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The current in an ideal, long solenoid ...

The current in an ideal, long solenoid is varied at a uniform rate of `0.01 As^(-1)`. The solenoid has 2000 turns/m and its radius 6.0 cm. (a) Consider a circle of radius 1.0 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 2.0 seconds. (b) find the electric field induced at a point on the circumference of the circle. (c) find the electric field induced at a point outside the solenoid at a distance 8.0cm from its axis.

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Verified by Experts

Here `di)/(dt)=0.01 A//s for 2s.`
`(di)/(dt)=0.02 A//s`
`n=2000turns/m,`
`R=6.0 cm =0.06m, `
`r=1cm=0.01m.`
`(a) phi=BA`
`implies (d phi) /(dt)= (mu_0 nA)(di)/(dt)`
`=4 pixx10^(-7)xx2xx10^3xxpi`
`xx1xx10^(-4)xx2xx10^(-2)`
`[A= pixx1xx10^(-4)]`
`=16 pi^2 xx10^(-10) Weber `
`=157.91xx10^(-10) Weber`
`=1.6xx10^(-5)Weber.`
`(d phi)/(dt) for 1s =0.785 Weber.`
`(b) int Edl=(d phi)/(dt)`
` implies Exx 2 pi r =(d phi)/(dt)`
` implies E=(0.785xx10^(-8))/(2 pi xx10^(-2))`
`=1.2xx10^(-7) V//m`
`(c ) (d phi)/(dt)=(mu_0)n(di)/(dt)A`
`=4pixx10^(-7)xx2000xx0.01`
`xxpixx(0.06)^2`
`E phi dl= (d phi)/(dt)`
` implies E= ((dphi)//(dt))/(2 pi r)`
`=(4pixx10^(-7)xx2000xx0.01`
`xxpixx(0.06)^2)/(pi xx 8xx10^(-2)) (dB)/(dt)`
`=5.64xx10^(-7) V//m.`
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