ can be obtained from the continuity condition for a field at the boundary between Regions I and II inside the material disk as

where the unit normal vector $n̂$ points from Region II to Region I. Thus, at the boundary between Regions I and II, where s = a, we obtain a charge density of

where we defined σ_U = ε₀ωBa (a uniform charge density) and $σNU(θ)=ε0ωBc cos θ$ (a non-uniform charge density) at this boundary. Note that the bulk charge density ρ_F and these surface charge densities make the cylinder electrically neutral.

There is also a charge density $σA(ϕ)$ at the boundary ρ = A between Regions I and III, which cannot be obtained from a relation such as that in Eq. (9), because the field $v→×B→$ is not known in vacuum. In the next section we calculate the electric potential from knowledge of the space charges $ρF(r→)$ and $σF(r→)$ alone, and $σA(ϕ)$ will be obtained from the boundary conditions for the potential at ρ = A.

We now solve the boundary-value problem for the potential in Regions I, II, and III. The boundary conditions are basically the continuity of the potential and the discontinuity of its derivative at the boundaries where the surface charge density is non-zero.

First, since ρ_F and σ_U are uniform charge densities on the cylinder of radius a, the electric potential due to them, $VU(r→)$⁠, is given by

From now on, we will assume that the potential due to the other space charges satisfies Laplace's equation everywhere. The electric potential $VNU(r→)$ due to the surface charge σ_NU(θ) can be easily obtained, because this problem is related to the standard one in which an infinitely long metal pipe is placed at right angles to an otherwise uniform electric field;² the solution is given by

Now, we have the condition that the radial component of the current density vanishes at the cylinder-vacuum boundary: $Jρ(ρ=A,ϕ)=0$⁠. From Ohm's law, this condition in terms of $V(r→)$ is given by

We start by searching for a potential $Vi(ρ,ϕ)$ that can be added to $VNU(ρ,ϕ)$⁠, such that $∂V(ρ,ϕ)/∂ρ=∂[VNU(ρ,ϕ)+Vi(ρ,ϕ)]/∂ρ=0$ at ρ = A. Since V_i is also a solution of Laplace's equation, we assume for V_i the same form as V_NU—Eq. (12) valid for Regions I and III—but with a change of sign and the constant αc in place of the constant c, where α is a dimensionless parameter to be determined. This gives

in Regions I and II.

Note that the simple transformation c → αc in Eq. (12) corresponds to a displacement of the cylinder with charge density σ_NU(θ) along the x-axis and to a change in the charge density from σ_NU(θ) to $ασNU(θ′)$⁠. Thus, $Vi(ρ,ϕ)$ is the potential of a cylinder with charge density $−ασNU(θ′)$ and center at $O′=(x=αc,y=0)$⁠, and with corresponding coordinates $θ′$ and $s′$⁠. The boundary of this cylinder, with α × c > A + a, is shown by the dashed circle in Fig. 1. We find that α = (A/c)² fits the boundary condition given by Eq. (13), and this result can be interpreted in terms of the method of images. The cylinder with center at $O′=[x=(A/c)2c,y=0]$ and charge density $−(A/c)2σNU(θ′)$ is the image of the cylinder with center at (x = c, y = 0) and surface charge density σ_NU(θ), and its potential inside the conducting cylinder is identical with that of the induced charges on the boundary ρ = A.

Thus, by adding V_U, V_NU, and V_i, and using Eqs. (1)–(4), we obtain the electric potential inside the cylinder to be

in Region II, and

in Region I. At the boundary ρ = A, the potential is given by

which is the open-circuit voltage measured by an ideal voltmeter connected between the axis and the point $(ρ=A,ϕ)$ on the periphery of the cylinder. For $ϕ=0$ we recover the result given in Eq. (27) of Ref. 1.

From the continuity of the potential at the boundary ρ = A, we find the potential outside the cylinder (in Region III) to be

The discontinuity in the derivative of the potential at ρ = A then gives the charge density at the cylinder surface

with a net surface charge of zero. The behaviors of the functions σ_F(θ)/ε₀ and $σA(ϕ)/ε0$ are shown in Fig. 2, with the values of the parameters ω, B, A, a, and c fixed as explained in Sec. II C.

From the electric potential we can calculate the conduction current density in the disk through Ohm's law to be

These expressions are quite long and will not be shown here. In order to compare our results with those obtained by Smythe, we calculate the current density from the magnetic potential $Φ(r→)$ given in Refs. 1 and 12. The expressions for $J(r→)$ obtained from the two approaches are identical, showing that the two methods, with different starting points and assumptions, have the same level of precision in their results.

The lines of force of the field $J→$ are all closed and were illustrated in Fig. 6 of Ref. 1. In a few words, $J→$ is nearly along the positive x-axis inside Region II, emerging from this Region on its curved face with s = a and –π/2 ≤ θ ≤ π/2, and returning through Region I to the opposite face of Region II. The current through any cross section of the cylinder given by a plane x = constant is zero. We can calculate the total conduction current flowing in the cylinder as the flux of the density $J(r→)$ through a cross section of Region II given by the plane x = c. The unit vector normal to this surface is $x̂$⁠, so we obtain

The external torque to maintain a constant ω, which was calculated in Ref. 12, can be obtained through

which results in

In order to obtain the electromotive force $E$ of this dynamo we set $EIC=Textω$ to get

Note that when c = 0 and a = A, our equations recover the results for the symmetrical dynamo.⁸ The space charges are

in Region I (= Region II), and

at the disk surface. The electric potential is given by

in Region I (= Region II), and

in Region III. In this case, the $E→$ field cancels the $v→×B→$ field inside the cylinder, and the conduction current is zero; no external torque is needed and there is no electromotive force.

For the experiments with the Faraday's dynamo reported in Ref. 1 (a circular copper disk with A ≃ 7.5 cm and h ≃ 0.86 mm, and a field with B ≃ 1.08 T and a ≃ 1.3 cm fixed at c ≃ 5 cm of the disk axis) we obtain, for ω ≃ 772 rpm, I_C ≃ 2144 A and $E≃88 mV$⁠.

The free charge densities induced on solid conductors rotating in a static magnetic field may or may not be accompanied by a steady-state electric current density. Faraday's first dynamo is a nontrivial example of a system where the rotation of the conductor induces charges and currents, for which we derived analytical expressions. Our results for $J→(r→)$ are equal to those obtained from the magnetic scalar potential deduced by Smythe in 1942. We believe that our analysis is more straightforward and based on less abstract concepts than Smythe's approach, and that it can help students understand that the voltages on the rotating cylinder can be associated with simple space charges within its volume and on its surfaces, as they are used to doing in the study of electrostatics.

The author would like to thank the anonymous reviewers for their suggestions for improving the manuscript.