Respuesta :
Answer:
Induced emf in the loop is 0.0603 volts.
Explanation:
It is given that,
Radius of the circular loop, r = 1.6 cm = 0.016 m
Magnetic field, B = 0.8 T
When released, the radius of the loop starts to shrink at an instantaneous rate of 75.0 cm/s, [tex]\dfrac{dr}{dt}=75\ cm/s=0.75\ m/s[/tex]
We need to find the magnitude of induced emf at that instant. Induced emf is given by :
[tex]\epsilon=\dfrac{-d\phi}{dt}[/tex]
Where
[tex]\phi[/tex] is the magnetic flux, [tex]\phi=B\times A[/tex]
[tex]\epsilon=\dfrac{-d(BA)}{dt}[/tex], A is the area of cross section
[tex]\epsilon=-\dfrac{-d(B(\pi r^2))}{dt}[/tex]
[tex]\epsilon=-2\pi r B(\dfrac{dr}{dt})[/tex]
[tex]\epsilon=-2\pi \times 0.016\times 0.8 \times 0.75\ m/s[/tex]
[tex]\epsilon=0.0603\ V[/tex]
So, the induced emf in the loop is 0.0603 volts. Hence, this is the required solution.
