Current Density Problems and Solutions2

 Problem #1

A small but measurable current of 1.2 x 10^─10 A exists in a copper wire whose diameter is 2.5 mm.The number of charge carriers per unit volume is 8.49 x 10²⁸ m^─3. Assuming the current is uniform, calculate the (a) current density and (b) electron drift speed.

Answer;
Known:
Current, i = 1.2 x 10^─10 A
diameter is d = 2.5 mm = 2.5 x 10^─3 m
number of charge carriers per unit volume is n = 8.49 x 10²⁸ m^─3

(a) current density is

J = i/A = nev

J = (1.2 x 10^─10 A)/[1/4 π(2.5 x 10^─3 m)²] = 2.44 x 10^─5 A/m²

(b) electron drift speed, given by

J = nev

2.44 x 10^─5 A/m² = (8.49 x 10²⁸ m^─3)v

v = 1.80 x 10^─15 m/s

Problem #2 
Near Earth, the density of protons in the solar wind (a stream of particles from the Sun) is 8.70 cm^─3, and their speed
is 470 km/s. (a) Find the current density of these protons. (b) If Earth’s magnetic field did not deflect the protons, what total current would Earth receive?

Answer;
Known:
the density of protons in the solar wind is n = 8.70 cm^─3 = 8.7 x 10⁶ m^─3
the speed of the protons, v = 470 km/s = 470 x 10³ m/s

(a) the current density of these protons is

J = nev

  = (8.7 x 10⁶ m^─3)(1.60 x 10^─19C)(470 x 10³ m/s)

J = 6.54 x 10^─7 A/m²

(b) If Earth’s magnetic field did not deflect the protons, total current would Earth receive is

I = JA

but, since the effective area is the cross section of the earth, the area is

A = πR²
Where R = 6.371 x 10⁶ m is the radius of the earth,

I = JπR²

Then,

I = (6.54 x 10^─7 A/m²)(π)(6.371 x 10⁶ m)²

I = 8.34 x 10⁷ A

Problem #3 
How long does it take electrons to get from a car battery to the starting motor? Assume the current is 300 A and the electrons travel through a copper wire with cross-sectional area 0.21 cm² and length 0.85 m. The number of charge carriers per unit volume is 8.49 x 10²⁸ m^─3.

Answer;
current is i = 300 A
cross-sectional area, A = 0.21 cm² = 0.21 x 10^─4 m²
length, L = 0.85 m
The number of charge carriers per unit volume is n = 8.49 x 10²⁸ m^─3

We use v = J/ne = i/Ane. Thus, the time required is given by

t = L/v = LAne/i

  = (0.85 m)(0.21 x 10^─4 m²)(8.49 x 10²⁸ m^─3)(1.60 x 10^─19C)/300 A

t = 8.1 x 10² s = 13.0 min

Share :