Answer:
[tex]6.3\cdot 10^{-4} V[/tex]
Explanation:
According to Faraday-Newmann-Lenz, the induced emf in the loop is given by:
[tex]\epsilon=-\frac{\Delta \Phi}{\Delta t}[/tex] (1)
where [tex]\frac{\Delta \Phi}{\Delta t}[/tex] is the rate of variation of the magnetic flux through the loop.
We know that the magnetic flux through the loop is given by
[tex]\Phi = BA[/tex]
where B is the magnetic field and A is the area of the loop. Since the magnetic field is constant, we can write the variation of flux as
[tex]\Delta \Phi = B \Delta A[/tex]
So eq.(1) becomes
[tex]\epsilon=-B\frac{\Delta A}{\Delta t}[/tex]
and the problem gives us:
[tex]B=0.50 T[/tex] is the magnetic field
[tex]\frac{\Delta \Phi}{\Delta t}=-1.26\cdot 10^{-3} m^2/s[/tex] is the rate at which the area changes
Substituting into the equation, we find
[tex]\epsilon=-(0.50 T)(-1.26\cdot 10^{-3} m^2/s)=6.3\cdot 10^{-4} V=0.63 mV[/tex]
