A static disc produces no emf between its center and circumference, whereas a rotating disc produces an emf. But as per Maxwells's equations, there should be no emf, since ${\cal E}=-d\phi/dt$, and as there is no change in area or current there, should be no change giving an emf. How we intuitively understand the reason for this emf according to Maxwell's equations?

  • $\begingroup$ Are we to assume that the axis of rotation is parallel to a magnetic field? $\endgroup$
    – R.W. Bird
    Aug 21, 2021 at 18:46
  • $\begingroup$ 17–2 Exceptions to the “flux rule and @JánLalinský's answer to this post Faraday's law with non-induced electric fields. $\endgroup$
    – Farcher
    Aug 21, 2021 at 22:10
  • $\begingroup$ Yes axis of rotation is parallel to magnetic field $\endgroup$
    – raj
    Aug 22, 2021 at 2:03
  • $\begingroup$ Even in absence of magnetic field, an emf will be generated in a rotating disc between centre and edge. This would be due to electron dirft $ \ \displaystyle m_e \omega ^2 \vec{r} - e\vec{E} = 0 \implies \vec{E} = \frac{\omega ^2 \vec{r}}{e/m_e } . \ $ This gives an emf $ \ \displaystyle \epsilon = - \frac{\omega ^2 r^2 }{e/m_e }. $ $\endgroup$ Apr 4, 2022 at 17:15

2 Answers 2


The emf induced in a conductor moving in a constant magnetic field is due to the magnetic Lorentz force acting on the charge carriers in the conductor. Thus for a charge carrier at a radius vector $\vec r$ from the centre of a disc rotating with angular velocity $\omega$ with a field $\vec B$ at right angles to its plane, $$\vec F=q\vec v \times \vec B\ = q(\vec \omega \times \vec r)\times \vec B=q\left[(\vec \omega.\vec B)\vec r - (\vec r.\vec B)\vec\omega \right]=±q\omega B\vec r.$$ (since $\vec r.\vec B=0$, and $\vec B$ is parallel or antiparallel $\vec\omega$.) The emf between the centre of the disc (radius $R$) and its outer edge is given by the 'work integral': $$\mathscr E=\frac1q \int_{r\ =\ 0}^R ±q\omega B \vec r.d\vec r=±\tfrac12\omega B R^2\ =±fBA.$$ in which $f$ is the number of revolutions per unit time and $A$ is the area of a face of the disc.

$fBA$ is the rate of cutting of flux by any radius, so we may be tempted to write $|\mathscr E| = \frac{d\Phi}{dt}$. This may be an abuse of notation, because $\frac{d\Phi}{dt}$ doesn't in this case represent a rate of change of flux. Nor does it give us insight into the origin of the emf – which is simply the magnetic Lorentz force!

Note that electromagnetic induction of this type (flux cutting by a moving conductor) arises in a different way from from the type due to varying flux density changing the flux linked with a stationary circuit. It is the latter type that is governed by the Maxwell equation $\text{curl} \vec E =-\frac {\partial \vec B}{dt}$.


Basically whenever a conductor 'cuts' magnetic field lines it is going develop emf in it. So you can consider your disc as collection infinitesimally small infinite rods which are rotating and swapping some area and then use formula for the emf across a rotating rod:- (1/2)(Br^2w) where B is magnetic field, r is radius which is the length of all the rods that make up the rotating disc and w is the angular velocity.


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