# Homopolar generator EMF

1. Feb 26, 2014

### fayled

I have proven the induced emf between the centre and rim in a circular disc of radius a and angular velocity w, with a magnetic field B parallel to its axis is 0.5wa2B.

I need to find the time taken for the disc to slow to half it's initial speed ignoring friction, given a resistance R is connected between centre and rim and all other circuit resistance is negligible. It has mass m.

So I know it's moment of inertia is 0.5ma2, and that energy is conserved such that the difference in the rotational KE initially and finally equals the energy dissipated in the resistance. This energy E is
E=0.5Iw2-0.5I(0.5w)2
E=0.5Iw2-0.125Iw2
E=(3/8)(0.5ma2)w2
E=(3/16)ma2w2

Now comes the problem in calculating the energy dissipated in the resistance. Obviously the induced emf and so induced current are time dependent. I'm not sure how I can get expressions for these in terms of time such that I could get the power dissipation with time and integrate for the energy. Any clues please? Thanks :)

2. Feb 26, 2014

### TSny

Hello.

How would you express the rate at which energy is "consumed" in the resistor in terms of the emf ε and the resistance R?

3. Feb 26, 2014

### fayled

dE/dt=ε2/R

Assuming this is correct, I can't see how to proceed to get the energy dissipated from there, because I don't know the dependence of ε with time.

4. Feb 26, 2014

### TSny

OK. Can you also find an expression for the rate at which rotational energy of the disk is changing?

5. Feb 26, 2014

### BvU

Hello Fayled,
As you indicate, things are time dependent. So ω is a function of time. If you take small steps in time your difference becomes a differential and you end up with a simple differential equation.

If you are comfortable with that, go ahead. If not, write ΔKE = ... Δt with a function of ω on the dots. Also write ΔKE = ... Δω with a function of ω on the dots.

And: for our sake, also use the template after your first post.

6. Feb 26, 2014

### fayled

dErot/dt=0.5ma2w(dw/dt)?

7. Feb 26, 2014

Yes.