Question

Question: A point charge$q$is a distance$a>R$from the axis of an infinite solenoid (radius R, n turns per unit length, current I). Find the linear momentum and the angular momentum (with respect to the origin) in the fields. (Put q on the x axis, with the solenoid along z; treat the solenoid as a nonconductor, so you don’t need to worry about induced charges on its surface.)

Short answer

The linear momentum and angular momentum is $p=\frac{{\mu }_{0}qnI{R}^2}{2a}\overbrace{y}$and $L=0$, respectively.

Step-by-step solution

Expression for the momentum density and linear momentum:

Write the expression for the momentum density.

$\rho ={\epsilon }_0\left(E\times B\right)$ ……. (1)

Here, ${\epsilon }_0$is the permittivity of free space, $E$is the electric field, and $B$is the magnetic field.

Write the expression for the linear momentum.

$p=\int \rho d\zeta$ …… (2)

Determine the expression for the linear momentum:

Substitute the known value in equation (2).

$$\begin{gathered} p=\int {\epsilon }_0\left(\frac{q}{4{\mathrm{\pi \epsilon }}_0}\frac{r}{r^3}\times {\mu }_{0}nI\overbrace{z}\right)d\zeta \\ =\frac{{\mu }_{0}nqI}{4\mathrm{\pi }}\int \frac{\left(r\times \overbrace{z}\right)}{r^3}d\zeta \end{gathered}$$

It is known that $r\times \overbrace{z}=y\overbrace{x}-\left(x-a\right)\overbrace{y}$.

The expression of linear momentum will be,

$p=\frac{{\mu }_{0}nqI}{4\mathrm{\pi }}\int \frac{y\overbrace{x}-\left(x-a\right)\overbrace{y}}{{\left({\left(x-a\right)}^2+y^2+z^2\right)}^{{\displaystyle \frac{3}{2}}}}dxdydz$

In the above expression, $\overbrace{x}$the term is odd in $y$then on integration, the $\overbrace{x}$will become zero. So,

$$\begin{gathered} p=\frac{-{\mu }_{0}nqI}{4\mathrm{\pi }}\overbrace{y}\int \frac{\left(x-a\right)}{{\left({\left(x-a\right)}^2+y^2+z^2\right)}^{{\displaystyle \frac{3}{2}}}}dxdydz \\ =-\frac{-{\mu }_{0}nqI}{2\mathrm{\pi }}\overbrace{y}\int \frac{x-a}{{\left(x-a\right)}^2+y^2}dxdy \end{gathered}$$ …… (3)

Change the coordinates into polar coordinates.

$$\begin{gathered} x=s\cos \varnothing \\ y=s\sin \varnothing \\ dxdy=sdsd\varnothing \\ {\left(x-a\right)}^2+y^2=s^2+a^2-2sa\cos \end{gathered}$$

Substitute all the known values in equation (3).

$p=-\frac{{\mu }_{0}nqI}{2\mathrm{\pi }}\overbrace{y}\int \frac{s\cos \varnothing -a}{s^2+a^2-2sa\cos \varnothing }sdsd\varnothing$ ….. (4)

Let,

${\int }_0^{2\mathrm{\pi }}\frac{\cos \varnothing d\varnothing }{A+B\cos \varnothing }=\frac{2\mathrm{\pi }}{B}\left(1-\frac{A}{\sqrt{A^2}-B^2}\right)$

Then, solve the integral as,

$$\begin{gathered} {\int }_0^{2\mathrm{\pi }}\frac{d\varnothing }{A+B\cos \varnothing }=\frac{2\mathrm{\pi }}{\sqrt{A^2}-B^2} \\ {A}^2-B^2={\left(s^2+a^2\right)}^2-4s^2{a}^2 \\ =s^4+a^4+2s^2{a}^2-4s^2{a}^2 \\ \sqrt{A^2}-B^2=s^2-a^2 \end{gathered}$$

From equation (4),

$$\begin{gathered} p=\frac{{\mu }_{0}nqI}{2}\overbrace{y}\left[1-\left(\frac{a^2+s^2}{a^2-s^2}\right)+\frac{2a^2}{a^2-s^2}\right]sds \\ =\frac{{\mu }_{0}nqI}{2}\overbrace{y}{\int }_0^{R}sds \\ =\frac{{\mu }_{0}qnI{R}^2}{2a}\overbrace{y} \end{gathered}$$

Therefore, the linear momentum is $p=\frac{{\mu }_{0}qnI{R}^2}{2a}\overbrace{y}$.

Determine the angular momentum:

Write the expression for the angular momentum.

$$\begin{gathered} I={\epsilon }_0\left(\times r\left(E\times B\right)\right) \\ =\frac{{\mu }_{0}nqI}{4\mathrm{\pi }}r\times \left[y\overbrace{x}-\left(x-a\right)\overbrace{y}\right] \\ =\frac{{\mu }_{0}nqI}{4\mathrm{\pi }}\left[z\left(x-a\right)\overbrace{x}+zy\overbrace{y}-\left[x\left(x-a\right)+y^2\right]\overbrace{z}\right] \end{gathered}$$

Here, $\overbrace{x}$and $\overbrace{y}$are odd in and integrated to zero.

The expression of angular momentum will be,

$$\begin{gathered} L=\frac{-{\mu }_{0}nqI}{4\mathrm{\pi }}\overbrace{z}\int \frac{\left(x^2+y^2-xa\right)dxdydz}{{\left[{\left(x-a\right)}^2+y^2+z^2\right]}^{{\displaystyle \frac{3}{2}}}} \\ =\frac{-{\mu }_{0}nqI}{4\mathrm{\pi }}\overbrace{z}\int \frac{\left(x^2+y^2-xa\right)}{\left[{\left(x-a\right)}^2+y^2\right]}dxdy \\ =\frac{-{\mu }_{0}nqI}{4\mathrm{\pi }}\overbrace{z}\int \frac{s-a\cos \varnothing }{s^2+a^2-2sa\cos \varnothing }{s}^{2}dsd\varnothing \\ =-{\mu }_{0}nqI\overbrace{z}\int \left[\frac{s^2}{a^2-s^2}+\left(1+\frac{a^2+s^2}{a^2-s^2}\right)\right]sds \end{gathered}$$

On further solving,

$$\begin{gathered} L=-{\mu }_{0}nqI\overbrace{z}\int \left[\frac{s^2-s^2}{a^2-s^2}\right]sds \\ =0 \end{gathered}$$

Therefore, the angular momentum is $L=0$.