= Solution
<Faraday's law> is
$$
\mathcal E=\oint_C\mathbf E\cdot d\mathbf l
=-\frac{d\Phi_B}{dt},
\qquad
\Phi_B=\int_S\mathbf B\cdot d\mathbf S.
$$
Choose the positive <circulation> to be anticlockwise as viewed from $+z$. Since $\Phi_B=\pi Br^2$,
$$
\boxed{I=\frac{\mathcal E}{R}
=-\frac{2\pi Br\dot r}{R}}.
$$
If $\dot r>0$, the current is clockwise and its field points in the $-z$ direction, reducing the field inside the loop. Since $d\mathbf F=I\,d\mathbf l\times\mathbf B$, the force due to the imposed field is radially inward. If $\dot r<0$, the current is anticlockwise, its field points in the $+z$ direction and increases the field inside, while the force is radially outward. In both cases the magnetic force opposes the radial motion, in accordance with <Lenz's law>.
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