Induced current in expanding metal ring

[arl146]

Homework Statement

A circular metal ring, as shown on the diagram below, is constructed so as to expand or contract freely. In a region with a constant magnetic field Bo oriented perpendicular to it, the ring expands, with its radius growing with time as r=r₀(1+\alphat²). As the ring expands and grows thinner, its resistance per unit length changes according to R=Ro(1+\betat²). Find the current induced in the ring as a function of time. To check your answer, suppose that B₀ = 7.30 mT, r₀ = 11.0 cm, R₀ = 3.00 m, \alpha= 0.245 × 10^(-4) s^(-2), and β = 0.500 × 10^(-2) s^(-2). What is the value of the induced current at t = 86.0 s? (Note: Give the direction of the current where when viewed from above a positive current will move counterclockwise.)

Homework Equations

I don't even know where to start ! Can you start with flux?
Flux=BA

The Attempt at a Solution

Again, no clue. help is definitely needed, so lost!

The picture doesn't show much, just the B0 points up ( guess you can just call it +y direction ) and the radius which of a circle is obvious. Help is greatly appreciated!

 

[bjd40@hotmail.com]

the induced emf may be given by e= dФ/dt, where Ф is the flux. now Ф can also can be given by Ф/area=B0. as the relation betwn r and t is given u can always find area as a function of time, hence also u can find dФ/dt and thus the expression for e ( B0 is const. e will be depending on rate change of area with respect to time). once u find e, u divide it by the instantenuous resistance (also a function of t) and get the instantenuous current. to check the answer put the given values and verify. while doing this always remember the total volume of the ring is constant, that means it becomes thinner with increasing length.

 

[arl146]

So, I have:

\frac{d\Phi}{dt}= B₀*pi*(r₀)²*(1+4\alphat+4\alpha²t³)

Is that correct?

For my final answer I got 3.01292E-6 A. I only have one try left on my homework and I'm afraid to try it. I'm not sure if the (1+..) should still be there?

Oh, and R0 is actually Ohm/m sorry, I didn't notice that until now. But for that, when I had to use the R equation, I just multiplied it by 2*pi*r (the total length)

 

[arl146]

it was -2.969845533E-08 A, thanks!