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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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