SME
University Physics III: Waves, Optics & Modern · Problem type
Drift speed in a wire
Use this for “A copper wire of 1 mm² carries 1 A. Find the electrons’ drift speed.”
This is an example. Type your own values to start a new problem.
I (A)
=
n (m⁻³)
×
q
×
A (mm²)
×
vd (m/s)
J
Current density
worked out (not typed) · A/mm²
Formulas
I
=
n
×
1.602 ×
10
-19
×
A
×
10
-6
×
vd
1
=
8.49 ×
10
28
×
1.602 ×
10
-19
×
1 ×
10
-6
×
(7.3524
×
10
-5
)
J
=
I
A
1
=
1
1
Assumptions
•
The current is carried by the free electrons, each of charge q = 1.602 × 10⁻¹⁹ C, spread evenly over the wire’s cross-section.
•
n is the free-electron density: one per atom for copper gives 8.49 × 10²⁸ m⁻³.
•
The electrons dart about near vF but drift slowly along the wire; the signal itself travels near light speed.
Step-by-step
WE KNOW
I = 1 A, A = 1 mm², n = 8.49 × 10²⁸ m⁻³
FIND
drift speed (vd), current density (J)
Step 1 · Find drift speed (vd)
Use I = n × 1.602 × 10⁻¹⁹ × A × 10⁻⁶ × vd
Each second the electrons in vd meters of wire pass any point; their charge nqAvd is the current, so divide I by nqA.
vd
=
I
n
× 1.602 ×
10
-19
×
A
×
10
-6
vd
=
1
8.49 ×
10
28
× 1.602 ×
10
-19
×
10
-6
vd
=
1
(1.36 ×
10
10
) ×
10
-6
vd
=
1
13,600
vd
=
7.3524 ×
10
-5
m/s
Step 2 · Find current density (J)
Use J = I ÷ A
The
current
through
each
square
millimeter;
J
=
nqvd,
so
the
drift
speed
follows
J
and
not
the
wire’s
size.
J
=
1
1
J
=
1
A/mm²
Check
Put the numbers back in. Both sides should match.
1
=
8.49 ×
10
28
×
1.602 ×
10
-19
×
10
-6
×
(7.3524
×
10
-5
)
✓
1
=
1
1
✓
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