# Induced electric field

1. Sep 13, 2015

### erisedk

1. The problem statement, all variables and given/known data
Charge Q is uniformly distributed on the rim of a thin insulating disc of mass m, radius R, which is initially at rest and placed on a smooth horizontal surface. What will be the angular velocity of the disk if a magnetic field B, perpendicular to the plane of the disk, is switched on?

2. Relevant equations
∫Eind.dl = -dφ/dt

3. The attempt at a solution
2πR. Eind = (B.πR2)/Δt
E=BR/2Δt
QE/m = QBR/2Δtm

QE/m is acceleration.

a = Rα (α is the angular acceleration)

α = QB/2Δtm

ω = ω0 + αΔt
ω = QB/2Δtm × Δt
= QB/2m

So, I tried another approach.
2πR. Eind = (B.πR2)/Δt
E=BR/2Δt
F=QE= QBR/2Δt
τ=Iα
QBR2/2Δt = (mR2)/2 × α
From here,
α = QB/Δtm
and
ω = QB/Δtm × Δt
= QB/m
And this is the correct answer.

What is the problem with the first answer???

2. Sep 13, 2015

### Staff: Mentor

Why? You don't have a linear acceleration.
If your disk would be a wheel this would work. Then your I would be mR^2 instead of mR^2/2, that's where the factor of 2 comes from.

3. Sep 13, 2015

### erisedk

Even in pure rotation, there is a certain linear acceleration. However, I think I have an inkling to where it's going wrong. Every circle on the disk would have a different linear acceleration, because a is proportional to r. So, I can't use a = Rα. That's why it would also work for a wheel.
But, I'm only considering the outer rim of charge right? Cos that's where all the charge is and that's where the induced electric field is acting. And even in writing torque (τ=QBR2/2Δt) I'm just considering the torque on the rim of the disk, i.e., R, because there isn't any charge on the inner part.
Therefore, since I'm only using the outer radius anyway, my answer shouldn't change? I mean it would make sense if the charge was all over it.
The thing is intuitively, I get that with what I'm doing in the first answer, the answer wouldn't change had there been a ring instead of the disk. I mean, I can feel that there's something inherently faulty because it's not incorporating moment of inertia and the angular velocity achieved is bound to depend on the MOI.
So, help?

4. Sep 13, 2015

### Staff: Mentor

The force is on the rim only, but the acceleration happens everywhere on the disk.
The outer parts will get accelerated more, the inner parts will see a smaller acceleration. If you integrate "a(r) dm" over the whole disk mass you get the force F, but that is a complicated way to find a.

5. Sep 13, 2015

### erisedk

Thanks!! That's the first thing that's made sense in days! D