Question

A tightly-wound solenoid of radius a and length l has n turns per unit length. It carries an electric current i. Consider a length dx of the solenoid at a distance x from one end. This contains n dx turns and may be approximated as a circular current i n dx. (a) Write the magnetic field at the centre of the solenoid due to this circular current. Integrate this expression under proper limits to find the magnetic field at the centre of the solenoid. (b) verify that if l >> a, the field tends to B = µ₀ni and if a >> l, the field tends to $B=\frac{{\mu }_{0}nil}{2a}$. Interpret these results.

Answer 1

(a) Given:
Current in the loop or circular current = indx
Radius of the loop having circular current = r
Distance of the centre of the solenoid from the circular current = $\frac{l}{2}-x$
Magnetic field at the centre due to the circular loop,
$B=\frac{{\mu }_0}{2}\frac{i{r}^2}{(x^2+r^2{)}^{3/2}}$
$$\begin{gathered} \mathrm{B}=\int d\mathrm{B} \\ =\underset{0}{\overset{1}{\int }}\frac{{\mathrm{\mu }}_0{a}^{2}nidx}{4\pi {\left[a^2+(l-2x{)}^2\right]}^{3/2}} \\ =\underset{0}{\overset{1}{\int }}\frac{{\mathrm{\mu }}_{0}ni{a}^{2}dx}{4\pi a^3{\left[1+{\left({\displaystyle \frac{l-2x}{a}}\right)}^2\right]}^{3/2}}=\frac{{\mu }_{0}ni}{4\pi a}\underset{0}{\overset{1}{\int }}\frac{dx}{{\left[1+{\left({\displaystyle \frac{l-2x}{a}}\right)}^2\right]}^{3/2}}=\frac{{\mu }_{0}ni}{4\pi a}.\frac{4\pi a}{\sqrt{1+\left({\displaystyle \frac{2a}{l}}\right)}}=\frac{{\mu }_{0}ni}{\sqrt{1+{\left({\displaystyle \frac{2a}{l}}\right)}^2}} \end{gathered}$$
(b) When a > > l,
$\mathrm{B}=\frac{{\mu }_{0}ni}{2a}$