Question

A beam contains$\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{8}}$doubly charged positive ions per cubic centimetre , all of which are moving north with a speed of $\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}\mathbf{}}\mathit{m}\mathbf{/}\mathit{s}$. What is the (a) magnitude of the current density$\overrightarrow{\mathbf{J}}$? (b) Direction of the current density$\overrightarrow{\mathbf{J}}$?(c) What additional quantity do you need to calculate the total current i of this ion beam?

Short answer

1. The magnitude of the current density is$6.4\frac{A}{m^2}$
2. The direction of the current density is the same as the direction of the beam particles i.e. to the north.
3. The additional quantity needed to calculate the total current i in this ion beam is cross-sectional area.

Step-by-step solution

The given data

1. Particle concentration,$n=2\times {10}^8/c{m}^3$
2. Drift speed,$V_d=1\times {10}^{5}m/sec$

Understanding the concept of the flow of current density

The term "current density" refers to the quantity of electric current moving across a certain cross-section. To calculate the current density, we have to substitute the given values of particle concentration and speed in the formula of current density. We can find the direction of current density by considering the direction of beam particles. We can find the additional quantity required to calculate the total current by using the relation between current and current density.

Formulae:

The magnitude of the current density related to the drift velocity, $J=nq{V}_d$ ...(i)

where, n is the particle concentration, q is the charge of each particle, and $V_d$is the drift velocity.

The current density flowing through the area, $J=\frac{i}{A}$x ...(ii)

(a) Calculation of the magnitude of the current density

Since, the particles are doubly charged positive ions,

$$\begin{gathered} q=2e \\ =2\times 1.6\times {10}^{-19}C \\ =3.2\times {10}^{-19}C \end{gathered}$$

The particle concentration is given as,

$\begin{array}{l}n=2\times {10}^8/{cm}^3\\ =\frac{2\times {10}^8}{{10}^{-6}{m}^3}\\ =2\times {10}^{14}/m^3\end{array}$

Substituting all the values in the formula of equation (i) for current density, we get the value as:

$$\begin{gathered} J=(2x{10}^{14}/m^3)(3.2x{10}^{-19}C)(1x{10}^{5}m/s) \\ =6.4A/m^2 \end{gathered}$$

Hence, the value of the current density is $6.4A/m^2$.

(b) Calculation of the direction of the current density

From equation (i), we get that the direction of the current density is same as the direction of the drift velocity for positive charge and opposite for the negative charge. As the beam particles are positively charged ions, so the direction of current density is same as the direction of the beam particles, that is, to the north.

(c) Calculation of the additional quantity needed to calculate the current of ion beam

From equation (ii), we can get the flow of the current formula as follows:

i=JA

where, A is the cross-sectional area of the beam.

Hence, if we want to calculate the total current, we must know the value of the cross-sectional area of the beam.