Fernando Garcia
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PHYS405 - Electricity & Magnetism I
Spring 2024
Problem 5.11
Find the magnetic field at a point $P $ on the axis of a tightly wound solenoid (helical coil) consisting on $n $ turns per unit length wrapped around a cylindrical tube of radius $a $ and carrying current $I $ (Fig. 5.24).
Express your answer in terms of $\theta _1 $ and $\theta _2 $. (It's tidiest 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 infinite solenoid? (Infinite in both directions)
As suggested by the problem, we can think of this as a collection of rings. Example 5.6 shows the $B $ field of a ring a distance $z $ above the point of interest:
\[B(z)=\frac{\mu _0 I}{2}\frac{R^2 }{(R^2 +z^2 )^{3/2}}\tag{Eq. 5.41} \]Now, we need to modify this slightly. We will replace
\[I\rightarrow nI \]So the integral we want to compute is of the form:
\[B=\frac{\mu _0 nI}{2}\int \frac{a^2 }{(a^2 +z^2 )^{3/2}}dz \]Instead of using a perpendicular distance to the rings $(z) $, it is of interest to use an angle as showcased in figure 5.24 above. We note that
\[\tan \left( \theta \right) =\frac{a}{z} \]So
\begin{align*} z &= \frac{a}{\tan \left( \theta \right) } \\ &= a\cot \left( \theta \right) \end{align*}Replacing this variable in the integral:
\begin{align*} dz &= a(-\csc ^2 (\theta ))d\theta \\ &= -\frac{a}{\sin ^2 \theta }d\theta \end{align*}Further, we note that
\begin{align*} \frac{1}{a^2 +a^2 \cot ^2 \theta } &= \frac{1}{a^2 \left( 1+\cot ^2 \theta \right) } \\ &= \frac{1}{a^2 \csc ^2 \theta }\\ &= \frac{\sin ^2 \theta }{a^2 } \end{align*}Therefore
\[\frac{1}{(a^2 +z^2 )^{3/2}}=\frac{\sin ^3 \theta }{a ^3} \]We see then that the integral reduces to:
\begin{align*} B &= \frac{\mu _0 nI}{2}\int_{\theta _1 }^{\theta _2 } \frac{a^2 \sin ^3 \theta }{a^3 \sin ^2 \theta }(-1)ad\theta \\ &= -\frac{\mu _0 nI}{2}\int_{\theta _1 }^{\theta _2 } \sin \theta d\theta \\ &= -\frac{\mu _0 nI}{2} (\cos \theta _2 - \cos \theta _1 ) \end{align*}If we now want to consider an infintie solenoid, consider what happens to the angles $\theta _1 $ and $\theta _2 $ as we stretch the solenoid to infinity:
\begin{align*} \theta _1 &= \pi \\ \theta _2 &= 0 \end{align*}Which reduces the above result to
\[B=\mu _0 nI \]