Explanation

(a) Initial magnetic flux linked with the loop
$\phi = B_{1}A_{1}\cos\phi$
$\ \ \ \ = 0.1 \times \left( 10 \times 10^{- 2} \right)^{2}\cos{45{^\circ}}$
$\ \ \ \ = \ \frac{0.1 \times 10^{- 2} \times 1}{\sqrt{2}} = \frac{10^{- 3}}{\sqrt{2}}$
Final magnetic flux linked with the loop,$\phi_{2} = 2$
Now, induced emf in the loop ,$e = \frac{- \ d\phi}{dt}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \ \frac{- \left\lbrack \frac{10^{- 3}}{\sqrt{2}} \right\rbrack}{0.7} = 10^{- 3}V$
$\therefore$ Induced current =$\frac{e}{R} = \frac{10^{- 3}}{1} = 1\ mA$