For the situation described in figure, the magnetic field changes with time according to
`B=(2.00t^3-4.00t^2+0.8)t` and `r_2=2R=5.0cm`

(a) Calculate the force on an electron located at `P_2` at `t = 2.00 s`
(b) What are the magnitude and direction of the electric field at `P_1` when `t = 3.00 s` and `r_1= 0.02 m`.

Explanation

(a) `B=2t^(3)-4t^(2)+0.8`
`(dB)/(dt)=6t^(2)-8t`
`(dB)/(dt)gt0` i.e. `6t^(2)-8tgt0 implies tgt4//3sec`
For `tgt(4)/(3)s` , is increasing.
`ointvec(E).vec(dl)=(dphi)/(dt)=(d)/(dt)(B.piR^(2))`
`e.2pir^(2)=pi^(2)(dB)/(dt)`
`E=(R^(2))/(2r_(2))/(dB)/(dt)=(R^(2))/(4R)(dB)/(dt)=(R)/(4)(dB)/(dt)`
`=(5xx10^(-2))/(8)[6(2)^(2)-8(2)]`
`(5xx10^(-2))/(8)xx8=5xx10^(-2)N//C`
Electric field will be tangential to circle of radius `r_(2)` at every point
Force on electron : `F=eE=1.6xx10^(-19)xx5xx10^(-2)`
`=8xx10^(-21)N`
Since `B` is increasing, `E` will be as shown
(b) `oinyvec(E).bvec(dt)=(dphi)/(dt)=(d)/(dt)(B.pir_(1)^(2)`
`E=(r_(1))/(2)(dB)/(dt)=(r_(1))/(2)(6t^(2)-8t)`
`=(0.02)/(2)[(3)^(2)-8(3)]`
`=0.01xx30=0.3V//m`