Calculate the direction and the magnitude of the induced current

An elastic circular conducting loop expands at a constant rate over time such that its radius is given by r(t) = r0 + vt, where r0 = 0.290 m and v = 0.0230 m/s. The loop has a constant resistance of R = 11.7 Ω and is placed in a uniform magnetic field of magnitude B0 = 1.95 T, perpendicular to the plane of the loop, as shown in the figure. Calculate the direction and the magnitude of the induced current, i, at t = 4.69 s. (Use positive numbers for a clockwise current and negative numbers for a counterclockwise current.)

http://img52.imageshack.us/i/p032figure.png/

First lets find out the direction of the induced current.
The magnetic field is upwards. As the radius of the loop is increasing, and so is its area, the magnetic flux in upward direction is increasing.
To counter this, a current will be induced so as to oppose this flux.
So the direction of the current will be such that the magnetic field is downwards, hence, the current will be clockwise when viewed from above.

Since we know the radius as a function of time, we need the area of the loop as a function of time.
A = πr²
A = π(rₒ + vt)²

By Faraday's law,
E = dɸ/dt
E = d(AB) / dt = B (dA / dt) (because B is constant)

dA / dt = d(πr²) / dt = 2πr (dr/dt) = 2π(rₒ + vt)(v) (because dr/dt = v)
At t = 4.69
dA / dt = 0.057 m²/sec

E = 1.95 x 0.057 = 0.11 V
I = V/R = 0.11 / 11.7 = 9.5 mA (Clockwise)