1. Faraday's law predicts that there will be zero EMF but there is a non-zero EMF.
2. Faraday's law predicts that there will be a non-zero EMF but there is a zero EMF.

Faraday deduced this law in 1831, after inventing the first electromagnetic generator or dynamo, but was never satisfied with his own explanation of the paradox.

## Paradoxes in which Faraday's law of induction predicts zero EMF but there is a non-zero EMF

These paradoxes are generally resolved by the fact that an EMF may be created by a changing flux in a circuit as explained in Faraday's law or by the movement of a conductor in a magnetic field. This is explained by Feynman as noted below.

### The equipment

Figure 1: Faraday's disc electric generator. The disc rotates with angular rate ω, sweeping the conducting disc circularly in the static magnetic field B due to a permanent magnet. The magnetic Lorentz force v × B drives the current radially across the conducting disc to the conducting rim, and from there the circuit path completes through the lower brush and the axle supporting the disc. Thus, current is generated from mechanical motion.

The experiment requires a few simple components (see Figure 1): a cylindrical magnet, a conducting disc with a conducting rim, a conducting axle, some wiring, and a galvanometer. The disc and the magnet are fitted a short distance apart on the axle, on which they are free to rotate about their own axes of symmetry. An electrical circuit is formed by connecting sliding contacts: one to the axle of the disc, the other to its rim. A galvanometer can be inserted in the circuit to measure the current.

### The procedure

The experiment proceeds in three steps:

1. The magnet is held to prevent it from rotating, while the disc is spun on its axis. The result is that the galvanometer registers a direct current. The apparatus therefore acts as a generator, variously called the Faraday generator, the Faraday disc, or the homopolar (or unipolar) generator.
2. The disc is held stationary while the magnet is spun on its axis. The result is that the galvanometer registers no current.
3. The disc and magnet are spun together. The galvanometer registers a current, as it did in step 1.

The experiment is described by some as a "paradox" as it seems, at first sight, to violate Faraday's law of electromagnetic induction, because the flux through the disc appears to be the same no matter what is rotating. Hence, the EMF is predicted to be zero in all three cases of rotation. The discussion below shows this viewpoint stems from an incorrect choice of surface over which to calculate the flux.

The paradox appears a bit different from the lines of flux viewpoint: in Faraday's model of electromagnetic induction, a magnetic field consisted of imaginary lines of magnetic flux, similar to the lines that appear when iron filings are sprinkled on paper and held near a magnet. The EMF is proposed to be proportional to the rate of cutting lines of flux. If the lines of flux are imagined to originate in the magnet, then they would be stationary in the frame of the magnet, and rotating the disc relative to the magnet, whether by rotating the magnet or the disc, should produce an EMF, but rotating both of them together should not.

In Faraday's model of electromagnetic induction, a circuit received an induced current when it cut lines of magnetic flux. According to this model, the Faraday disc should have worked when either the disc or the magnet was rotated, but not both. Faraday attempted to explain the disagreement with observation by assuming that the magnet's field, complete with its lines of flux, remained stationary as the magnet rotated (a completely accurate picture, but maybe not intuitive in the lines-of-flux model). In other words, the lines of flux have their own frame of reference. As we shall see in the next section, modern physics (since the discovery of the electron) does not need the lines-of-flux picture and dispels the paradox.

### Modern explanations

#### Using the Lorentz force

After the discovery of the electron and the forces that affect it, a microscopic resolution of the paradox became possible. See Figure 1. The metal portions of the apparatus are conducting, and confine a current due to electronic motion to within the metal boundaries. All electrons that move in a magnetic field experience a Lorentz force of F = qv × B, where v is the velocity of the electrons and q is the charge on an electron. This force is perpendicular to both the velocity of the electrons, which is in the plane of the disc, and to the magnetic field, which is normal (surface normal) to the disc. An electron at rest in the frame of the disc moves circularly with the disc relative to the B-field, and so experiences a radial Lorentz force. In Figure 1 this force (on a positive charge, not an electron) is outward toward the rim according to the right-hand rule.

Of course, this radial force, which is the cause of the current, creates a radial component of electron velocity, generating in turn its own Lorentz force component that opposes the circular motion of the electrons, tending to slow the disc's rotation, but the electrons retain a component of circular motion that continues to drive the current via the radial Lorentz force.

This mechanism agrees with the observations: an EMF is generated whenever the disc moves relative to the magnetic field, regardless of how that field is generated.

The use of the Lorentz equation to explain the Faraday Paradox has led to a debate in the literature as to whether or not a magnetic field rotates with a magnet. Since the force on charges expressed by the Lorentz equation depends upon the relative motion of the magnetic field to the conductor where the EMF is located it was speculated that in the case when the magnet rotates with the disk but a voltage still develops, that the magnetic field must therefore not rotate with the magnetic material as it turns with no relative motion with respect to the conductive disk.

However, careful thought showed if the magnetic field was assumed to rotate with the magnet and the magnet rotated with the disk that a current should still be produced, not by EMF in the disk (there is no relative motion between the disk and magnet) but in the external circuit linking the brushes[1] which is in fact in relative motion with respect to the rotating magnet. In fact it was shown that so long as a current loop was used to measure induced EMFs from the motion of the disk and magnet it is not possible to tell if the magnetic field does or does not rotate with the magnet.

Several experiments have been proposed using electrostatic measurements or electron beams to resolve the issue, but apparently none have been successfully performed to date.

However, In case 2, since there is no current observed—the magnetic field did not rotate with the rotating magnet.

#### Relation to Faraday's law of induction

The flux through the portion of the path from the brush at the rim, through the outside loop and the axle to the center of the disc is always zero because the magnetic field is in the plane of this path (not perpendicular to it), no matter what is rotating, so the integrated emf around this part of the path is always zero. Therefore, attention is focused on the portion of the path from the axle across the disc to the brush at the rim.

Faraday's law of induction can be stated in words as:[2]

The induced electromotive force or EMF in any closed circuit is equal to the time rate of change of the magnetic flux through the circuit.

Mathematically, the law is stated:

$\mathcal{E} = - \frac {d \Phi_B} {dt} = -\frac {d}{dt}\iint_{\Sigma (t)} d \boldsymbol{A} \cdot \mathbf{B} (\mathbf{r},\ t) \ ,$

where ΦB is the flux, and d A is a vector element of area of a moving surface Σ(t) bounded by the loop around which the EMF is to be found.

Figure 2: Two possible loops for finding EMF: the geometrically simple path is easy to use, but the other provides the same EMF. Neither is intended to imitate any line of physical current flow.

How can this law be connected to the Faraday disc generator, where the flux linkage appears to be just the B-field multiplied by the area of the disc?

One approach is to define the notion of "rate of change of flux linkage" by drawing a hypothetical line across the disc from the brush to the axle and asking how much flux linkage is swept past this line per unit time. See Figure 2. Assuming a radius R for the disc, a sector of disc with central angle θ has an area:

$A = \frac {\theta}{2\pi} \pi R^2 \ ,$

so the rate that flux sweeps past the imaginary line is

$\mathcal{E} = - \frac {d \Phi_B} {dt} = B \frac{dA} {dt} = B\ \frac {R^2}{2}\ \frac {d \theta}{dt} =B\ \frac {R^2}{2}\omega \ ,$

with ω = d θ / dt the angular rate of rotation. The sign is chosen based upon Lenz's law: the field generated by the motion must oppose the change in flux caused by the rotation. For example, the circuit with the radial segment in Figure 2 according to the right-hand rule adds to the applied B-field, tending to increase the flux linkage. That suggests that the flux through this path is decreasing due to the rotation, so d θ / d t is negative.

This flux-cutting result for EMF can be compared to calculating the work done per unit charge making an infinitesimal test charge traverse the hypothetical line using the Lorentz force / unit charge at radius r, namely |v × B | = B v = B r ω:

$\mathcal{E} = \int_0^R dr Br \omega = \frac {R^2}{2} B \omega \ ,$

which is the same result.

The above methodology for finding the flux cut by the circuit is formalized in the flux law by properly treating the time derivative of the bounding surface Σ ( t ). Of course, the time derivative of an integral with time dependent limits is not simply the time derivative of the integrand alone, a point often forgotten; see Leibniz integral rule and Lorentz force.

In choosing the surface Σ ( t ), the restrictions are that (i) it has to be bounded by a closed curve around which the EMF is to be found, and (ii) it has to capture the relative motion of all moving parts of the circuit. It is emphatically not required that the bounding curve corresponds to a physical line of flow of the current. On the other hand, induction is all about relative motion, and the path emphatically must capture any relative motion. In a case like Figure 1 where a portion of the current path is distributed over a region in space, the EMF driving the current can be found using a variety of paths. Figure 2 shows two possibilities. All paths include the obvious return loop, but in the disc two paths are shown: one is a geometrically simple path, the other a tortuous one. We are free to choose whatever path we like, but a portion of any acceptable path is fixed in the disc itself and turns with the disc. The flux is calculated though the entire path, return loop plus disc segment, and its rate-of change found.

Figure 3: Mapping of the Faraday disc into a sliding conducting rectangle example. The disc is viewed as an annulus; it is cut along a radius and bent open to become a rectangle.

In this example, all these paths lead to the same rate of change of flux, and hence the same EMF. To provide some intuition about this path independence, in Figure 3 the Faraday disc is unwrapped onto a strip, making it resemble a sliding rectangle problem. In the sliding rectangle case, it becomes obvious that the pattern of current flow inside the rectangle is time-independent and therefore irrelevant to the rate of change of flux linking the circuit. There is no need to consider exactly how the current traverses the rectangle (or the disc). Any choice of path connecting the top and bottom of the rectangle (axle- to-brush in the disc) and moving with the rectangle (rotating with the disc) sweeps out the same rate-of-change of flux, and predicts the same EMF. For the disc, this rate-of-change of flux estimation is the same as that done above based upon rotation of the disc past a line joining the brush to the axle.

### Some observations

Whether the magnet is "moving" is irrelevant in this analysis, as it does not appear in Faraday's law. In fact, rotating the magnet does not alter the B-field. Likewise, rotation of the magnet and the disc is the same as rotating the disc and keeping the magnet stationary. The crucial relative motion is that of the disk and the return path, not of the disk and the magnet.

This becomes clearer if a modified Faraday disk is used in which the return path is not a wire but another disk. That is, mount two conducting disks just next to each other on the same axle and let them have sliding electrical contact at the center and at the circumference. The current will be proportional to the relative rotation of the two disks and independent of any rotation of the magnet.

### Configuration without a return path

A Faraday disk can also be operated with neither a galvanometer nor a return path. When the disk spins, the electrons collect along the rim and leave a deficit near the axis (or the other way around). It is possible in principle to measure the distribution of charge, for example, through the electromotive force generated between the rim and the axle (though not necessarily easy). This charge separation will be proportional to the magnetic field and the rotational velocity of the disk. The magnetic field will be independent of any rotation of the magnet. In this configuration, the polarisation is determined by the absolute rotation of the disk, that is, the rotation relative to an inertial frame. The relative rotation of the disk and the magnet plays no role.

Figure 4: An example, based on one of Feynman's examples, where Faraday's law does not work. A rectangle of photoconductive material slides along a pair of wires. At a fixed location a strong light and a strong magnetic field create a narrow immovable strip of conducting material subject to a Lorentz force.

Figure 4 shows a translating rectangle of material with a narrow conducting strip subject to a magnetic field. This strip of material is rendered conducting at a fixed location by, for example, a strong light beam at this location. The magnetic field also is confined to the same strip. The Lorentz force drives a current from the top rail to the bottom rail through this strip, and the circuit is completed through leads attached to the top and bottom conducting rails. In this example, the circuit does not move, and the magnetic flux through the circuit is not changing, so Faraday's law suggests no current flows. However, the Lorentz force law suggests a current does flow. This example is based upon one devised by Richard Feynman to illustrate the inapplicability of Faraday's law of induction to certain situations (that is, the version of Faraday's law of induction which relates EMF to magnetic flux, which he terms the "flux rule"). Referring to his example, Feynman said:[3]

The "flux rule" does not work in this case. It must be applied to circuits in which the material of the circuit remains the same. When the material of the circuit is changing, we must return to the basic laws. The correct physics is always given by the two basic laws

$\mathbf{F}=q(\mathbf{E}+\mathbf{v}\times\mathbf{B})$
$\nabla \times \mathbf{E}=- \begin{matrix}\frac {\partial} {\partial t}\end{matrix}\mathbf{B} \ .$

— Richard P Feynman  The Feynman Lectures on Physics

Accordingly, he explains the phenomenon using the Lorentz force law, as described above. The point is that the flux law applies only to some situations, albeit some very practical ones.

#### Using the special theory of relativity

There is no paradox or difficulty if one invokes the special theory of relativity. It tells us that the observer sitting on the frame S of the stationary guide rail and the fixed-location strip sees the photoconductive rectangle moving through the magnetic induction field $B\,$ with a velocity $v\,$ . He sees an electric field in his frame of reference of $E=vB\,$ . The Lorentz force experienced by the charged particle at the beamed photoconductive area is $F_{Lorentz}=qvB\,$ .

For the observer resting on the frame S′ of the photoconductive rectangle, he sees $E\,'=\gamma vB$ . Thus the Lorentz force experienced by the charged particle is $F\,'_{Lorentz}=q\gamma vB$ .

Noting the fact that from frame S to frame S′, the force perpendicular to the relative velocity transforms as follows:

$F'= \frac{F}{(1-v^2/c^2)\gamma}=\gamma F$

Thus, the charged particle experiences the same force in frame S or frame S′.[4] [5]

Note: The value of gamma has to be larger than 1, therefore F is not equal to F', the charged particle does not experience the same force in frame S or frame S′ as the relativistic explanation shows it, while the Volt meter would read the same value.

## Paradoxes in which Faraday's law of induction predicts non-zero EMF but there is a zero EMF

These paradoxes are generally resolved by determining that the apparent motion of the circuit is actually deconstruction of the circuit followed by reconstruction of the circuit on a different path.

Now that it has been proven that the magnetic field rotates with the magnet[citation needed] as discussed in the observation section, what is really causing the paradox? Before addressing this question, we will discuss when Faraday's Law is valid and when it breaks down as in the disk experiment. In the case when the disk alone spins there is no change in flux through the circuit, however, there is an electromotive force induced contrary to Faraday's law. We can also show an example when there is a change in flux, but no induced voltage. Figure 5 (near right) shows the setup used in Tilley's experiment.[6] It is a circuit with two loops or meshes. There is a galvanometer connected in the righthand loop, a magnet in the center of the lefthand loop, a switch in the lefthand loop, and a switch between the loops. We start with the switch on the left open and that on the right closed. When the switch on the left is closed and the switch on the right is open there is no change in the field of the magnet, but there is a change in the area of the galvanometer circuit. This means that there is a change in flux. However the galvanometer did not deflect meaning there was no induced voltage, and Faraday's law does not work in this case. According to A. G. Kelly this suggests that an induced voltage in Faraday's experiment is due to the "cutting" of the circuit by the flux lines, and not by "flux linking" or the actual change in flux. This follows from the Tilley experiment because there is no movement of the lines of force across the circuit and therefore no current induced although there is a change in flux through the circuit. Nussbaum suggests that for Faraday's law to be valid work must be done in producing the change in flux.[7]
To understand this idea, we will step through the argument given by Nussbaum.[7] We start by calculating the force between two current carrying wires. The force on wire 1 due to wire 2 is given by:

$\mathbf{F}_{21} = \frac {\mu_0} {4 \pi} I_1 I_2 \oint_{C_1} \oint_{C_2} \frac {d \mathbf{l_1}\ \mathbf{ \times} \ (d \mathbf{l_2} \ \mathbf{ \times } \ \hat{\mathbf{r}}_{21} )} {r_{21}^2} \$

The magnetic field from the second wire is given by:

$\mathbf{B}_{2} = \frac {\mu_0} {4 \pi} I_2 \oint_{C_2} \frac { (d \mathbf{l_2} \ \mathbf{ \times } \ \hat{\mathbf{r}}_{21} )} {r_{21}^2} \$

So we can rewrite the force on wire 1 as:

$\mathbf{F}_{21} = I_1 \oint_{C_1} d \mathbf{l_1}\ \mathbf{ \times} \mathbf{B}_{2}$

Now consider a segment $d \mathbf{l}$ of a conductor displaced $d \mathbf{r}$ in a constant magnetic field. The work done is found from:

$d W = d \mathbf{F} \cdot d \mathbf{r}$

If we plug in what we previously found for $d \mathbf{F}$ we get:

$d W = (I d \mathbf{l} \mathbf{ \times} \mathbf{B}) \cdot d \mathbf{r}$

The area covered by the displacement of the conductor is:

$d \mathbf{S}= d\mathbf{r} \mathbf{ \times} d\mathbf{l}$

Therefore:

$d W = I \mathbf{B} \cdot d\mathbf{S} = I d \Phi_B$

The differential work can also be given in terms of charge $dq$ and potential difference $V$:

$d W = V dq = V I dt$

By setting the two equations for differential work equal to each other we arrive at Faraday's Law.

$d \Phi_B = V dt$

Furthermore, we now see that this is only true if $d W$ is nonvanishing. Meaning, Faraday's Law is only valid if work is performed in bringing about the change in flux.

## References

1. ^ A. G. Kelly, Monographs 5 & 6 of the Institution of Engineers of Ireland, 1998, ISBN 1-898012-37-3 and ISBN 1-898012-42-3]
2. ^ See, for example, M N O Sadiku (2007). Elements of Electromagnetics (Fourth ed.). NY/Oxford UK: Oxford University Press. pp. §9.2 pp. 386 ff. ISBN 0-19-530048-3.
3. ^ Richard Phillips Feynman, Leighton R B & Sands M L (2006). The Feynman Lectures on Physics. San Francisco: Pearson/Addison-Wesley. Vol. II, pp. 17–2, 17–3. ISBN 0-8053-9049-9.
4. ^ Griffiths, David J. (1998). Introduction to Electrodynamics (3rd ed.). Prentice Hall. pp. 516–532. ISBN 0-13-805326-X.
5. ^ Hughes, William; Young, Frederick (1966). The electromagnetodynamics of fluids. Wiley. p. 31.
6. ^ Tilley, D. E., Am. J. Phys. 36, 458 (1968)