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Fernando Garcia

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PHYS405 - Electricity & Magnetism I

Spring 2024

Problem 5.11

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:

$$\begin{array}{}\text{(Eq. 5.41)}& B(z)=\frac{{\mu }_{0}I}{2}\frac{R^2}{(R^2+z^2{)}^{3/2}}\end{array}$$

Now, we need to modify this slightly. We will replace

$$I\to nI$$

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

$$\begin{array}{rl}z& =\frac{a}{\tan \left(\theta \right)}\\ & =a\cot \left(\theta \right)\end{array}$$

Replacing this variable in the integral:

$$\begin{array}{rl}dz& =a(-{\csc }^2(\theta ))d\theta \\ & =-\frac{a}{{\sin }^2\theta }d\theta \end{array}$$

Further, we note that

$$\begin{array}{rl}\frac{1}{a^2+a^2{\cot }^2\theta }& =\frac{1}{a^2(1+{\cot }^2\theta )}\\ & =\frac{1}{a^2{\csc }^2\theta }\\ & =\frac{{\sin }^2\theta }{a^2}\end{array}$$

Therefore

$$\frac{1}{(a^2+z^2{)}^{3/2}}=\frac{{\sin }^3\theta }{a^3}$$

We see then that the integral reduces to:

$$\begin{array}{rl}B& =\frac{{\mu }_{0}nI}{2}{\int }_{{\theta }_1}^{{\theta }_2}\frac{a^2{\sin }^3\theta }{a^3{\sin }^2\theta }(-1)ad\theta \\ & =-\frac{{\mu }_{0}nI}{2}{\int }_{{\theta }_1}^{{\theta }_2}\sin \theta d\theta \\ & =-\frac{{\mu }_{0}nI}{2}(\cos {\theta }_2-\cos {\theta }_1)\end{array}$$

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:

$$\begin{array}{rl}{\theta }_1& =\pi \\ {\theta }_2& =0\end{array}$$

Which reduces the above result to

$$B={\mu }_{0}nI$$

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