Question

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 solemoid has 1000 turns ans is wound from wire of diameter `1.0 mm`.
b. What is the cross sectional area of the solenoid?

Answer 1

Given `L=8.xx10^-5H, i=3.0A` and `N=1000 turns`
a. From the relation `L=(Nphi)/i`
The flux with one turn
`phi=(Li)/N=((8.0xx10^-5)(3.0))/1000`
`=2.4xx10^-7Wb`
b. this `phi=BS=(mu_0ni)(S)`

Here `n=`number of turns per unit length
`N/l=N/(Nd)=1/d`
`:. phi=(mu_0iS)/d`
`or S=(phid)/(mu_0i) =((2.4xx10^-7)(1.0xx10^-3))/((4pixx10^I-7)(3.0))`
`=6.37xx10^-5m^2`