On Lenz's law, Wikipedia says:

Faraday's law states that the EMF is also given by the rate of change of the magnetic flux where epsilon is the electromotive force (EMF) and $\phi_B$ is the magnetic flux.

$$\varepsilon = - \frac{-d \phi_B}{dt}$$

But is there also an influence of the size/shape of the receiving circuit on the EMF produced? I'm trying to recreate this at home using aluminium powder suspended in water but not seeing any movement, and many of the things I read on the internet seem to suggest that a larger conductor will work better, which would explain why I can't get this to work?


1 Answer 1


(a) "But is there also an influence of the size/shape of the receiving circuit on the EMF produced?"

$\Phi_B$ takes account of the size of the loop. For example, if the magnetic flux density has magnitude $B$ all over the loop, and is directed at angle $\theta$ to the loop, then $$\Phi_B=BA\cos \theta$$ in which $A$ is the area of the loop. Clearly a given change in $B$ will produce a larger change in $\Phi_B$ in a loop of larger area.

(b) You were expecting to see moment of the aluminium powder due to induced currents in the individual grains.

I believe that the grains were too small for this to happen. Here is an attempt to show this by the method of dimensions. Assume that the current is proportional to $\frac{dB}{dt} =\dot B$, and to unknown powers ($\alpha$ and $\beta$) of the resistivity, $\rho$, and radius, $r$. Thus $$I=\dot B\rho^\alpha r^\beta$$ Equating SI units: $$\text A=\text {T s}^{-1} (\Omega\ \text m)^\alpha\ \text m^\beta$$

Working towards expressing in SI base units:

$$\text A=\text {N A}^{-1}\text m^{-1} \text s^{-1}(\text{V A}^{-1} \text m)^\alpha\ (\text m)^\beta $$ So $$\text A=\text {N A}^{-1}\text m^{-1} \text s^{-1}(\text{N m s}^{-1} \text A^{-2} \text m)^\alpha\ (\text m)^\beta $$ So $$\text A=\text {kg m s}^{-2} \text{A}^{-1}\text m^{-1} \text s^{-1}(\text{kg m s}^{-2} \text{m s}^{-1} \text A^{-2} \text m)^\alpha\ (\text m)^\beta $$ Equating powers of $\text{kg}$: 0 = 1 +$\alpha$, so $\alpha$ = –1

Equating powers of $\text m$: 0 = 1 +3$\alpha$ + $\beta$, so $\beta$ = 2

Equating powers of $\text s$: 0 = –3 –3$\alpha$, so $\alpha$ = –1

Equating powers of $\text A$: 1 = –1 –2$\alpha$, so $\alpha$ = –1

We see that the current is proportional to the square of the radius for a given rate of change of flux density. Therefore the currents in very small spheres will be very small indeed, and no doubt too small for the spheres to experience significant magnetic forces – which, incidentally, they would only do if the field were non-uniform.

  • $\begingroup$ I put a small amount of ultrafine Al powder (usually used as filler in casting resins) in a small test tube of water, swirled it around so its cloudy, and put a spinning halbach array up against the outside of the tube - I expected it to clear the cloudiness close to that side of the tube? $\endgroup$ Oct 1, 2021 at 8:05
  • $\begingroup$ So you produced a changing magnetic field within the test tube and this would produce emfs within loops in the water. But I wouldn't expect any effects to be seen unless there were currents ('eddy currents') in these loops. There would be currents only if the water were electrically conducting, and the presence of suspended aluminium powder would not make it conduct well enough. That, I think, is why you saw no effects. $\endgroup$ Oct 1, 2021 at 10:57
  • $\begingroup$ Tbh i hadnt thought it would have much effect on the water at all - i was hoping it would exert a repulsive force on the al dust in the water, and this would manifest as the water becoming clearer near the magnets as the al was pushed towards the other side? $\endgroup$ Oct 1, 2021 at 18:52
  • $\begingroup$ I got the idea from videos of levitating quadcopters, which typically use such spinning Halbach arrays above a copper or aluminium plate - I get the impression it only works though if this plate is quite thick - using them on a really thin sheet sounds like it doesnt work as well - I want to know if that's true - if the thickness of the conductor being repulsed matters? As in that case maybe to make my aluminium dust move about in the water I'm going to need much stronger magnets? $\endgroup$ Oct 1, 2021 at 19:06
  • $\begingroup$ If I've understood your set-up aright, you are expecting to induce eddy currents in water. So my original comment stands: the resistivity of the water is too high for there to be significant currents. Without the currents you won't get the forces you are looking for. $\endgroup$ Oct 1, 2021 at 20:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.