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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`?

Text Solution

Verified by Experts

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