91Ó°ÊÓ

There is a tightly wound solenoidconsisting of nturns per unit length wrapped around a cylindrical tube of radius aand carrying current I.The angle made by the ends of the coil with the point P are ${\mathrm{θ}}_1$ and${\mathrm{θ}}_2$ .

The magnetic field at a distance $z$ on the axis of a circular coil of radius $a$ and carrying current $I$ is

$\mathbf{B}\mathbf{=}\frac{{\mathbf{μ}}_{\mathbf{0}}\mathbf{I}}{\mathbf{2}}\frac{{\mathbf{a}}^{\mathbf{2}}}{{\mathbf{(}{\mathbf{a}}^{\mathbf{2}}\mathbf{+}{\mathbf{z}}^{\mathbf{2}}\mathbf{)}}^{\mathbf{3}\mathbf{/}\mathbf{2}}}$ …… (1)

Here, ${\mathrm{μ}}_0$ is the permeability of free space.

Current in a small length $dz$ of the coil = $nIdz$

From equation (1), magnetic field at P from this small length of coil is

$\mathrm{dB}=\frac{{\mathrm{μ}}_0\mathrm{nI}}{2}\frac{{\mathrm{a}}^2}{{({\mathrm{a}}^2+{\mathrm{z}}^2)}^{3/2}}\mathrm{dz}$ …… (2)

But $z=a\cot \mathrm{θ}$

Here, $\mathrm{θ}$ is the angle made by the small coil of length $dz$at P.

Therefore,

Substitute these values in equation (2) and integrate from ${\mathrm{θ}}_1$to ${\mathrm{θ}}_2$,

$\begin{array}{c}B=−\frac{{\mathrm{μ}}_{0}nI}{2}\underset{{\mathrm{θ}}_1}{\overset{{\mathrm{θ}}_2}{∫}}\sin \mathrm{θ}d\mathrm{θ}\\ =\frac{{\mathrm{μ}}_{0}nI}{2}(\cos {\mathrm{θ}}_2−\cos {\mathrm{θ}}_1)\end{array}$

For an infinite ring,

$\begin{array}{l}{\mathrm{θ}}_2=0\\ {\mathrm{θ}}_1=\mathrm{π}\end{array}$

The formula for the field becomes:

$\begin{array}{c}B=−\frac{{\mathrm{μ}}_{0}nI}{2}(\cos 0−\cos \mathrm{π})\\ =\frac{{\mathrm{μ}}_{0}nI}{2}\end{array}$

Thus, field for a finite solenoid is $\frac{{\mathrm{μ}}_{0}nI}{2}(\cos {\mathrm{θ}}_2−\cos {\mathrm{θ}}_1)$ and the field for an infinite solenoid is $\frac{{\mathrm{μ}}_{0}nI}{2}$.