(a) Find magnetic field inside a long solenoid, carrying current `(1//pi)` `A` and having number of turns per unit length `10`. Also calculate magnetic field at ends.
(b) A long solenoid is fabricated by closely winding a wire of diameter `10 mm` over a cylinderical non-magnetic frame so that the successive turns nearly touch each other. Find magnetic field at the centre and ends of solenoid if it carriers a current `(4//pi)` `A`?
(c) A single-layer coil (solenoid) has length `l` and cross-section radius `R`. The number of turns per unit length is equal to `n`. Find the magnetic induction at the centre of the coil when a current `i` flows through it.
(d) A solenoid of length `0.4 m` and diameter `0.6 m` consists of a single layer of `1000` turns of the fine wire carrying a current of `5 mA`. Calculate the magnetic field on the axis at the middle and at the ends of the solenoid.

Explanation

(a) `i=1/piA, n=10`
`B_(centre)=mu_(0)ni=(4pixx10^(-7))(10)(1/pi)=4xx10^(-6) T`
`B_(end)=1/2 mu_(0) ni=2xx10^(-6) T`
(b)
Number of turns in `10 mm` length `=1`
Number of turns in `1m` length `n=100`
`i=4/(pi) A`
`B_(centre)=mu_(0)ni=(4pixx10^(-7))(100)(4/pi)`
`=16xx10^(-5)T`
`B_(end)=1/2 mu_(0) ni =8xx10^(-5) T`
(c)
`cos theta_(1)=cos theta_(2)=(l//2)/(sqrt(R^(2)+(l//2)^(2)))=l/(sqrt(l^(2)+4R^(2)))`
`B_(C)=1/2mu_(0)ni(cos theta_(1)+cos theta_(2))`
`=1/2 mu_(0)nixx(2l)/(sqrt(l^(2)+4R^(2)))`
`=(mu_(0)nil)/(sqrt(l^(2)+4R^(2)))=(mu_(0)ni)/(sqrt(l+(2R//l)^(2)))`
If `l to oo, B_(C)=mu_(0)ni`
(d) `i=5mA=5xx10^(-3) A, m=N//l=1000/0.4=2500`

`cos theta_(1)=0.2/(sqrt((0.2)^(2)+(0.3)^(2)))=0.2/(sqrt(0.13))=2/(sqrt(13))`
`B_(C)=1/2mu_(0)ni(cos theta_(1)+cos theta_(2))`
`=1/2(4pixx10^(-7))(2500)(5xx10^(-3))xx2xx2/(sqrt(13))`
`=(pixx10^(-5))/(sqrt(13)) T`
At ends

`cos theta=0.4/0.5=4/5=0.8`
`B_(A)=1/2mu_(0)ni(cos 90^(@)+cos theta)`
`=1/2(4pixx10^(-7))(2500)(5xx10^(-3))xx(0.8)`
`=2pixx10^(-6) T`