The angle $\theta$ made by the area vector of the coil with the magnetic field is 45°.

Initial magnetic flux is $\Phi = BA \cos \theta$

$\Phi = \large \frac{(0.1 \times 10^{-2}}{\sqrt 2} $$Wb$

Final Flux $\Phi_{min} = 0$

The change in flux is brought about in 0.70 s.

The magnitude of the induced emf is given by $\varepsilon = \large \frac{|\Delta \Phi_B|}{\Delta T} $

$\varepsilon = \large \frac{|\Delta \Phi_B|}{\Delta T} $$ = \large \frac{|\Delta \Phi - 0|}{\Delta T} $

$\varepsilon = \large\frac{10^{-3}}{\sqrt 2 \times 0.7}$$ = 1.0 mV$

Magnitude of Current $I = \large\frac{\varepsilon}{R}$

Therefore, Magnitude of Current $I = \large\frac{1.0mV}{0.5 \Omega} $$ = 2.0 mA$