Induced emf in a circular conducting wheel

Consider a conducting wheel with $N \in \mathbb{N}$ spokes which is completely in a homogenous magnetic field $\vec{B}$ perpendicular to the wheel plane. Then the Lorenz force on a charge $q$ on a spoke at distance $r$ from the axis is $F(r) = qvB = q\omega rB$. Thus the electromotoric force is

$$\mathcal{EMF} = \frac{1}{q} \int_0^R F(r) \, \mathrm{d}r = \omega B \int_0^R r \, \mathrm{d}r = \frac{1}{2} \omega BR^2$$

However the flux at time $t$ is

$$\Phi(t) = \int_A \vec{B} \,\mathrm{d}\vec{A} = BA$$

Since the field is homogenous, where the area enlosed by the circle is denoted by $A$.

Thus $\Phi$ is independent of $t$ and $\dot \Phi = 0$ which would mean that the induced $\mathcal{EMF}$ is zero.

How can one resolve this (apparent) contradiction in detail? Especially the second method should work since it is more general than using the Lorentz force.

• isn't this the same problem as the homopolar dynamo en.wikipedia.org/wiki/Homopolar_generator ? – hyportnex Oct 1 '14 at 18:26
• The Flux Method is for a loop. You're looking at the EMF across a spoke, which is not a loop. What you've found via the flux method is that the EMF along two spokes and an arc is zero, which it should be as the spokes contribute opposing EMF there is no EMF across the arc. – Michael Oct 1 '14 at 18:50
• @Michael: Why not make that an answer? – DanielSank Oct 1 '14 at 19:35
• @DanielSank I felt a full answer should involve a derivation of the flux law but I didn't have time. – Michael Oct 2 '14 at 13:40