Calculating a Homopolar Generator's EMF

[ElPimiento]

Homework Statement

Homework Equations

$\omega = v/r$
$q\vec v \times \vec B\ =\ \vec F_B$
$|\vec F|d\ =\ |\vec E|$
$q \Delta V\ =\ E$

The Attempt at a Solution

(So, my answer is twice and large as the correct answer. This is the second time this has happened on a question like this . . .)
Since a negative charge would collect on the outside of the disk I began by analyzing one negative charge and treated it in equilibrium on the edge of the disk,
$$\begin{align}
q\vec v \times \vec B- \frac{|\vec E|}{d}\ &=\ 0 \\
qvB \ &=\ \frac{q \Delta V}{r} \\
vBr \ &=\ \Delta V
\end{align}$$
And since $v\ =\ \omega r$:
$$\begin{align}
(\omega r)Br \ &=\ \Delta V \\
\Delta V \ =\ \epsilon \ &\approx\ 48.3\ V
\end{align}$$
Any help would be wonderful c:

 

[rude man]

Start with the differential form for emf:
d(emf) = Bv(r)dr
and show off your powers of integration! :-)

 

[ElPimiento]

I'm back tracking the work a little to make sure i understand where your getting that form of emf:
$\begin{align} F_B - F_E \ &=\ 0 \\ F_B \ &=\ F_E\ =\ q|\vec E|\ =\ q\frac{d(\epsilon)}{dr} \\ vB\ =\ \omega r B\ &=\ \frac{d(emf)}{dr} \\ \int (\omega r B)dr\ &=\ \epsilon \\ \frac{\omega B r^2}{2} \ &=\ \epsilon\ \approx 24.1\ V \end{align}$
thanks so much!

 

[rude man]

I'm back tracking the work a little to make sure i understand where your getting that form of emf:
$\begin{align} F_B - F_E \ &=\ 0 \\ F_B \ &=\ F_E\ =\ q|\vec E|\ =\ q\frac{d(\epsilon)}{dr} \\ vB\ =\ \omega r B\ &=\ \frac{d(emf)}{dr} \\ \int (\omega r B)dr\ &=\ \epsilon \\ \frac{\omega B r^2}{2} \ &=\ \epsilon\ \approx 24.1\ V \end{align}$
thanks so much!

Looking good!

d(emf) = B(r) v(r) dr is just the generalized (differential) form of emf = Bvr. Just basic calculus. It allows B and/or v to be functions of r. If B and v are not functions of r then it's just Bvr. But in your case B = constant but v = v(r) so then
d(emf) = B v(r) dr = B ωr dr
and when integrated over r=0 to r= R you get emf = ωBR²/2
where R is the disk's radius.