Answer:
Induced emf, [tex]\epsilon=0.584\ V[/tex]
Explanation:
It is given that,
Radius of the circular loop, r = 10.9 cm = 0.109 m
Magnetic field, B = 0.797 T
When released, the radius of the loop starts to shrink at an instantaneous rate of 107 cm/s, [tex]\dfrac{dr}{dt}=107\ cm/s=1.07\ m/s[/tex]
Due to change in loop of the loop, an emf is induced in the loop. It is given by :
[tex]\epsilon=\dfrac{d\phi}{dt}[/tex]
[tex]\phi[/tex] = magnetic flux
[tex]\epsilon=\dfrac{d(BA)}{dt}[/tex]
[tex]\epsilon=B\dfrac{d(\pi r^2)}{dt}[/tex]
[tex]\epsilon=2\pi rB\dfrac{dr}{dt}[/tex]
[tex]\epsilon=2\pi \times 0.109 \times 0.797\times 1.07[/tex]
[tex]\epsilon=0.584\ V[/tex]
So, the emf is induced in the loop at that instant is 0.584 volts. Hence, this is the required solution.