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What is the drift velocity of electrons if the current flowing through a copper wire of \(1 \mathrm{~mm}\) diameter is \(1.1 \mathrm{~A} ?\) Assume that each atom of copper contributes one electron: (Given : density of \(\mathrm{Cu}=9 \mathrm{~g} / \mathrm{cm}^{3}\) and atomic weight of \(\mathrm{Cu}=63\) ) (a) \(0.3 \mathrm{~mm} / \mathrm{s}\) (b) \(0.5 \mathrm{~mm} / \mathrm{s}\) (c) \(0.1 \mathrm{~mm} / \mathrm{s}\) (d) \(0.2 \mathrm{~mm} / \mathrm{s}\)

Short Answer

Expert verified
(d) 0.2 mm/s

Step by step solution

01

Calculate Cross-sectional Area

The diameter of the copper wire is given as \(1 \, \text{mm}\). Convert this into meters: \(1 \, \text{mm} = 0.001 \, \text{m}\). The cross-sectional area (\(A\)) of the wire can be calculated using the formula for the area of a circle: \(A = \pi r^2\), where \(r\) is the radius of the wire. The radius \(r\) is half the diameter, so \(r = 0.0005 \, \text{m}\). Thus, \(A = \pi \times (0.0005)^2 \, \text{m}^2 = 7.85 \times 10^{-7} \, \text{m}^2\).
02

Calculate Number Density of Atoms

The number density of atoms (\(n\)) can be found using the formula \(n = \frac{\rho \times N_A}{M}\), where \(\rho\) is the density of copper \(9 \times 10^3 \, \text{kg/m}^3\), \(N_A = 6.022 \times 10^{23} \text{ atoms/mol}\) is Avogadro's number, and \(M = 0.063 \, \text{kg/mol}\) is the atomic mass of copper. Thus, \(n = \frac{9 \times 10^3 \times 6.022 \times 10^{23}}{0.063} \approx 8.6 \times 10^{28} \, \text{atoms/m}^3\).
03

Calculate Drift Velocity

The drift velocity (\(v_d\)) is given by the formula \(v_d = \frac{I}{nAe}\), where \(I = 1.1 \, \text{A}\) is the current, \(A = 7.85 \times 10^{-7} \, \text{m}^2\) is the cross-sectional area, \(e = 1.6 \times 10^{-19} \, \text{C}\) is the charge of an electron, and \(n = 8.6 \times 10^{28} \, \text{atoms/m}^3\). Substituting these values gives: \(v_d = \frac{1.1}{8.6 \times 10^{28} \times 7.85 \times 10^{-7} \times 1.6 \times 10^{-19}} \approx 2.0 \times 10^{-4} \, \text{m/s}\).
04

Convert Drift Velocity to mm/s

To compare the drift velocity with the given options, convert \(v_d = 2.0 \times 10^{-4} \, \text{m/s}\) to millimeters per second. Since \(1 \, \text{m} = 1000 \, \text{mm}\), \(v_d = 2.0 \times 10^{-4} \, \text{m/s} \times 1000 = 0.2 \, \text{mm/s}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Drift Velocity Derivation
Drift velocity is a concept used to describe the average velocity of charged particles, like electrons, in a conductor when a current flows. It's different from their random thermal motion. The drift velocity, denoted as \(v_d\), is derived from the relationship between current, the number of charge carriers, and their velocity.
The formula to find drift velocity is:
  • \(v_d = \frac{I}{nAe}\)
where \(I\) is the current flowing through the wire, \(n\) is the number density of electrons, \(A\) is the cross-sectional area of the conductor, and \(e\) is the elementary charge of an electron (\(1.6 \times 10^{-19} \text{ C}\)). By substituting the known values into this equation, you can calculate the drift velocity. This derivation helps in understanding that the drift velocity of electrons is significantly smaller compared to their random thermal speeds.
Current Flow in Conductors
Current flow in conductors is the movement of electrons through a conductive material. In the context of metals, which include copper wires, electrons flow due to the electric field applied across the material.
  • Copper is highly conductive because it has free electrons available due to its atomic structure.
  • When a potential difference is applied (like a battery), it creates an electric field, motivating these free electrons to move towards the positive terminal.
This movement creates what we call an electric current. It's essential to understand here that electrons themselves move quite slowly, but the electric field propagates at near-light speed, hence the almost immediate response of the current when the circuit is completed.
Number Density of Electrons
The number density of electrons, indicated by \(n\), refers to the number of charge carriers (electrons) present per unit volume of the conductor. For copper, this can be calculated knowing the material's density and its atomic structure.
To calculate number density, use:
  • \(n = \frac{\rho \times N_A}{M}\)
where \(\rho\) is the density of copper (in kg/m³), \(N_A\) is Avogadro's number \( (6.022 \times 10^{23} \text{ atoms/mol})\), and \(M\) is the molar mass of copper (kg/mol).
This formula gives us the number of copper atoms per cubic meter and since each copper atom contributes one electron (as assumed here), it directly relates to the number density of electrons. This parameter is crucial for understanding the potential for current flow in a material.
Cross-sectional Area Calculation
Cross-sectional area of a conductor, specifically its shape and size, affects the current's ability to flow through it. For a cylindrical wire like copper, this cross-sectional area can be computed using the formula:
  • \(A = \pi r^2\)
Here, \(r\) is the radius of the wire, which is half its diameter. For a given diameter, converting it to meters ensures consistency with other measurements used in calculations.
This cross-sectional area is pivotal in calculating the drift velocity, as it directly influences how many electrons can pass through the wire at once. A larger area allows more electrons to flow concurrently, reducing individual drift velocity for the same total current.

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Most popular questions from this chapter

When one ampere current flows for one minute through a silver voltameter, it deposites \(0.067 \mathrm{~g}\) of silver on the cathode, then how much charge will flow to deposite \(108 \mathrm{~g}\) of silver? (a) \(10.6 \times 10^{4} \mathrm{C} / \mathrm{g}_{\mathrm{eq}}\) (b) \(9.67 \times 10^{4} \mathrm{C} / \mathrm{geq}\) (c) \(8.7 \times 10^{4} \mathrm{C} / \mathrm{g}_{\mathrm{eq}}\) (d) \(4.3 \times 10^{4} \mathrm{C} / \mathrm{geq}\)

A silver and copper voltameters are connected in parallel to a \(12 \mathrm{~V}\) battery of negligible resistance. At what rate is energy being delivered by the battery, if in 30 minutes, \(1 \mathrm{~g}\) of silver and \(1.8 \mathrm{~g}\) of copper are deposited ? (Assume electrochemical equivalent of silver \(=11.2 \times 10^{-7} \mathrm{~kg} / \mathrm{C}\), electrochemical equivalent of copper \(\left.=6.6 \times 10^{-7} \mathrm{~kg} / \mathrm{C}\right)\) (a) \(42.2 \mathrm{~J} / \mathrm{s}\) (b) \(40.4 \mathrm{~J} / \mathrm{s}\) (c) \(24.1 \mathrm{~J} / \mathrm{s}\) (d) \(20.4 \mathrm{~J} / \mathrm{s}\)

To convert a galvanometer into an ammeter, we should connect: (a) a low resistance in series with it (b) a high resistance in series with it (c) a low resistance in parallel with it (d) a high resistance in parallel with it

The temperature of cold junction of thermo-couple is \(0^{\circ} \mathrm{C}\). If the neutral temperature is \(270^{\circ} \mathrm{C}\), then the inversion temperature is: (a) \(540^{\circ} \mathrm{C}\) (b) \(520^{\circ} \mathrm{C}\) (c) \(640^{\circ} \mathrm{C}\) (d) \(580^{\circ} \mathrm{C}\)

For accurate measurements, the resistance of a voltmeter should be : (a) as large as possible (b) equal to the resistance across which the potential difference is to be measured (c) as small as possible (d) infinity

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