A magnetic field directed into the page changes with time according to `B = (0.0300t^(2) + 1440)T`, where `t` is in seconds. The field has a circular cross section of radius `R = 2.50 cm`. What are the magitude and direction of the electric field at point `P_(1)` when `t = 3.00 s` and `r_(1) = 0.0200 m`?

Explanation

`e = oint E.dl = (+d phi)/(dt)`
`E = (2pir) = A.(dB)/(dt) = pi r^(2) xx (d)/(d t)(0.03t^(2)+1.4)`
`E = (pi r^(2))/(2pi r) xx (0.06 t) = (r )/(2)(0.06 t)`
`|E| = (0.02)/(2) xx 0.06 xx 3 = 18 xx 10^(-4) N//C`