Q70P - Fundamentals Of Physics

Question

A caterpillar of length 4.0cm crawls in the direction of electron drift along a 5.2-mm-diameter bare copper wire that carries a uniform current of 12A. (a) What is the potential difference between the two ends of the caterpillar? (b) Is its tail positive or negative relative to its head? (c) How much time does the caterpillar take to crawl 1.0cm if it crawls at the drift speed of the electrons in the wire? (The number of charge carriers per unit volume is$\mathbf{8}\mathbf{.}\mathbf{49}\mathbf{\times }{\mathbf{10}}^{\mathbf{28}}\mathbf{/}{\mathit{m}}^{\mathbf{3}}$.)

Step-by-Step Solution

Answer

a) The potential difference between the two ends of the caterpillar is$3.8\times {10}^{-4}V$

b) Its tail is negative relative to its head.

c) The time the caterpillar takes to crawl 1.0cm if it crawls at the drift speed of the electrons in the wire is 240s or 4 min.

Step 1: The given data

a) Length of the caterpillar,$L=4\mathrm{cm} \mathrm{or} 0.04\mathrm{m}$

b) Diameter of the copper wire,$d=5.2 mm or 5.2\times {10}^{-3}m$

c) Uniform current flow,i=12 A

d) Distance of crawl,$x=1 cm or 0.01m$

e) Number of charge carriers per unit volume,$n/V=8.49\times {10}^{28}/m^3$

Step 2: Understanding the concept of the potential difference and current flow

Using Ohm’s law, we can write the relation between the potential, current, and resistance. Also using equations 26-16, we can write resistance in terms of resistivity. From that, we can find the potential difference between the two ends of the caterpillar. As per the given condition, the caterpillar crawls in the direction of the electron. It means it is travelling against the current, so the tail is negative as compared to its head. By using the relation between drift speed and time of travel, we can calculate the total time.

 

Formulae:

The voltage equation due to Ohm’s law,$V=iR$                                                 (i)

The resistance value of a material,$R=\frac{\rho L}{A}$                                                       (ii)

The time taken by the electrons in terms of drift speed,$t=\frac{L}{V_d}$                          (iii)

The cross-sectional area of the circle, $A=\pi r^2$                                                 (iv)

The current density in terms of drift velocity,$J=\left(ne\right)v_d$                                    (v)

The current density due to uniform current flow,$J=\frac{i}{A}$                                       (vi)

Step 3: a) Calculation of the potential difference between the ends of caterpillar

The radius of the end can be calculated as follows:

$$\begin{gathered} r=\frac{5.2\times {10}^{-3}}{2} \\ =2.6\times {10}^{-3}m \end{gathered}$$

Now, the value of the area of the end can be calculated using the radius value in equation (iv) as follows:

$$\begin{gathered} A=3.14\times {\left(2.6\times {10}^{-3}\right)}^2 \\ =2.12\times {10}^{-5}{m}^2 \end{gathered}$$

For copper, the resistivity is

$\rho =1.69\times {10}^{-8}\Omega .m$.

Now, we can get the potential difference can be given using the resistance value of equation (ii) in equation (i) and using the above values as follows:

$$\begin{gathered} V=i\frac{\rho L}{A} \\ =\frac{12\times 1.69\times {10}^{-8}\times 0.04}{2.12\times {10}^{-5}} \\ =3.8\times {10}^{-4}V \end{gathered}$$

Hence, the value of the potential difference is $3.8\times {10}^{-4}V$.

Step 4: b) Calculation to know the position of tail relative to head

According to the given condition, the caterpillar crawls in the direction of electron that means against the current.

Hence, the tail is negative as compared to its head.

Step 5: c) Calculation of the time taken by the caterpillar

Using the formulae of current density from equations (v) and (vi), we can get the value of the drift speed as follows:

$$\begin{gathered} \left(ne\right)v_d=\frac{i}{A} \\ v_d=\frac{ine}{A} \end{gathered}$$

Now, substituting the above value in equation (iii), we can get the time taken the caterpillar to crawl 1 cm as follows:

$$\begin{gathered} t=\frac{LAne}{i} \\ =\frac{0.04\times 2.12\times {10}^{-5}\times 8.49\times {10}^{28}\times 1.6\times {10}^{-19}}{4\times 12} \\ =240s or 4\min \end{gathered}$$

Hence, the value of the time is 240s or 4 min.