Answer:
[tex]E = 2 \times 10^{-3} V[/tex]
Explanation:
As we know that rate of change in flux will induce EMF
So here we can
[tex]EMF = \frac{d\phi}{dt}[/tex]
now we have
[tex]EMF = \pi r^2\frac{dB}{dt}[/tex]
now we also know that induced EMF is given by
[tex]\int E. dL = \pi r^2\frac{dB}[dt}[/tex]
[tex]E (2\pi r) = \pi r^2\frac{dB}{dt}[/tex]
[tex]E = \frac{r}{2}(\frac{dB}{dt})[/tex]
now plug in all values in it
[tex]E = \frac{0.0133}{2}(0.12 t)[/tex]
[tex]E = 8 \times 10^{-4} (2.50) = 2 \times 10^{-3} V/m[/tex]
