The drift velocity of free electrons through a conducting wire of radius \(\mathrm{r}\), carrying current \(\mathrm{I}\), is \(\mathrm{V}_{\mathrm{d}}\) if the same current is passed through a conductor of radius \(2 \mathrm{r}\) what will be the drift velocity? (A) \(\left(\mathrm{V}_{\mathrm{d}} / 4\right)\) (B) \(\mathrm{V}_{\mathrm{d}}\) (C) \(2 \mathrm{~V}_{\mathrm{d}}\) (D) \(24 \mathrm{~V}_{\mathrm{d}}\)

Short Answer

Expert verified
The drift velocity of the conductor with radius 2r is \(\frac{1}{4}V_{d}\), which corresponds to answer choice (A).

Step by step solution

01

Find the relationship between drift velocity, radius, and current

To find the relationship, we must first make use of the formula for current (I) in a wire, which is given by: \[I = nAvq\] where, - I: Current in the wire - n: the number density of electrons - A: cross-sectional area of the wire - v: drift velocity of the electrons - q: charge of an electron The cross-sectional area of a conducting wire is given by: \[A = \pi r^2\] Now, we can rewrite the current formula as follows: \[I = n(\pi r^2)vq\]
02

Find the drift velocity of the wire with radius r

We are given the drift velocity, Vd, for the wire with radius r. So, for this wire, the current formula can be written as: \[I = n(\pi r^2)V_{d}q\]
03

Find the drift velocity of the wire with radius 2r

We have to find the drift velocity, Vd', for the wire with radius 2r. Both wires carry the same current, I. So, we can write the current formula for the wire with radius 2r: \[I = n(\pi (2r)^2)V_{d'}q\]
04

Solve for the new drift velocity

Now we must simply solve for \(V_{d'}\), the drift velocity of the electrons in the wire with the radius 2r. Divide the current equation of wire 2r by the current equation of wire r: \[\frac{n(\pi (2r)^2)V_{d'}q}{n(\pi r^2)V_{d}q} = 1\] Notice that the number density n, electron charge q, and π will cancel each other out: \[\frac{(2r)^2V_{d'}}{r^2V_{d}} = 1\] Solve for \(V_{d'}\): \[V_{d'} = \frac{r^2V_{d}}{(2r)^2} = \frac{1}{4}V_{d}\]
05

Identify the correct answer

After solving for the new drift velocity, we have: \[V_{d'} = \frac{1}{4}V_{d}\] Therefore, the correct answer is (A) \(\left(\mathrm{V}_{\mathrm{d}} / 4\right)\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Potentiometer wire of length \(1 \mathrm{~m}\) is connected in series with $490 \Omega\( resistance and \)2 \mathrm{v}\( battery If \)0.2 \mathrm{mv} / \mathrm{cm}$ is the potential gradient, then resistance of the potentiometer wire is. (A) \(4.9 \Omega\) (B) \(7.9 \Omega\) (C) \(5.9 \Omega\) (D) \(6.9 \Omega\)

The potential difference through the \(3 \Omega\) resistor shown in fig is.... (A) Zero (B) \(1 \mathrm{~V}\) (C) \(3.5 \mathrm{~V}\) (D) \(7 \mathrm{~V}\)

In each of the following questions, match column \(\mathrm{I}\) and column II and select the correct match out of the four given choicesColumn I Column II (a) The series combination of cells is for (p) More current (b) The parallel combination of cell is for (q) More voltage (c) In series combination of n cells, each of (r) \(\varepsilon\) \(\mathrm{emf}_{\mathrm{e}}\) the effective voltage is (d) In parallel combination of n cells, each of emf \(\varepsilon\) (s) ne the effective voltage is (A) \(a-p, b-q, c-r, d-s\) (B) a - q, b-p, c-r, d - s (C) \(a-q, b-p, c-s, d-r\) (D) \(a-p, b-q, c-s, d-r\)

Area of cross-section of a copper wire is equal to area of a square of $2 \mathrm{~mm}\( length It carries a current of \)8 \mathrm{~A}$ Find drift velocity of electrons (Density of free electrons in copper $\left.=8 \times 10^{28} \mathrm{~m}^{-3}\right)$ (A) \(1.56 \times 10^{-2} \mathrm{~ms}^{-1}\) (B) \(1.56 \times 10^{-4} \mathrm{~ms}^{-1}\) (C) \(3.12 \times 10^{-2} \mathrm{~ms}^{-1}\) (D) \(3.12 \times 10^{-3} \mathrm{~ms}^{-1}\)

Which is the dimensional formula for conductance from the give below? (A) \(\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{2}\) (B) \(\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^{3} \mathrm{~A}^{2}\) (C) \(\mathrm{M}^{1} \mathrm{~L}^{-3} \mathrm{~T}^{-3} \mathrm{~A}^{-2}\) (D) \(\mathrm{M}^{1} \mathrm{~L}^{-3} \mathrm{~T}^{3} \mathrm{~A}^{2}\)

See all solutions

Recommended explanations on English Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free