Introductory Physics Volume Two

68 Circuits 3.9
Each sodium ion has a charge +1.6 × 10 −19 C, so for 10,000 ions, ∆Q =
10, 000 × (1.6 × 10 −19 C) = 1.6 × 10 −15 C. Since it takes 1ms:
I = ∆Q
∆t = 1.6 × 10−15 C
= 1 × 10 −12 A = 1pA.
.001s
The ions flow through the sides of the cylinder, so the magnitude of
the current density is
J = I S = 1.6 × 10 −12 A
(2π)(10µm)(500µm) = 5.1 × A 10−5 m 2 = 51µA m 2
Example
A copper wire, with a diameter of 1mm, has a current of 5A flowing
through it. What is the electric field in the wire?
Comment: How can E ≠ 0? Earlier it was stated that E = 0 inside a
conductor, however this was for the case of electrostatics. In the case
of a current flowing, since electric charges are not static, the electric
field will be nonzero.
For our wire, the current density is
J = I A = 5A
π(0.0005m) 2 = 6.4 × A 106 m 2
From Ohm’s law:
J = 1 ρ E
−→ E = ρJ = (1.7 × 10 −8 Ωm)(6.4 × 10 6 A m 2 )
= 0.11 Ω·A
= 0.11 V m
m
Example
A light bulb with a resistance of 3Ω is connected to a 9V battery. In
one second how many electrons flow through the bulb?
I
R = 3Ω
-
+
9V
Use Ohm’s law to find the current:
I = ∆V
R
= 9V
3Ω = 3A