A solenod of length 0.4m and diametre 0.6m consists of a single layer of 1000 turns of fine wire carrying a current of `5.0xx10^-3A` . Calculate the magnetic field intensity on the axis at the middle and at the ends of the solenoid.

Explanation

`B=1/2mu_(0)ni[cos theta_(1)-cos theta_(2)]`
`rArr n=1000/0.4=2500` per meter
`i=5xx10^(-3)`
(i) `cos theta_(1)=0.2/sqrt((0.3)^(2)+sqrt(0.2)^(2))=0.2/sqrt0.13`
`cos theta_(2)=0.2/sqrt0.13`
`rArr B=1/2xx(4xxpixx10^(-7))xx2500xx5xx10^(-3)(2xx02)/sqrt(0.13)`
`=(pixx10^(-5))/sqrt13`
(ii) At the end
`costhera_(1)=2.4/sqrt((0.3)^(2)+(0.4)^(2))=0.8`
`cos theta_(2)=cos 90^(@)=0`
`B=1/2xx(4xxpixx10^(-7))xx2500xx5xx10^(-3)xx0.8`
`rArr B=2pi xx 10^(-6) Wb//m^(2)`