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Chapter 18: Problem 18

A gold wire of 0.50 mm diameter has \(5.90 \times 10^{28} \mathrm{con}\) duction electrons/m". If the drift speed is \(6.5 \mu \mathrm{m} / \mathrm{s}\), what is the current in the wire?

Short Answer

Expert verified
Answer: The current in the gold wire is approximately \(1.22 \times 10^{-3}\) A.

Step by step solution

01

Calculate the cross-sectional area

First, we need to find the cross-sectional area (A) of the gold wire. For a cylindrical wire, the area is computed using the formula: \(A = \pi r^2\), where r is the radius. The diameter of the wire is given as 0.50 mm, so the radius would be \(r = \frac{0.50}{2}\) mm = \(0.25\) mm =\(2.50 \times 10^{-4}\) m. Now calculating the area: \(A = \pi r^2 = \pi (2.50 \times 10^{-4})^2 = \pi (6.25 \times 10^{-8}) = 1.9635 \times 10^{-7} \, \mathrm{m}^2\)
02

Charge of one electron

The charge of one electron (q) is a constant value, which is approximately \(1.6 \times 10^{-19}\) Coulombs.
03

Calculate the number of conduction electrons

We are given that the concentration of conduction electrons is \(5.90 \times 10^{28}\, \mathrm{m}^{-3}\). We will use this value directly as the number of conduction electrons (n) per unit volume.
04

Calculate the current in the wire

Now that we have all the required values, we can use the formula for current in terms of drift speed, charge of an electron, cross-sectional area of the wire, and concentration of conduction electrons, which is: \(I = nqAv\) where: I = current n = number of conduction electrons per unit volume q = charge of one electron A = cross-sectional area of the wire v = drift speed Substituting the known values: \(I = (5.90 \times 10^{28} \, \mathrm{m}^{-3})(1.6 \times 10^{-19} \, \mathrm{C})(1.9635 \times 10^{-7} \, \mathrm{m}^2)(6.5 \times 10^{-6} \, \mathrm{m/s})\) \(I = 1.22 \times 10^{-3}\, \mathrm{A}\) So the current in the gold wire is approximately \(1.22 \times 10^{-3}\) A.

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Most popular questions from this chapter

A 6.0 -pF capacitor is needed to construct a circuit. The only capacitors available are rated as \(9.0 \mathrm{pF} .\) How can a combination of three 9.0 -pF capacitors be assembled so that the equivalent capacitance of the combination is \(6.0 \mathrm{pF} ?\)
(a) Given two identical, ideal batteries (emf \(=\mathscr{E}\) ) and two identical light-bulbs (resistance \(=R\) assumed constant), design a circuit to make both bulbs glow as brightly as possible. (b) What is the power dissipated by each bulb? (c) Design a circuit to make both bulbs glow, but one more brightly than the other. Identify the brighter bulb.
A copper wire and an aluminum wire of the same length have the same resistance. What is the ratio of the diameter of the copper wire to that of the aluminum wire?
Two copper wires, one double the diameter of the other, have the same current flowing through them. If the thinner wire has a drift speed \(v_{1},\) and the thicker wire has a drift speed \(v_{2},\) how do the drift speeds of the charge carriers compare?
A defibrillator passes a brief burst of current through the heart to restore normal beating. In one such defibrillator, a 50.0 - \(\mu \mathrm{F}\) capacitor is charged to \(6.0 \mathrm{kV}\). Paddles are used to make an electric connection to the patient's chest. A pulse of current lasting 1.0 ms partially discharges the capacitor through the patient. The electrical resistance of the patient (from paddle to paddle) is \(240 \Omega\) (a) What is the initial energy stored in the capacitor? (b) What is the initial current through the patient? (c) How much energy is dissipated in the patient during the 1.0 ms? (d) If it takes 2.0 s to recharge the capacitor, compare the average power supplied by the power source with the average power delivered to the patient. (e) Referring to your answer to part (d), explain one reason a capacitor is used in a defibrillator.
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