Question

There must be Homogenously concentrated fixed singly ionized (+ve) ions having concentration 'n' within a parallel homogenous electron beam of a circular cross section having concentration '$n_0$' so that radius of the beam remains constant if the beam moves with constant velocity v along its axis. Find $(\frac{n}{n_0})$?

• A. $1-\frac{v^2}{c^2}$
• B. $1+\frac{v^2}{c^2}$
• C. $\frac{v^2}{c^2}$
• D. $\frac{c^2}{v^2}$

Answer 1

Answer: (B)

$1+\frac{v^2}{c^2}$

Answer 2

Let the concentration of electrons in the beam be $n_0$ and their velocity be $\vec{v}$ along the axis of the beam (say, the z-axis). The electrons have charge $-e$.
The concentration of fixed singly ionized positive ions is $n$. The ions have charge $+e$.

Consider an electron at a radial distance $r$ from the axis of the beam. This electron experiences two forces:

1. The electrostatic force due to the net charge density inside the beam.
2. The magnetic force due to the current created by the moving electrons.

The charge density of the electrons is $\rho_e = -e n_0$.
The charge density of the fixed positive ions is $\rho_i = +e n$.
The net charge density is $\rho = \rho_e + \rho_i = e(n - n_0)$.

Using Gauss's law for a cylindrical charge distribution, the electric field at a radial distance $r < R$ (where $R$ is the beam radius) is given by $E_r = \frac{\rho r}{2\epsilon_0}$.
The electric force on an electron (charge $-e$) at radius $r$ is $\vec{F}_e = -e \vec{E}_r = -e \frac{\rho r}{2\epsilon_0} \hat{r} = \frac{-e^2(n - n_0)r}{2\epsilon_0} \hat{r}$.
The radial component of the electric force is $F_{e,r} = \frac{-e^2(n - n_0)r}{2\epsilon_0}$. This force is radially inward if $n > n_0$ (net charge is positive) and radially outward if $n < n_0$ (net charge is negative).

The moving electrons constitute a current. The current density is $\vec{J} = \rho_e \vec{v} = (-e n_0) \vec{v}$. If $\vec{v} = v \hat{z}$, then $\vec{J} = -e n_0 v \hat{z}$. The conventional current is in the opposite direction, $I = e n_0 v \pi R^2$ in the $+\hat{z}$ direction.
Using Ampere's law for a cylindrical current distribution, the magnetic field at a radial distance $r < R$ is given by $B_\theta = \frac{\mu_0 J r}{2}$ (for current density $J$ in the z-direction, B is in the $\theta$ direction). Using the magnitude of the current density $J = e n_0 v$, the magnetic field is $B_\theta = \frac{\mu_0 e n_0 v r}{2}$. The direction is azimuthal.

The magnetic force on an electron (charge $-e$) moving with velocity $\vec{v} = v \hat{z}$ at radius $r$ is $\vec{F}_m = -e (\vec{v} \times \vec{B})$.
$\vec{v} \times \vec{B} = (v \hat{z}) \times (B_\theta \hat{\theta}) = v B_\theta (\hat{z} \times \hat{\theta}) = v B_\theta (-\hat{r})$.
$\vec{F}_m = -e v B_\theta (-\hat{r}) = e v B_\theta \hat{r}$.
The radial component of the magnetic force is $F_{m,r} = e v B_\theta = e v \frac{\mu_0 e n_0 v r}{2} = \frac{\mu_0 e^2 n_0 v^2 r}{2}$. This force is always radially outward.

For the radius of the beam to remain constant, the net radial force on the electrons at the edge of the beam ($r=R$) must be zero.
$F_{net,R} = F_{e,R} + F_{m,R} = 0$.
$\frac{-e^2(n - n_0)R}{2\epsilon_0} + \frac{\mu_0 e^2 n_0 v^2 R}{2} = 0$.
Since $e \neq 0$ and $R \neq 0$, we can divide by $\frac{e^2 R}{2}$:
$\frac{-(n - n_0)}{\epsilon_0} + \mu_0 n_0 v^2 = 0$.
$\mu_0 n_0 v^2 = \frac{n - n_0}{\epsilon_0}$.
$\mu_0 \epsilon_0 n_0 v^2 = n - n_0$.

Using the relationship between the speed of light $c$, permittivity of free space $\epsilon_0$, and permeability of free space $\mu_0$: $c^2 = \frac{1}{\mu_0 \epsilon_0}$, or $\mu_0 \epsilon_0 = \frac{1}{c^2}$.
Substituting this into the equation:
$\frac{1}{c^2} n_0 v^2 = n - n_0$.
$\frac{n_0 v^2}{c^2} = n - n_0$.

We need to find the ratio $\frac{n}{n_0}$. Divide the equation by $n_0$:
$\frac{v^2}{c^2} = \frac{n}{n_0} - 1$.
$\frac{n}{n_0} = 1 + \frac{v^2}{c^2}$.

This result shows that the concentration of positive ions $n$ must be greater than the concentration of electrons $n_0$ (since $v^2/c^2$ is positive). This makes sense because the electric force (due to net positive charge) needs to be inward to counteract the outward magnetic force and the outward space-charge repulsion between electrons (if $n \le n_0$).