Question

Find the magnetic field at point Pon the axis of a tightly woundsolenoid(helical coil) consisting of nturns per unit length wrapped around a cylindrical tube of radius aand carrying current I(Fig. 5.25). Express your answer in terms of ${\mathbf{\theta }}_{\mathbf{1}}$and ${\mathbf{\theta }}_{\mathbf{2}}$ (it's easiest that way). Consider the turns to be essentially circular, and use the result of Ex. 5.6. What is the field on the axis of an infinitesolenoid (infinite in both directions)?

Short answer

The magnetic field at point Pon the axis of a tightly wound solenoid consisting of nturns per unit length wrapped around a cylindrical tube of radius aand carrying current Iis$\frac{{\mathrm{\mu }}_0\mathrm{nI}}{2}({\mathrm{cos\theta }}_2-{\mathrm{cos\theta }}_1)$.

The field on the axis of an infinitesolenoid is $\frac{{\mu }_{0}nI}{2}$.

Step-by-step solution

Given data

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 ${\theta }_1$ and${\theta }_2$ .

Determine the formula for the magnetic field on the axis of a ring

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

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

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

Determine the magnetic field on the axis of a solenoid

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

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

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

But $z=a\cot \theta$

Here, $\theta$ is the angle made by the small coil of length $dz$at P.

Therefore,

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

$\begin{array}{c}B=-\frac{{\mu }_{0}nI}{2}\underset{{\theta }_1}{\overset{{\theta }_2}{\int }}\sin \theta d\theta \\ =\frac{{\mu }_{0}nI}{2}(\cos {\theta }_2-\cos {\theta }_1)\end{array}$

For an infinite ring,

$\begin{array}{l}{\mathrm{\theta }}_2=0\\ {\mathrm{\theta }}_1=\mathrm{\pi }\end{array}$

The formula for the field becomes:

$\begin{array}{c}B=-\frac{{\mu }_{0}nI}{2}(\cos 0-\cos \pi )\\ =\frac{{\mu }_{0}nI}{2}\end{array}$

Thus, field for a finite solenoid is $\frac{{\mu }_{0}nI}{2}(\cos {\theta }_2-\cos {\theta }_1)$ and the field for an infinite solenoid is $\frac{{\mu }_{0}nI}{2}$.