Question

Derive an expression of magnetic field due to long current carrying solenoid?

Answer 1

Consider a solenoid 2L,radius a and having n turns per unit length,carrying a current I.

Considering a circular elementl section of thickness dx which is at a distance x from centre O.

$dB=\frac{{\mu }_{0}dxnI{a}^2}{2[{(r-x)}^2+a^2{]}^{\frac{3}{2}}}$

Now integrating the above equation from limit x = -l to x=l

$B=\frac{{\mu }_{0}nI{a}^2}{2}{\int }_{-1}^{+1}\frac{dx}{2{\left[\right({r-x)}^2+a^2]}^{\frac{3}{2}}}$

Considering far axial field of the solenoid , so r>>a &>>a&r>>1,

We get

$[{{(r-x)}^2+a^2]}^{\frac{3}{2}}\approx r^2$

Thus,

$B=\frac{{{\mu }_{0}nIa}^2}{2r^3}{\int }_{-l}^{l}dxB=\frac{{{\mu }_{0}nIa}^2}{2r^3}2l=\frac{{\mu }_{0}nI{a}^{2}l}{r^3}$