Q34P - Fundamentals Of Physics

Question

Figure shows wire section 1 of diameter ${\mathit{D}}_{\mathbf{1}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{00}\mathit{R}$ and wire section 2 of diameter ${\mathit{D}}_{\mathbf{2}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{ }\mathit{R}$, connected by a tapered section. The wire is copper and carries a current. Assume that the current is uniformly distributed across any cross-sectional area through the wire’s width. The electric potential change V along the length L=2.00m  shown in section 2 is $\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{ }\mathit{\mu }\mathit{V}\mathbf{ }$. The number of charge carriers per unit volume is $\mathbf{8}\mathbf{.}\mathbf{49}\mathbf{\times }{\mathbf{10}}^{\mathbf{2}}\mathbf{8}{\mathbf{m}}^{\mathbf{-}\mathbf{3}}$. What is the drift speed of the conduction electrons in section 1?

Step-by-Step Solution

Answer

The drift speed of the conduction electrons in section 1 is $5.44\times {10}^{-9}\mathrm{m}/\mathrm{s}$ .

Step 1: Identification of given data

a) The diameter of the section 1,$D_1=4.00R$ 

b) The diameter of the section 2,$D_2=2.00R$ 

c) The length of the wire, $L=2.00 m$

d) The electric potential change in section ,$V=10.0\mu V or 10.0\times {10}^{-6}V$ 

e) The number of charge carriers per unit volume, $n=8.49\times {10}^{28}c$

Step 2: Significance of flow of current and its density

The current is the rate of flow of charges with respect to time. The current density is equal to the rate of flow of charge per unit time per unit cross-section area. 

We can find the electric field and current density in the second section by using the corresponding formulae. By using the law of conservation of charge, we can find the current density in the first section. Finally, we can find drift velocity by using its expression in terms of current density.

 

Formulae:

The electric field strength of the material,$E=\frac{V}{L}$                                                         …(i)

Here, E is the electric field, V is voltage, and L is separation.

The current density is related to the resistivity of the material, $J=\frac{E}{\rho }$                        …(ii)

Here, E is the electric field, J is the current density, and $\rho$ is the resistivity of the material.

The current density due to the drift velocity of the electrons,$J=\left(ne\right)v_d$                  …(iii)

Here, J is the current density, $V_d$ is the drift velocity, n is the number of electrons, e is the charge on the electron.

The cross-sectional area of the wire, $A={\mathrm{\pi r}}^2$                                                           …(iv)

Here,   is the area of the cross-section,   radius of the conductor.

Step 3: Determining the drift speed of the electron

The strength of the electric field in section 2 can be calculated using equation (i) as follows:

$$\begin{gathered} E_2=\frac{10.0\times {10}^{-6}V}{2.00 m} \\ =5.00\times {10}^{-6}V/m \end{gathered}$$ 

The resistivity of copper;$\rho =1.69\times {10}^{-8}\Omega .m$  

The current density in the section   can be calculated using the given data in equation (ii) as follows:

$$\begin{gathered} J_2=\frac{5.00\times {10}^{-6}V/m}{1.69\times {10}^{-8}\Omega .m} \\ =296A/m^2 \end{gathered}$$ 

According to the conservation law of electric charge, we can say that the current from both the sections is the same. Thus, we get the equation using equations (iv) and the value of the current density of section 1 as:

$$\begin{gathered} J_1=\left({\mathrm{\pi R}}_1^2\right)=J_2\left({\mathrm{\pi R}}_2^2\right) \\ {J}_1=\left(4R^2\right)=J_2\left(R^2\right) \\ J_1=\frac{296A/m^2}{4} \\ =74A/m^2 \end{gathered}$$ 

Now, the drift speed of conduction of electrons in section   can be calculated using the given data in equation (iii) as follows:

$$\begin{gathered} v_d=\frac{74 A/m^2}{8.49\times {10}^{28}{m}^{-3}\times 1.6\times {10}^{-19}C} \\ =5.44\times {10}^{-9}\mathrm{m}/\mathrm{s} \end{gathered}$$ 

Hence, the value of the drift speed is $5.44\times {10}^{-9}\mathrm{m}/\mathrm{s}$.