Resistance and Resistivity Problems and Solutions2

  Problem#1

Kiting during a storm. The legend that Benjamin Franklin flew a kite as a storm approached is only a legend—he was neither stupid nor suicidal. Suppose a kite string of radius 2.00 mm extends directly upward by 0.800 km and is coated with a 0.500 mm layer of water having resistivity 150 Ωm. If the potential difference between the two ends of the string is 160 MV, what is the current through the water layer? The danger is not this current but the chance that the string draws a lightning strike, which can have a current as large as 500 000 A (way beyond just being lethal).

Answer;
Known:
a kite string of radius r_string = 2.00 mm = 2.00 x 10^─3 m
Kite string length, L = 0.800 km = 800 m
Water having resistivity is ρ_water = 150 Ωm
Potential difference between the two ends of the string is V = 160 MV = 160 x 10⁶ V
thick layer of air is r_water = 0.500 mm = 0.500 x 10^─3 m

To approach this problem, note that we want the current through only the water layer and not the string. We can assume that the radii given for the string originates from the center of the string, but they only give the thickness of the water layer as opposed to its actual radius.

Then, the current through the water layer given by

i = VπA/ρL

i = Vπ[(r_water + r_string)² ─ (r_string)²]/ρL

  = (160 x 10⁶ V)π[(0.500 x 10^─3 m + 2.00 x 10^─3 m)² + (2.00 x 10^─3 m)²]/( 150 Ωm x 800 m)

i = 9.42 x 10^─5 A

Problem#2
Earth’s lower atmosphere contains negative and positive ions that are produced by radioactive elements in the soil and cosmic rays from space. In a certain region, the atmospheric electric field strength is 120 V/m and the field is directed vertically down.This field causes singly charged positive ions, at a density of 620 cm^-3 , to drift downward and singly charged negative ions, at a density of 550 cm^-3, to drift upward (Fig. 1). The measured conductivity of the air in that region is 2.70 x 10^-14 (Ωm)^-1 . Calculate (a) the magnitude of the current density and (b) the ion drift speed, assumed to be the same for positive and negative ions.

Fig.1

Answer:
Known
That electric field strength is E = 120 V/m
density of the positive ions is n_p = 620 cm^-3 = 620 x 10^-6 cm^-3
density of the negative ions is n_e = 550 cm^-3 = 550 x 10^-6 cm^-3
net density is n = n_p – n_e = 70 x 10^-6 m^-3

conductivity of the air region is σ = 2.7 x 10^-14 Ω^-1m^-1

(a) the magnitude of the current density given by

J = σE

  = (2.7 x 10^-14 Ω^-1m^-1)(120 V/m)

J = 3.24 x 10^-12 A/m²

(b) the ion drift speed given by is

J = nqv

v = J/nq = (3.24 x 10^-12 A/m²)/(70 x 10^-6 m^-3 x 1.6 x 10^-19 C)

v = 2.89 x 10¹¹ m/s

Problem#3
Figure 2 shows wire section 1 of diameter D₁ = 4.00R and wire section 2 of diameter D₂ = 2.00R, connected by a tapered section. The wire is copper and carries a current. Assume that the current is uniformly distributed across any cross-sectional area through the wire’s width. The electric potential change V along the length L = 2.00 m shown in section 2 is 10.0 μV. The number of charge carriers per unit volume is 8.49 x 10²⁸ m^-3 . What is the drift speed of the conduction electrons in section 1?

Fig.2

Answer:
Known:
wire section 1 of diameter D₁ = 4.00R
wire section 2 of diameter D₂ = 2.00R
the length L = 2.00 m
charge carriers per unit volume is n = 8.49 x 10²⁸ m^-3

The electric  field in section 2 is

E₂ = V/L₂ = 10μV/2 m = 5 μV/m

a current density vector of magnitude is (for wire section 2)

J₂ = E₂/ρ = (5 x 10^-6 V/m)/1.69 x 10^-8 Ωm = 296 A/m²

Conservation  of  electric  current  from  section  1  into  section  2 implies

J₁A₁ =J₂A₂

J₁r₁² = J₂r₂² or J₁D₁² = J₂D₂², when

J₁ = 296 A/m²(2R/4R)² = 74 A/m²

So that the drift speed of the conduction electrons in section 1 is

v = J₁/nq =  74 A/m²/(8.49 x 10²⁸ m^-3 x 1,6 x 10^-19 C) = 5.45 x 10^-9 m/s