Failling to understand why a rotating disk inside an uniform magnetic field generates a potential difference

Question

Suppose we have a conducting disk of radius $R$, rotating about its axis, with rotational velocity $\omega$. If surrounding the disk is a constant magnetic field parallel to its normal vector, a non-zero voltage appears between the center and the border of the disk.

I don't understand how this is possible: given that $\mathcal{E}=-\frac d {dt}\iint_{\Sigma} B\cdot dS$, and since (taking $\Sigma$= rotating disk)

$$ \frac d {dt}\iint_{\Sigma} B\cdot dS=\frac d{dt}(\|B\|\ A) = 0 $$

Where $A=\pi R^2$ is the area of the disk. It should follow that $\mathcal E = 0$ (which is the same as $V_{\text {border}}-V_{\text {center}}$, right? And so $V=0$.

I believe the answer should be $V= \|B\| \ A\ f$, with $f=\frac \omega {2\pi}$, but I have no idea where this comes from, or why what I did yields an incorrect answer.

Also, could anyone illustrate the differences between $\mathcal E$, the induced EMF, and $V$ here? They always seem to be the same thing...

Answer 1

What bjorne said is right but incomplete. The effect is not due to the flux.

Due to Lorentz force electrons which are moving at a distance $r$ from the center with tangential velocity $v=r\omega $ (due to rotation of the disk) experience a radial force (do the vector product to see this) $$F_L=er\omega B $$ where B is the field and $e $ the electrons charge.

This leads to electrons moving towards the periphery and creates a charge discontinuity (also because they leave their nuceli behind) which generates a radial electric field (it is radial because it is due to a radial gradient of charge) $E(r)={dV\over dr} $ and thus a voltage $V (r) $. The fact that $E={dV\over dr}$ is due to the symmetry of the problem: all other components of the gradient vanish as the field only variates in the radial component.

In the steady state the electric ($F_E=eE=e{dV\over dr} $) and Lorentz forces cancel out as nothing should move at equilibrium.

So $$er\omega B=e{dV\over dr} $$ and this gives us an equation for the voltage which when integrated leads to: $$V(r)=B\omega {r^2 \over 2} $$ with respect to the center of the disk where $V=0$.

Between the edge of the disk the voltage is $$V (R)=B\omega { R^2 \over 2}$$

which is the same formula you propose.

EDIT: Notice that in this answer I have been neglecting centrifugal force $F_C=m_e\omega^2 r$ acting on the electrons. We can add it to show that the effect, the fact that a voltage arises, is not a purely magnetical effect.

Once again the trick is saying that in the steady state nothing moves, so the total force must be zero. That means: $$e{dV\over dr} =er\omega B+m_e\omega^2 r$$ or rather $${dV\over dr} =\left(B+{m_e\over e}\omega\right)\omega r$$ which when integrated leads to $$V(r)=\left(B+{m_e\over e}\omega\right)\omega {r^2\over 2}$$

Thus even if $B=0$ we still get a voltage $V(R)={m_e\over e}\omega^2{R^2\over 2}$ across the disk.

Yet notice that the effect is small as for an electron ${m_e\over e}\sim 10^{-11}$ (which by the way is in principle measurable from this device).

Answer 2

I think the formula you're using is $\mathcal{E}=-d\Phi/dt$ where $\Phi$ is the magnetic flux through your conducting loop, or in your case, conducting material. Without the algebra you can tell that the disk is always in the same magnetic field, so the flux through it is unchanging. The motional emf is then indeed zero. The force must be coming from somewhere else - not Faraday's law of induction.

The part of electromagnetism that explains this phenomenon is the Lorentz force. A charge moving with velocity $v$ in a magnetic field $B$ is acted upon by a force per unit charge of $v\times B$. From the set up, $v$ is always tangential to the disk and of magnitude $\omega r$, while $B$ is always vertical, so the force will always be radial. The conducting disk has positively charged nucleii that are stuck in the lattice, but the electrons subject to the force will move and induce a current (like in a homopolar dynamo), or "pile up" with a voltage between different radii of the disk.

A steady state will quickly be reached where the Lorentz force is balanced by the electric field, which allows us to calculate the voltage between the centre and a point of radius $R$ (including the edge),

$$ V(R) = -\int_0^R E \cdot dl = \int_0^R B\omega r\ dr = B\omega R^2/2 $$

Answer 3

I think that in the original question there is a confusion between e.m.f. $\mathcal{E}$ and (electrostatic) Voltage $V$. As others have pointed out, if enough time passes a static radial electric field $\vec{\mathbf E}$ will appear due to Lorentz forces acting on the mobile conducting electrons, caused by a charge separation in the disk. In this limit we have an electrostatic situation, the field $\vec{\mathbf E}$ can be considered the gradient of a potencial and the potencial difference is as shown above.

However, if a circuit is closed by connecting a point $P$ in the perifery of the disk to its conducting axis via a wire with resistance $R$, then the radius $OP$ can be seen as a moving segment of the closed circuit, and as it moves it changes the circuit area at a rate $\frac{d A}{d t}=\frac{1}{2}\omega\, R^2$ (assuming everything else remains in place) and thus the flux rate is indeed $$\mathcal{E}=-B\, \frac{d A}{d t}=-\frac{1}{2}\omega\, R^2B.$$ This e.m.f will induce a current $i= \mathcal{E}/R$ in the circuit (assume the disk is a perfect conductor) but we cannot associate an electrostatic field $\vec{\mathbf{E}}$ to each point in the disk (there is no charge separation now), although a tension can de defined between $P$ and $O$ as $V_{PO}=-\mathcal{E}$ by integration of the stationary electric field in the wire.