Question

Figure 26-22 gives the drift speed v_dof conduction electrons in a copper wire versus position xalong the wire. The wire consists of three sections that differ in radius. Rank the three sections according to the following quantities, greatest first: (a) radius, (b) number of conduction electrons per cubic meter, (c) magnitude of electric field, (d) conductivity

Short answer

1. The rank of the three sections according to the radius is$r_c>r_B>r_A$
2. The rank of the three sections according to the number of conduction electrons per cubic meter is$n_A=n_B=n_C$
3. The rank of the three sections according to the magnitude of the electric field is $E_A>E_B>E_C$.
4. The rank of the three sections according to the conductivity is${\sigma }_A={\sigma }_B={\sigma }_C$

Step-by-step solution

The given data

The figure which gives the drift speed $V_d$of conduction versus position x is given.

Understanding the concept of the field, conductivity and drift speed

The electric field $\overrightarrow{\mathbf{E}}$at any point is defined in terms of the electrostatic force that would be exertedon a positive test charge ${\mathit{q}}_{\mathbf{0}}$placed there:

$\overrightarrow{\mathbf{E}}\mathbf{=}\frac{\overrightarrow{\mathbf{F}}}{{\mathbf{q}}_{\mathbf{0}}}$

Drift speed is the average speed achieved by the electron due to the electric field.

The conductivity of the material is its ability to carry the electric current.

We use the relation between the drift speed and the radius to rank the three sections according to the radius. By using the relation between the electric field and the drift speed, we can rank the three sections according to the magnitude of the electric field. Then, we use the relation between the electric field and the drift speed to rank the three sections according to the magnitude of the electric field. As conductivity is a characteristic of the material of the wire, we can rank the three sections according to their conductivity.

Formulae:

The drift speed of the electrons,$V_d=\frac{J}{ne}=\frac{l}{neA}$ …(i)

$V_d$is the drift speed, J is current density, is the number of electrons, e is the charge of the electron, l is current, and A is the area of cross-section.

The electric field magnitude due to the drift speed,$E=pne{V}_d$ …(ii)

Here, is the electric field, is the resistivity of the material.

The conductivity of a material,$\sigma =1/\mathrm{p}$ …(iii)

Here, $\sigma$is the conductivity of the material.

(a) Calculation of the rank according to their radius

Substituting the area value in equation (i), we can get the drift speed of the electrons as follows:

$$\begin{gathered} V_d=\frac{l}{ne{\mathrm{\pi r}}^2} \\ {V}_d\alpha \frac{1}{r^2} \\ \mathrm{So},r_C>r_B>r_A \end{gathered}$$

Therefore, the rank of the three sections according to the radius is $r_C>r_B>r_A$.

(b) Calculation of the rank according to their number of conduction electrons per cubic meter

Since the three sections of the wire are made up of the same material, the number of electrons per cubic meter is same in all three sections.

Therefore, rank of the three sections according to the number of conduction electrons per cubic meter is $n_C=n_B=n_A$.

(c) Calculation of the rank according to their electric field

From equation (ii), the relation of field to that of drift speed is given as:

$E\alpha V_d$

Again, we know that${\left(V_d\right)}_A>{\left(V_d\right)}_B>{\left(V_d\right)}_C$

Hence, magnitude of the electric field is,role="math" localid="1661407887588" $E_A>E_B>E_C$

Therefore, the rank of the three sections according to the magnitude of electric field is $E_A>E_B>E_C$.

(d) Calculation of the rank according to their conductivity

As all the three sections are made up of the same material, using equation (iii), we can say that all three sections have the same conductivity for same resistivity. Thus,${\sigma }_A={\sigma }_B={\sigma }_C$

Therefore, the rank of the three sections according to the conductivity is${\sigma }_A={\sigma }_B={\sigma }_C$