a. What is the magnetic flux through one turn of a solenoid of self inductance `8.0xx10^-5H` when a current of `3.0 A` flows throgh it? Assume that the solenoid has 1000 turns are wound from wire of diameter `1.0 mm`.
b. What is the cross sectional area of the solenoid?

Explanation

Here, `L = 8.0 xx 10^(-5) H, I = 3.0 A`
N = 1000 , diameter, `d = 1.0 mm = 10^(-3) m`.
If `phi` is magnetic flux linked with each turn of sloenoid, then `N ohi = LI`
`phi = (LI)/(N) = (8 xx 10^(-5) xx 3.0)/(1000) = 2.4 xx 10^(-7) Wb`
Now, `phi = BA = (mu_(0) xx (Ni)/(l)) A` where `l = N xx d`
`:. phi = mu_(0) (ai)/(d)` or `A = (phi d)/(mu_(0) i)`
`:. A = (2.4 xx 10^(-7) xx 10^(-3))/((4 pi xx 10^(-7)) xx 3) = 6.37 xx 10^(-5) m^(2)`