Answer:
Ein: 2.75*10^-3 N/C
Explanation:
The induced electric field can be calculated by using the following path integral:
[tex]\int E_{in} dl=-\frac{\Phi_B}{dt}[/tex]
Where:
dl: diferencial of circumference of the ring
circumference of the ring = 2πr = 2π(5.00/2)=15.70cm = 0.157 m
ФB: magnetic flux = AB (A: area of the loop = πr^2 = 1.96*10^-3 m^2)
The electric field is always parallel to the dl vector. Then you have:
[tex]E_{in}\int dl=E_{in}(2\pi r)=E_{in}(0.157m)[/tex]
Next, you take into account that the area of the ring is constant and that dB/dt = - 0.220T/s. Thus, you obtain:
[tex]E_{in}(0.157m)=-A\frac{dB}{dt}=-(1.96*10^{-3}m^2)(-0.220T/s)=4.31*10^{-4}m^2T/s\\\\E_{in}=\frac{4.31*10^{-4}m^2T/s}{0.157m}=2.75*10^{-3}\frac{N}{C}[/tex]
hence, the induced electric field is 2.75*10^-3 N/C
