Question

Which integral gives the axial field after summing the contributions over the full length of the solenoid?

पर्याय

• \[B=\frac{\mu_0 nIa^2}{2}\int_{-l}^{+l}\frac{dx}{\left[(r-x)^2+a^2\right]^{3/2}}\]

• \[B=\frac{\mu_0 nIa^2}{2}\int_{-l}^{+l}\frac{dx}{\left[r^2+a^2\right]^{3/2}}\]

• \[B=\frac{\mu_0 nIa^2}{2}\int_{0}^{+l}\frac{dx}{\left[(r-x)^2+a^2\right]^{3/2}}\]

• \[B=\frac{\mu_0 nIa^2}{2}\int_{-l}^{+l}\frac{dx}{\left[(r-x)^2+a^2\right]}\]

Answer 1

Because the solenoid is centred at \[O\] and has total length \[2l\], integration runs from \[x=-l\] to \[x=+l\]. The integrand retains the axial distance \[r-x\] for each small element.