I am currently studying the textbook Practical Electronics for Inventors, Fourth Edition, by Scherz and Monk. In section 2.4.1 Applying a Voltage, the authors have written the following:

In the case of alternating current, the field reverses directions in a sinusoidal fashion, causing the drift velocity component of electrons to swish back and forth. If the alternating current has a frequency of 60 Hz, the velocity component would be vibrating back and forth 60 times a second. If our maximum drift velocity during an ac cycle is 0.002 mm/s, we could roughly determine that the distance between maximum swings in the drift distance would be about 0.00045 mm. Of course, this doesn’t mean that electrons are fixed in an oscillatory position. It means only that the drift displacement component of electrons is — if there is such a notion. Recall that an electron’s overall motion is quite random and its actual displacement quite large, due to the thermal effects.

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I'm wondering how the authors concluded that the distance between maximum swings in the drift distance would be about 0.00045 mm? What is the calculation that was done here?

I would appreciate it if someone would please take the time to clarify this.

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    \$\begingroup\$ The information you have provided in your question is not sufficient to verify those results, in order to calculate the drift velocity in a wire you need to know the current in the wire, the diameter of the wire and the material of the wire, as the drift velocity is determined by the current in the wire in amps and the number of free electrons in the wire per length unit, which can be calculated if you know the diameter of the wire and the material. All these are parameters that you have not given in your question or in the picture. \$\endgroup\$
    – user173292
    Commented Jan 1, 2020 at 22:26
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    \$\begingroup\$ @ThePointer You can see some thoughts here. Just FYI. Not an answer. \$\endgroup\$
    – jonk
    Commented Jan 2, 2020 at 6:53
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    \$\begingroup\$ According to my calculations for 0.002 mm/s maximum velocity at 60Hz the amplitude should be 0.002 / 2pi*60 = 0.0000053mm. I have no idea how they got a value 85 times higher. \$\endgroup\$ Commented Jan 2, 2020 at 10:17
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    \$\begingroup\$ Use \$ for inline mathjax electronics.meta.stackexchange.com/questions/5565/… (they want us to use mathjax, but this is all the help they give us...) \$\endgroup\$ Commented Jan 5, 2020 at 16:03
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    \$\begingroup\$ @ThePointer I'm glad Bruce got you to an answer. It's worth learning. If you have ever used Latex, it's like a "some stuff removed to make it safer" version. You can use the simpler portions of Latex with some confidence. \$\endgroup\$
    – jonk
    Commented Jan 5, 2020 at 20:24

2 Answers 2


Recall that displacement \$d\$ is the area under the velocity curve. For a sinusoidal drift velocity \$v_d\$ having radian frequency \$\omega=2\pi f\$ where \$f=60\,\text{Hz}\$, the magnitude of maximum displacement over one half cycle can be calculated as the integral of \$v_d\$ with respect to time, during the time interval \$(0 \le t \le \pi/\omega)\,\text{s}\$:

$$ \begin{align*} d &= \int_{0}^{\pi/\omega}v_d\,dt,\;\;v_d(t) = J(t) / (\rho_e\,e)\\ &= \frac{1}{\rho_e\,e}\int_{0}^{\pi/\omega}J(t)\,dt,\;\;J(t) = I(t)/A\\ &= \frac{1}{\rho_e\,e\,A}\int_{0}^{\pi/\omega}I(t)\,dt,\;\;I(t) = k\,\sin (\omega t)\\ &= \frac{k}{\rho_e\,e\,A}\int_{0}^{\pi/\omega}\sin(\omega t)\,dt\\ &= \frac{2\,k}{\rho_e\,e\,A\,\omega} \end{align*} $$

where \$k=0.1\,\text{A}\$ (as specified in the book example).

For what it's worth, when I crunch the numbers with MATLAB (see Listing 1 and Figure 1 below) the calculated displacement—i.e., drift distance—is approximately 12 nm; so I'm not sure how the authors arrived at the value 450 nm for the drift distance.

See also:

Listing 1. MATLAB source code

%% Housekeeping

%% Givens
d = 2.05e-3;            % wire diameter, m
r = d/2;                % wire radius, m
A = pi*(r^2);           % wire cross-sectional area, m^2

q = 1.602e-19;          % electron charage, C
                        % (NB: This is 'e' in the equation above).

n = 8.46e28;            % estimate of the number of charge-conducting 
                        % electrons per cubic meter in solid copper
                        % (NB: This is 'rho_e' in the equation above).

k = 0.1;                % Sinusoidal current amplitude, peak
f = 60;                 % Sinusoidal current frequency, Hz
w = 2 * pi * f;         % Sinusoidal current frequency, rad/sec

%% Equations
% Current in the wire, C/s
I = @(t)  k * sin(w*t);

% Current density in the wire at time t, C s^-1 m^-2
% J = I/A = k*sin(w*t)/A = k/A * sin(w*t)
% Let k2 = k/A
k2 = k/A;
J = @(t)  k2 * sin(w*t);

% Average electron drift velocity at time t, m/s
% vd = J/n/q = I/n/q/A = k*sin(w*t)/n/q/A
% Let k3 = k/n/q/A
k3 = k/n/q/A;
vd = @(t)  k3 * sin(w*t);

% Average electron displacement at time t, m
% displacement = k/n/q/A/w * (1 - cos(w*t))
% Let k4 = k/n/q/A/w
k4 = k/n/q/A/w;
displacement = @(t)  k4 * (1 - cos(w*t));

%% Solutions
% For sin(w*t), max drift velocity occurs at w*t == pi/2 -> t = pi/2/w
vd_max = vd( pi/2/w )
    % 2.2355e-06 -> ~2.2 um/s

% Maximum average displacement of an electron during 1/2 cycle of 60 Hz 
% can be calculated as the area under the drift velocity curve during 
% the time interval (0 <= t <= pi/w) sec
% NB: For sin(w*t), 1/2 cycle occurs at w*t == pi -> t = pi/w
displacement_max = integral(vd, 0, pi/w )
    % 1.1860e-08 -> ~12 nm

%% Plot the velocity and displacement curves vs time

% NB: For sin(w*t), 1/2 cycle occurs at w*t == pi -> t = pi/w
t_ = linspace( 0, pi/w );

% drift velocity in micrometers/sec at time t
vd_t = vd(t_) * 1e6;
yyaxis left
plot(t_, vd_t)

% displacement in nanometers at time t
displacement_t = displacement(t_) * 1e9;
yyaxis right
plot(t_, displacement_t)

yyaxis left
title('Velocity and Displacement vs time')
xlabel('Time (sec)')
ylabel('Velocity (um/s)')
yyaxis right
ylabel('Displacement (nm)')
grid on

MATLAB plot of electron velocity and displacement vs. time

Figure 1. MATLAB plot of electron velocity and displacement vs. time.

  • \$\begingroup\$ Thanks for the answer, Jim. Since no one seems to be able to get the same result as the textbook authors, unless the physics guys come up with something, I think it's likely that it is an error. \$\endgroup\$ Commented Jan 5, 2020 at 23:05
  • \$\begingroup\$ User Philip Wood from physics.stackexchange just answered that he thinks that Bruce Abbott's answer is the correct one physics.stackexchange.com/a/523322/141502 \$\endgroup\$ Commented Jan 5, 2020 at 23:40
  • \$\begingroup\$ Jim, Phillip says that you seem to be working from a peak current, \$k\$, of \$0.1A\$, which doesn't seem to be relevant to the problem in hand. \$\endgroup\$ Commented Jan 6, 2020 at 0:19
  • \$\begingroup\$ The current value is relevant. Current is a function of voltage, and it's the electric field through the wire (caused by the potential difference across the wire) that determines the force that's applied to the electrons, and therefore the drift velocity of the electrons, and therefore the displacement of the electrons. \$\endgroup\$ Commented Jan 6, 2020 at 2:14
  • \$\begingroup\$ Phillip said the following: "The stuff in the shaded box headed "12 gauge copper.." in the page that you reproduce is about a wire that carries a constant current. It doesn't seem relevant to your question." \$\endgroup\$ Commented Jan 6, 2020 at 2:17

I just saw the answer to this question by user freecharly. They claim that the mean drift velocity of electrons is $$v = \dfrac{j}{n e},$$


$$j = I/A$$

is the current density for current \$I\$ and cross-sectional area \$A\$, and the electron density is \$n\$. If I'm not mistaken, applying this gives us

$$\dfrac{\frac{3.02 \times 10^4 \text{A}}{\text{m}^2}}{\frac{8.5 \times 10^{28}}{\text{m}^3} \times 1.602 \times 10^{-19} \text{coulombs}} = 0.00000221781\text{m/s},$$

since \$1 \text{A} = 1 \ \frac{\text{coulomb}}{\text{s}}\$

EDIT: Ok, I just noticed that this calculation is for the maximum drift velocity, and for a direct current cycle, whereas what we're trying to find is the distance between maximum swings in the drift distance for alternating current. I seem to have confused myself and forgot precisely which result it was that I was trying to find. My apologies for any confusion caused.

  • \$\begingroup\$ That's the drift velocity for DC current, right? How does 'distance between maximum swings in the drift distance' of an AC signal relate to that? \$\endgroup\$ Commented Jan 5, 2020 at 16:15
  • \$\begingroup\$ @BruceAbbott Uhh, is it? Haha, I honestly have no idea; after all, I'm the one who asked the question. But it produces the same value, right? So either we're misunderstanding something, or the author made the incorrect calculation. Given how confusing this has been, I'm going to ask the people on physics.stackexchange for clarification on this. \$\endgroup\$ Commented Jan 5, 2020 at 16:20
  • \$\begingroup\$ I think the downvote is unreasonable. The result is the same value as that of the textbook authors'. \$\endgroup\$ Commented Jan 5, 2020 at 16:36
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    \$\begingroup\$ It maybe half-answers the question, but not the most important part. I couldn't explain how they got the 0.00045 mm, and I looked up the source to check that there wasn't something missing or taken out of context. But my physics knowledge is not deep so I didn't risk putting up a wrong answer. Hopefully the physics guys will be able to clarify... \$\endgroup\$ Commented Jan 5, 2020 at 17:01
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    \$\begingroup\$ Ok, according to Phillip Wood's answer physics.stackexchange.com/a/523322/141502, it seems that my calculation of \$0.00000221781m/s\$ is actually for direct current, which is precisely what Bruce said, and not relevant to the calculation of the distance between maximum swings in the drift distance for alternating current. \$\endgroup\$ Commented Jan 5, 2020 at 23:45

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