Question

A beam contains $2 \times {10^8}$ doubly charged positive ions per cubic centimetre, all of which moving with a speed of ${10^5}{\rm{ }}{{\rm{m}} {\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}$. The current density is
A. $6.4{\rm{ }}{{\rm{A}} {\left/{\vphantom {{\rm{A}} {{{\rm{m}}^2}}}} \right.} {{{\rm{m}}^2}}}$
B. ${\rm{3}}{\rm{.2 }}{{\rm{A}} {\left/{\vphantom {{\rm{A}} {{{\rm{m}}^2}}}} \right. } {{{\rm{m}}^2}}}$
C. $1.6{\rm{ }}{{\rm{A}} {\left/{\vphantom {{\rm{A}} {{{\rm{m}}^2}}}} \right.} {{{\rm{m}}^2}}}$
D. None of these

Answer 1

Current density is the current value when it is divided per unit area of the beam. Mathematically, the current density is the product of charge density and velocity with which ions are moving. We know that charge density is equal to the product of the number of ions and charged on them.

Complete step by step answer:
Given:
The number of positively charged ions in the given beam is $n = 2 \times {10^8}{\rm{ /c}}{{\rm{m}}^3}$.
The speed with which positive ions are moving is $v = {10^5}{\rm{ }}{{\rm{m}} {\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}$.

Let us write the expression for charge density of the given positive ions.
$\rho = nq$……(1)
Here q is the charge on the given ions.
We know that the charge on an electron is given by $1.6 \times {10^{ - 19}}{\rm{ C}}$. We also know that the charge on positive ions and negative ions is the same in magnitude but different in sign, so we can write it as the value of charge q as below.
$q = 1.6 \times {10^{ - 19}}{\rm{ C}}$

It is given the positive ions are doubly charged so we can write the charge on them as twice the value of q.
$q = 2 \times 1.6 \times {10^{ - 19}}{\rm{ C}}$
On substituting $2 \times {10^8}{\rm{ /c}}{{\rm{m}}^3}$ for n and $2 \times 1.6 \times {10^{ - 19}}{\rm{ C}}$ for q in equation (1), we can write:

$$\rho = \left( {2 \times {{10}^8}{\rm{ /c}}{{\rm{m}}^3}} \right)\left( {2 \times 1.6 \times {{10}^{ - 19}}{\rm{ C}}} \right)\\\ \Rightarrow \rho = 6.4 \times {10^{ - 11}}{\rm{ }}{{\rm{C}} {\left/{\vphantom {{\rm{C}} {{\rm{c}}{{\rm{m}}^3}}}} \right. } {{\rm{c}}{{\rm{m}}^3}}}$$

Let us write the expression for current density for the given beam
$J = \rho v$
On substituting $4 \times {10^{ - 11}}{\rm{ }}{{\rm{C}} {\left/ {\vphantom {{\rm{C}} {{\rm{c}}{{\rm{m}}^3}}}} \right. } {{\rm{c}}{{\rm{m}}^3}}}$ for $\rho$ and ${10^5}{\rm{ }}{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. } {\rm{s}}}$ for v in the above expression, we get:

$$J = \left( {6.4 \times {{10}^{ - 11}}{\rm{ }}{{\rm{C}} {\left/ {\vphantom {{\rm{C}} {{\rm{c}}{{\rm{m}}^3} \times \dfrac{{{\rm{1}}{{\rm{0}}^6}{\rm{c}}{{\rm{m}}^3}}}{{{{\rm{m}}^3}}}}}} \right. } {{\rm{c}}{{\rm{m}}^3} \times \dfrac{{{\rm{1}}{{\rm{0}}^6}{\rm{c}}{{\rm{m}}^3}}}{{{{\rm{m}}^3}}}}}} \right)\left( {{{10}^5}{\rm{ }}{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. } {\rm{s}}}} \right)\\\ \therefore J = 6.4{\rm{ }}{{\rm{A}} {\left/ {\vphantom {{\rm{A}} {{{\rm{m}}^2}}}} \right. } {{{\rm{m}}^2}}}$$

Therefore, the current density due to the given number of positive ions in the beam is $6.4{\rm{ }}{{\rm{A}} {\left/ {\vphantom {{\rm{A}} {{{\rm{m}}^2}}}} \right. } {{{\rm{m}}^2}}}$.

So, the correct answer is “Option A”.

Note:
We have to carefully understand the question, especially the statement “the doubly charged positive ions,” which means that the value of charge on each positive ion is double to their actual value. Also, do not forget to convert the units into the same system to follow the homogeneity law of units.