Q8P - Fundamentals Of Physics

Question

A small but measurable current of $\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}\mathit{A}$ exists in a copper wire whose diameter is 2.5 mm .The number of charge carriers per unit volume is $\mathbf{8}\mathbf{.}\mathbf{49}\mathbf{\times }{\mathbf{10}}^{\mathbf{28}}{\mathit{m}}^{\mathbf{-}\mathbf{3}}$ . Assuming the current is uniform, calculate the (a) current density and (b) electron drift speed.

Step-by-Step Solution

Answer

a) The current density is $2.4\times {10}^{-5}A/m^2$

b) The electron drift velocity is $1.8\times {10}^{-15}m/s$

Step 1: The given data

a) Measured current,$i=1.2\times {10}^{-10}A$

b) Diameter of copper wire, $d=2.5 mm or 2.5\times {10}^{-3}m$

c) Electron concentration, $n=8.49\times {10}^{28}{m}^{-3}$ 

Step 2: Understanding the concept of the current density

The term "current density" refers to the quantity of electric current moving across a certain cross-section. We have to use the formula of current density to find the current density and then we have to use the relation between current density and drift velocity, to find the electron drift speed.

 

Formulae:

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

The current density in relation to the drift velocity, $J=ne{V}_d$                                  ...(ii)

The cross-sectional area of the circle, $A={\mathrm{\pi r}}^2$                                                       ...(iii)

 

where, n is the particle concentration, i is the current, and $V_d$ is the drift velocity, r is the radius of wire and e is the charge of electron  

= $1.6\times {10}^{-9}C$

Step 3: (a) Calculation of the magnitude of the current density

Substituting the given data in the formula of the area value from equation (iii) in equation (i), we can get the magnitude value of the current density as follows:

$$\begin{gathered} J=\frac{i}{{\mathrm{\pi r}}^2} \left(\because radius in relation to diameter ,r =\frac{d}{2}\right) \\ =\frac{4i}{\pi d^2} \\ =\frac{4\left(1.2\times {10}^{-10}A\right)}{\mathrm{\pi }{\left(2.5\times {10}^{-3}\mathrm{m}\right)}^2} \\ =2.44\times {10}^{-5}A/m^2 \\ \approx 2.44\times {10}^{-5}A/m^2 \end{gathered}$$ 

Hence, the value of the current density is $2.44\times {10}^{-5}A/m^2$ .

Step 4: (b) Calculation of the drift velocity of the electron

Using the given data in equation (ii), we can get the required drift velocity of the electron as follows:

$$\begin{gathered} V_d=\frac{J}{ne} \\ =\frac{2.44\times {10}^{-5 }A/m^2}{\left(8.49\times {10}^{28}{m}^{-3}\right)\left(1.6\times {10}^{-19}C\right)} \\ =1.796\times {10}^{-15}m/s \\ \approx 1.796\times {10}^{-15}m/s \end{gathered}$$

Hence, the value of the drift velocity is $1.796\times {10}^{-15}m/s$ .