Induced Voltage in a Radial Magnetic Field We have a radial magnetic field, say
$$B= \frac{k}{\sqrt{x^2+y^2}}(\cos{\theta}\cdot\hat{i}+\sin\theta\cdot\hat{j})$$
and we have a conducting ring, with its axis along the $z$ axis. What happens if we give this ring a velocity in the $z$ direction?
From Faraday's Law, there is no change in flux, and hence there should be no induced EMF. But on each element, there is a  magnetic Lorentz Force acting on the electrons along the ring, so they should move, and there should be a current which implies that there is an EMF.
 A: 
From Faraday's Law, there is no change in flux

The magnetic field you propose violates Maxwell's equation
$$
\nabla \cdot \mathbf B = 0
$$
on the line $(x=0,y=0)$, which means flux cannot be uniquely assigned to the ring, only to some specific choice of surface attached to the ring. If that surface is plane disk, then flux is zero, but if that surface is a long cylinder hat, then flux is not zero and is increasing as the ring moves along $z$.

and hence there should be no induced EMF.

Correct, but for different reason: induced EMF in general is due to induced electric field, which is not present in this case at all (vanishes everywhere). What you meant to say is that due to zero change in flux, there should be no EMF at all . That is not true (because we can't say flux is not changing, because there isn't unique way to assign flux). In the hypothetical case the magnetic field was as you proposed (which would be a major discovery contradicting standard EM theory), there would be motional EMF
$$
\oint_{ring} (\mathbf v \times \mathbf B) \cdot d\mathbf l \neq 0
$$
and Faraday's law wouldn't be obeyed (because there would be no unique flux).
