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Question:

a) The amount of charge that passes through a copper wire in 5.4 s is 2.37 C. Find the current in the wire.
The wire has a charge carrier density of 7.5 × 10^26 /3
and a radius of 1.0 mm.
b) What is the cross sectional area of the wire?
c) What is the charge of the charge carriers in the copper wire?
d) Using all of the above, calculate the drift speed of the electrons in this wire

User Revliscano
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1 Answer

4 votes

Answer:

Step-by-step explanation:

a) To find the current in the wire, use the formula:

I = Q / t

Where I is the current, Q is the amount of charge that passes through the wire (2.37 C), and t is the time (5.4 s).

I = 2.37 C / 5.4 s = 0.44 A

So the current in the wire is 0.44 A.

b) To find the cross-sectional area of the wire, use the formula:

A = π * r^2

Where A is the cross-sectional area, r is the radius of the wire (1.0 mm = 0.001 m), and π is Pi (3.14).

A = π * (0.001 m)^2 = 3.14 * 10^-6 m^2

So the cross-sectional area of the wire is 3.14 * 10^-6 m^2.

c) To find the charge of the charge carriers in the wire, use the formula:

Q = n * e

Where Q is the charge, n is the number of charge carriers, and e is the elementary charge (1.60 × 10^-19 C).

The wire has a charge carrier density of 7.5 × 10^26 /3, so the number of charge carriers can be calculated as:

n = (charge carrier density * cross-sectional area of the wire)

n = (7.5 × 10^26 / 3) * (3.14 * 10^-6 m^2)

n = 7.5 × 10^26 * 3.14 * 10^-6 / 3

n = 2.35 × 10^20

So the charge of the charge carriers in the wire is:

Q = n * e = 2.35 × 10^20 * 1.60 × 10^-19 C = 3.72 × 10^-19 C

d) To find the drift speed of the electrons in the wire, use the formula:

v = I / (n * e * A)

Where v is the drift speed, I is the current (0.44 A), n is the number of charge carriers (2.35 × 10^20), e is the elementary charge (1.60 × 10^-19 C), and A is the cross-sectional area of the wire (3.14 * 10^-6 m^2).

v = 0.44 A / (2.35 × 10^20 * 1.60 × 10^-19 C * 3.14 * 10^-6 m^2)

v = 0.44 / (2.35 × 10^20 * 1.60 × 10^-19 * 3.14 * 10^-6)

v = 0.44 / (3.74 × 10^-5)

v = 11766 m/s

So the drift speed of the electrons in this wire is 11766 m/s.

User Michael Allen
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7.7k points