# Electrodynamics for a rotating ring

1. May 11, 2014

### scotshocker

1. The problem statement, all variables and given/known data
A thin copper (resistivity 1.7 x 10-8 Ωm, density 8.9 g/cm3) ring rotates about an axis perpendicular to a uniform magnetic field B0. Its initial frequency of rotation is ω0. Calculate the time it takes the frequency to decrease to 1/e of its initial value, under the assumption that the energy goes into joule heat, if the magnitude of the field is 0.002 T.

2. Relevant equations

3. The attempt at a solution
I had found a version similar to this but unfortunately when I try to convert it into SI units the units do not work out.
Kinetic Energy of rotation:=.5Izω2
Iz=.5Mr^2 M=2πArρ Iz=πAr3ρ
r= radius A=cross section ρ=mass density
magnetic flux $\Phi$ = B0r2πcos(Θ)
Θ= angle between field and plane of the ring
rotation Θ=ωt
Lenz's law: V=$\frac{d\phi }{dt}$ r2πB0ωsin(ωt)
The rest can be found here but as I said when I convert into SI I am not ending up with the correct units( the problem in question uses different numbers): http://www.physics.fsu.edu/QualifyingExam/Spring07_solutions.pdf [Broken] Solution 10

Last edited by a moderator: May 6, 2017
2. May 12, 2014

### scotshocker

Any and all suggestions would be greatly appreciated.

3. May 16, 2014

### Hazzattack

Hi, thought I'd give your question a bash. Hopefully this helps or at the very least offers you some possible routes to experiment with.

So, we have a ring that is presumably rotating on a frictionless pivot. We can therefore say that the total energy of the system (since we are neglecting any friction terms) is given solely by the rotational energy is possesses,

E = $\frac{1}{2}$I ω(t)^2, (1)

We now apply a B-field in an orthogonal plane, since the ring is rotating there will be a change in flux. We know from Gauss' law that the flux is given by,

$\Phi$ = BA.

However, we just noted that the ring is rotating and as such the amount of flux the ring 'sees' will be oscillating between a maximum (when the surface is perpendicular to the field) and minimum values (when its parallel). Hence, the flux as a function of time is given by,

$\Phi$ = BAcos(ωt),

where A is the area of the ring $\pi$r2.

This change in flux induces a electromotive force in the wire that causes a current to move around it. The emf induced by the changing flux is given by Lenz' law,

-$\frac{\Phi}{dt}$ = $\upsilon$. (2)

Hence, we can see that power is being supplied to the ring in order to induce this current. The total power being provided by the B-field will be given by,

P= $\frac{v^2}{R}$. (3)

Due to the wire possessing a finite resistance as the the current moves around it heat will be given off (since the wire will increase in temperature). Since we assumed prior to the B-field the total energy of the system was given by the total rotational energy (1), we can conclude that any reduction in this energy is due to heat being dissipated from the ring. This naturally leads us to the conclusion that the ring will begin to slow down as the wire heats up. We first note that power may be written,

P = $\frac{dE}{dt}$,

however, we noted that the loss of energy through heat dissipation will serve to reduce the the rotational energy. Hence, the gradient will be negative,

P = -$\frac{dE}{dt}$ = -$\frac{I}{2}\frac{dω(t)}{dt}$.

This expression can be equated to (3) since it is the power being provided by the B-field that is causes the reduction in rotational energy (causing the wire to slow). Therefore, going back to equation (2), v is given by,

v = ω(t)Bπr2sin(ωt),

squaring this gives,

v2 = ω(t)2B2π2r4sin2(ωt).

We take the average of sin2(ωt) = $\frac{1}{2}$. We therefore have a simple case of separation of variables by equating the powers,

$\frac{1}{2}$$\frac{ω(t)^2 B^2 π^2 r^4}{R}$ =$\frac{I}{2}$ $\frac{dω(t)^2}{dt}$.

We now move variables to the respective sides and obtain the following,

$\frac{dω(t)^2}{ω(t)^2}$ = -$\frac{B^2 π^2 r^4 dt}{IR}$.

Integrate both sides, where ω is from ω0 to ωf and t is from 0 to t. This yields,

ln[ωf2] - ln[ω02] = -$\frac{B^2 π^2 r^4 t}{IR}$ (4),

of course, ln[ωf] - ln[ω0], can be re-written in terms of a fraction. We then turn to the original statement that when the final angular frequency is 1/e of the original frequency, that is,

ωf = $\frac{ω0}{e}$ => $\frac{ωf}{ω0}$ = $\frac{1}{e}$.

You can then square both sides and log it and we have the same expression given by (4) (once you re-write the log). I'll stop here and let you do the rest but you should find that the - signs cancel due to the log of $\frac{1}{e^2}$ being -2. You can also sub for everything else, since you are given the resistivity and density of the wire.

Hope this helped,

Harry.

NB: apologies for any sloppy notation, I was struggling with the syntax (in particular the fractions kept messing up) so in the end I scrapped subscripts.