The cruuent in an ideal, long solenoid is varied at a uniform rate of `0.02A//s`. The solenoid has 1000 turns/s and its radius is 8 cm.
(i) Consider a circle of radius 2 cm inside the solenoid with its axis coinciding with the axis of the solenoid. Write the change in the magnetic flux through this circle in 4 s.
(ii) Find the electric field induced at a point on the circumference of the circle.
(iii) Find the electric field induced at a point outside the solenoid at a distance 9 cm from its axis.

Explanation

Given, for solenoid, n = 1000 m/s
`and" "(di)/(dt)=0.02A//s`
`rArr" "di=0.02dt`….(i)
(i)Magnetic field due to solenoid is given as
`B=mu_(0)ni`
Flux through circle of radius `r,phi_(B)=Bpir^(2)`
`" "(becausephi_(B)=B.S)`
`=mu_(0)nipir^(2)`
`rArr" "dphi_(B)=mu_(0)npir^(2)i`
`=mu_(0)npir^(2)(0.02)dt" "[because"using Eq.(i)"]`
`rArr" "phi=mu_(0)npir^(2)(0.02)int_(0)^(4)dt`
`=4pixx10^(-7)xx1000xxpi(0.02)^(2)(0.02xx4)`
`1.26xx10^(-7)Wb`
(ii) Electrcic field induced is related to emf as, `ointE.dl=(dphi)/(dt)`
`rArr" "E.2pir=mu_(0)npir^(2)(di)/(dt)`
`E=(mu_(0)nr)/(2)(di)/(dt)`
`=(4pixx10^(-7)xx1000xx0.02)/(2)xx0.02`
`-8pixx10^(-8)V//m`
(iii)

As discussed in (i), similarly for radius outside the solenoid,
`ointE.dl=(d)/(dt)(mu_(0)nipiR^(2))`
`rArrE.2pir=mu_(0)n(di)/(dt)piR^(2)rArrE=(mu_(0)nR^(2))/(2r)(di)/(dt)`
`=(4pixx10^(-7)xx1000xx(0.08)^(2)xx0.02)/(2xx0.09)`
`=28pixx10^(-8)V//m`