A solenoid of length 0.6 m has a radius of 2 cm and is made up of 600 turns If it carries a current of 4 A, then the magnitude of the magnetic field inside the solenoid is

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`