1.1 Current in different situations

  Current Electricity

1. If $n$ particles each having charge $q$, passes through any cross-sectional area in time $t$ then,$${I= \frac{{nq}}{t}}$$

2. If a point charge $q$ moving in a circle of radius $r$ with speed $v$ and frequency $f$ then

current at any point on the circle is, $$I = qf = \frac{{qv}}{{2\pi r}}$$

3. When voltage $V$ is applied across resistor of resistance $R$ then the current flowing through the resistor is $${I = \frac{V}{R}}$$

Current density $(j)$

The current density is defined as the amount of charge $(q)$ flowing through a unit cross-section $(A)$ area in a unit time $(t)$ is known as current density.

$${j = \frac{{\frac{q}{t}}}{A} = \frac{I}{A}}$$

where $I$ is the current flowing uniformly and normally through an area of cross-section $A$ of conductor.

SI unit is Ampere/${m^2}$

Dimensional formula [$A{L^{ - 2}}$] where $A$ is Ampere. (Don't get confused with area $A$)

When a plane of small area $\Delta A$ makes an angle $\theta $ with the direction of current($\Delta I$) , then

Component of $\Delta A$ normal to direction of current flow will be

$${(\Delta A)_n} = \Delta A\cos \theta $$

Current density, $$\begin{equation} \begin{aligned} j = \frac{{\Delta I}}{{(\Delta A)n}} = \frac{{\Delta I}}{{\Delta A\cos \theta }} \\ \Delta I = j\Delta A\cos \theta \\ \Delta I = \vec j.\Delta \vec A \\\end{aligned} \end{equation} $$ or $$I = \vec j.\vec A$$ So, current density is defined as the dot product of current density $\left( {\overrightarrow j } \right)$ and the cross-sectional area $\left( {\overrightarrow A } \right)$.

Question 1. An electron moves in a circle of radius 20 $cm$ with a constant speed of $8 \times {10^6}m{s^{ - 1}}$. Find the electric current at a point on the circle.

Solution: Consider a point $A$ on the circle as shown in the figure.

Distance travelled by electron in one revolution = $2\pi \times 20 \times {10^{ - 2}}m$

Number of revolutions made by the electron in one second, $$\frac{{8 \times {{10}^6}m}}{{2\pi \times 20 \times {{10}^{ - 2}}m}} = \frac{4}{\pi } \times {10^7}s^{-1}$$

Charge on electron, $$e = 1.6 \times {10^{ - 19}}C$$

The charge crossing the point $A$ per second is $$\frac{4}{\pi } \times {10^7} \times 1.6 \times {10^{ - 19}}C$$

Thus, electric current at point $A$ = $$2 \times {10^{ - 12}}A$$

Question 2. An electron beam has an aperture $2m{m^2}$. A total of $3 \times {10^{16}}$ electrons go through any prependicular cross-section per second. Find

(a) Current $(I)$

(b) Current density $(j)$ in the beam.

Solution:

(a)

Given Number $(N)$ of electrons passing through any cross-sectional area per second $=3 \times {10^{16}}$

Change on a electron $=1.6 \times {10^{ - 19}}C$

The total charge crossing the prependicular cross-section in one second is $$q = Ne$$

$$ = 3 \times {10^{16}} \times 1.6 \times {10^{ - 19}}$$

$$ = 4.8 \times {10^{ - 3}}C$$

Current is, $$i = \frac{q}{t} = \frac{{4.8 \times {{10}^{ - 3}}C}}{{1s}} = 4.8 \times {10^{ - 3}}A$$

(b)

Current density is $$j = \frac{I}{A} = \frac{{4.8 \times {{10}^{ - 3}}}}{{2 \times {{10}^{ - 6}}}} = 2.4 \times {10^3}A{m^{ - 2}}$$

Question 3. Current flows through a constricted conductor as shown in figure. The diameter ${d_1} = 1.0mm$ and the current density ${j_1}$ is $2.54 \times {10^6}A{m^{ - 2}}$.

(i). What current flows into the construction?

(ii) What will be the diameter ${d_2} $ if the current density is doubled as it emerges from the right side of construction?

Solution:

Given, $$\begin{equation} \begin{aligned} {d_1} = 2.0mm, \\ {j_1} = 2.54 \times {10^6}A{m^{ - 2}}, \\ {j_2} = 2{j_1} \\\end{aligned} \end{equation} $$

(i) Current flowing into the constriction,$${I_1} = {j_1} \times \pi {\left( {\frac{{{d_1}}}{2}} \right)^2} = 2.54 \times {10^6} \times 3.14 \times {\left( {0.5 \times {{10}^{ - 3}}} \right)^2} = 1.993A$$

(ii) For steady flow of current, $$\begin{equation} \begin{aligned} \Rightarrow {I_1} = {I_2} \\ \Rightarrow {j_1}{A_1} = {j_2}{A_2} \\\end{aligned} \end{equation} $$ From question, $$ \Rightarrow {j_2} = 2{j_1}$$ Also, $$\begin{equation} \begin{aligned} \Rightarrow {A_1} = \pi {\left( {\frac{{{d_1}}}{2}} \right)^2} \\ \Rightarrow {A_2} = \pi {\left( {\frac{{{d_2}}}{2}} \right)^2} \\\end{aligned} \end{equation} $$ So, $$\begin{equation} \begin{aligned} \Rightarrow {j_1} \times \pi {\left( {\frac{{{d_1}}}{2}} \right)^2} = 2{j_1} \times \pi {\left( {\frac{{{d_2}}}{2}} \right)^2} \\ \Rightarrow {d_2} = \frac{{{d_1}}}{{\sqrt 2 }} = 0.707mm \\\end{aligned} \end{equation} $$