michaelw said:
argh so that means that they will change the magnitude of their velocity?
By Fardaday's Law, there will be an electric field induced by the time varying magnetic field that has a component aligned with the direction of motion of the charge. So yes, the magnitude of the velocity will change. Gokul43201's comment that this problem is not the same as a current carrying loop is entirely correct, but the same electric field that induces an emf in a coil affects the velocity of the charge in this problem.
There is another form of Faraday's Law that I assumed you have not yet seen. Perhaps I should not assume that. In differential form Faraday's Law does not rely on calculations of "flux". It is
\nabla \times E = - \frac{\partial B}{\partial t}
The left hand side of this equation is called the "curl" of the electric field. If we take the direction of the magnetic field in this problem to be the z direction, then its derivative with respect to time is also in the z direction and this equation reduces to
\frac{\partial E_y}{\partial x} - \frac{\partial Ex}{\partial y} = - \frac{\partial B}{\partial t}
If you have seen this before, great. If not, all you need to know about it is that it tells you that there is an electric field generated in the area of your charge because the magnetic field at every point in space is changeing with time. The equation shows that the field has "circulation", which means the Faraday path integral in the integral form of Faraday's Law is not zero. Work will be done on the charge by this electic field.
This is in contrast to your new problem, where you have a charge moving through a magnetic field that varies from one point in space to another, but is independent of time. The moving charge will experience different forces due to its motion through the magnetic field, but there will be no induced electric field that will do work on the charge. The magnetic force will always be perpendicular to the motion, so no work will be done on the charge.