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Chapter 18: Problem 13

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?

Short Answer

Expert verified
Answer: The drift speed in the thinner wire is 4 times greater than the drift speed in the thicker wire.

Step by step solution

01

Recalling the formula for current related to drift speed

Recall the formula to calculate the current (I) in a wire: \(I = nAveq\) where n is the number density of the charge carriers, A is the cross-sectional area of the wire, v is the drift speed, and e is the charge of one carrier (atomic charge value 1.6 x 10^{-19} C). Remember that the current in both wires is the same, so: \(I_1 = I_2\)
02

Considering wire diameters

Let's call the diameter of the thinner wire d_1, and the diameter of the thicker wire d_2. Since the thicker wire has twice the diameter of the thinner one, we can write: \(d_2 = 2d_1\) The cross-sectional area, A, of a wire can be calculated using its diameter d, as a function of the radius (r): \(A = \pi r^2 = \pi (\frac{d}{2})^2\)
03

Finding Cross-sectional area ratio

Let A_1 and A_2 be the cross-sectional areas of the thinner and thicker wires, respectively. To find the relationship between the areas: \(A_2 = \pi (\frac{d_2}{2})^2 = \pi (\frac{2d_1}{2})^2 = 4\pi (\frac{d_1}{2})^2 = 4A_1\)
04

Writing the current formulas

Now, we can write the current formulas for both wires as a function of drift speed: \(I_1 = nA_1v_{1}eq\) \(I_2 = nA_2v_{2}eq\)
05

Equating the currents

Since the currents in both wires are the same, we can equate the equations in Step 4: \(nA_1v_{1}eq = nA_2v_{2}eq\)
06

Solving for the drift speed ratio

Now, we can solve for the ratio of the drift speeds: \(\frac{v_1}{v_2} = \frac{nA_2v_{2}eq}{nA_1v_{1}eq}\) Since n and e are constants, we can cancel: \(\frac{v_1}{v_2} = \frac{A_2}{A_1}\) We found in Step 3 that \(A_2 = 4A_1\), so: \(\frac{v_1}{v_2} = \frac{4A_1}{A_1}\)
07

Final Answer

Finally, we find the relationship between drift speeds: \(\frac{v_1}{v_2} = 4\) This means that the drift speed in the thinner wire is 4 times greater than the drift speed in the thicker wire.

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Most popular questions from this chapter

A string of 25 decorative lights has bulbs rated at \(9.0 \mathrm{W}\) and the bulbs are connected in parallel. The string is connected to a \(120-\mathrm{V}\) power supply. (a) What is the resistance of each of these lights? (b) What is the current through each bulb? (c) What is the total current coming from the power supply? (d) The string of bulbs has a fuse that will blow if the current is greater than 2.0 A. How many of the bulbs can you replace with 10.4-W bulbs without blowing the fuse?
Refer to Problem \(34 .\) With the copper wire connected to an ideal battery, the current increases greatly when the wire is immersed in liquid nitrogen. Ignoring changes in the wire's dimensions, state whether each of the following quantities increases, decreases, or stays the same as the wire is cooled: the electric field in the wire, the resistivity, and the drift speed. Explain your answers.
The starter motor in a car draws \(220.0 \mathrm{A}\) of current from the \(12.0-\mathrm{V}\) battery for \(1.20 \mathrm{s}\). (a) How much charge is pumped by the battery? (b) How much electric energy is supplied by the battery?
An ammeter with a full scale deflection for \(I=10.0 \mathrm{A}\) has an internal resistance of \(24 \Omega .\) We need to use this ammeter to measure currents up to 12.0 A. The lab instructor advises that we get a resistor and use it to protect the ammeter. (a) What size resistor do we need and how should it be connected to the ammeter, in series or in parallel? (b) How do we interpret the ammeter readings?
A \(500-\) W electric heater unit is designed to operate with an applied potential difference of \(120 \mathrm{V}\). (a) If the local power company imposes a voltage reduction to lighten its load, dropping the voltage to $110 \mathrm{V}$, by what percentage does the heat output of the heater drop? (Assume the resistance does not change.) (b) If you took the variation of resistance with temperature into account, would the actual drop in heat output be larger or smaller than calculated in part (a)?
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