Answer:
Induced EMF,

Step-by-step explanation:
Given that,
Radius of the circular loop, r = 1 m
Time, t = 2 s
Initial magnetic field,

Final magnetic field,

The expression for the induced emf within the circular loop is given by :

= magnetic flux

Here,





So, the induced emf in the loop is 6.28 volts. Hence, this is the required solution.