A solenoid of length `0.4 m` and diameter `0.6 m` consists of a single layer of `1000` turns of fine wire carrying a current of `5.0xx10^(-3)` ampere.The magnetic field strength on the axis at the middle is `pi/Nxx10^(-5)T`, then find value of `N`.

Options

A. `2.775pixx10^-7T` `,`
`2pixx10^-7T`
B. `2.775pixx10^-4T` `,`
`2pixx10^-6T`
C. `2.775pixx10^-6T` `,`
`2pixx10^-6T`
D. `2.775pixx10^-6T` `,`
`2pixx10^-6T`

Answer

C

Explanation

the solenoid is shown in Fig
The magnetic field at the middle point P on the axis of a current
carrying solenoide is given by
`B=(mu_0 ni)/2(cos alpha_1-cos alpha_2)`
where n=number of turns`//`meter and i=current in the solenoid
Here `n=1000/0.4=2500` per metre and `i =5.0xx10^-3A`
`cos alpha_2=cos(180-alpha_1)=-cos alpha_1`
`:. B=(mu_0 ni)/2 2cos alpha_1`
`cos alpha_1=0.2/(sqrt((0.2)^2+(0.3)^2))=0.2/(sqrt(0.13))=0.555`
`:. B=(4pixx10^-7)xx2500xx5.0xx10^-3xx0.555`
`=2.775pixx10^-6T`
At the end of the solenoid
`B=(mu_0 ni)/2[cos alpha'-cos alpha'']`
`cos alpha'= 0.4/ (sqrt({(0.3)^2 +(0.4)^2})) = 0.4/0.5 =0.8`
`cos alpha''=cos90^@=0`
`:. B=((4pixx10^-7)(2500)(5.0xx10^-3))/2xx0.8=2pixx10^-6T`