Question

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

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

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: \[I = n(\pi r^2)vq\]

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: \[I = n(\pi r^2)V_{d}q\]

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: \[I = n(\pi (2r)^2)V_{d'}q\]

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 \(V_{d'}\): \[V_{d'} = \frac{r^2V_{d}}{(2r)^2} = \frac{1}{4}V_{d}\]

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) \(\left(\mathrm{V}_{\mathrm{d}} / 4\right)\).

Key concepts

Current in a wire

Electric current through a wire is the flow of electrons from one point to another, like water through a pipe. In metal wires, this current is carried by electrons which move through the conducting material.

Current, often represented by the symbol \(I\), is measured in amperes (A), and it quantifies how much charge passes through a point in the wire per unit time.
The formula to calculate current in a wire is given as:

• \(I = nAvq\)

Where:

• \(I\) is the current,
• \(n\) is the number density of electrons,
• \(A\) is the cross-sectional area of the wire,
• \(v\) is the drift velocity of the electrons,
• \(q\) is the charge of an electron.

The current directly depends on these variables, meaning any change in them can alter the current flowing through the wire.

Cross-sectional area of wire

The cross-sectional area of a wire is crucial in determining how much current it can carry. It's like the width of a highway — the wider the highway, the more cars it can accommodate.
In mathematical terms, for a cylindrical wire with radius \(r\), the cross-sectional area \(A\) is calculated using the formula:

• \(A = \pi r^2\)

Here, \(\pi\) (pi) is a constant approximately equal to 3.14159, and \(r\) is the radius of the wire.
A larger cross-sectional area allows more electrons to pass through simultaneously, which in turn can allow for a higher current to flow through the wire. In our exercise, doubling the radius increases the cross-sectional area by a factor of four \((A = \pi (2r)^2)\), affecting the current and the drift velocity of the electrons.

Number density of electrons

Number density, commonly denoted by \(n\), refers to the concentration of free electrons available to carry a current in a material.
It's defined as the number of conduction electrons per unit volume.
Different materials have differing number densities, which affects their ability to conduct electricity.

For instance, metals generally have high electron number densities, which makes them excellent conductors.
The number density remains constant for a specific material and determines how many electrons are available to drift and hence contribute to the current when a voltage is applied across the wire.
In our context, as the wire's cross-sectional area changes, the number density ensures an adequate supply of electrons to maintain current.

Charge of an electron

The charge of an electron, symbolized as \(q\), is a fundamental constant that plays a key role in electricity.
An electron carries a negative charge of \(-1.602 \times 10^{-19}\) coulombs.

This charge is incredibly small, yet because there are so many electrons present in conductive materials, even a small amount of movement results in a measurable electric current.
In the formula for current, \(I = nAvq\), the electron charge \(q\) helps quantify the amount of electricity being carried.
Although individual electrons have a minuscule charge, this cumulative effect allows for the calculation of electric current in practical terms.
In terms of drift velocity, the charge of electrons is one of the factors contributing to how quickly an electron can "drift" through a conductor when a current is applied.