Worked examples · Sheet 1

Currents, transported charges and drift velocity

Theory used: Electric current Current density Current density and drift velocity Steady current and conservation of charge Ohm's law in local form

Go to: Exercise 1 Exercise 2 Exercise 3 Exercise 4

Exercise 1

The battery of a phone delivers a constant current $I = 0.25 A$ for $Δ t = 2.0 h$ . Compute the charge that flows through the circuit and the corresponding number of electrons.

Q = I Δ t, N = \frac{Q}{e}

Q N = 0.25 \cdot 2.0 \cdot 3600 C = 1.8 \cdot 1 0^{3} C = \frac{1800}{1.60 \cdot 1 0 ^{- 19}} \approx 1.1 \cdot 1 0^{22}

The current is constant, therefore by the definition of electric current the transported charge is the product of the current and the time interval, expressed in seconds: $2.0 h = 7200 s$ . Each electron carries the charge $e$ in absolute value, hence the number of electrons is obtained by dividing the total charge by $e$ . The charge of $1800 C$ corresponds to $0.25 A \cdot 2.0 h = 0.50 Ah$ , that is, $500 mAh$ , the unit usually used to state the capacity of batteries.

Exercise 2

In a conductor the current grows linearly in time according to the law $i (t) = (2.0 + 0.50 t) A$ , with $t$ in seconds, for $0 \leq t \leq 4.0 s$ . Compute the charge that crosses a section of the conductor in this interval and the average current.

Current

i (t) = (2.0 + 0.50 t) A

, with

t

in seconds: the charge transported between

0

and

4.0 s

is the area of the colored trapezoid,

12 C

q = \int_{0}^{4.0} i (t) d t, I_{m} = \frac{q}{Δ t}

q I_{m} = [2.0 t + 0.25 t^{2}]_{0}^{4.0} C = (8.0 + 4.0) C = 12 C = \frac{12}{4.0} A = 3.0 A

The current varies in time, therefore the transported charge is the integral of $i (t)$ over the interval, that is, the area under the graph of $i$ . The region is a trapezoid with bases $2.0 A$ and $4.0 A$ and height $4.0 s$ , and the geometric calculation, $\frac{( 2.0 + 4.0 ) \cdot 4.0}{2} C = 12 C$ , confirms the integral. The average current is the ratio of the charge to the duration. Since $i$ is a linear function, the average coincides with the arithmetic mean of the values at the endpoints, $3.0 A$ .

Exercise 3

A copper wire of diameter $d = 1.6 mm$ carries a current $I = 5.0 A$ . Copper has $n = 8.5 \cdot 1 0^{28}$ conduction electrons per cubic meter. Compute the current density, the drift velocity of the electrons and the time an electron takes on average to travel $1.0 m$ of wire.

A = \frac{π d ^{2}}{4}, J = \frac{I}{A}, v_{d} = \frac{J}{n e}, Δ t = \frac{ℓ}{v _{d}}

A J v_{d} Δ t = π \cdot (0.80 \cdot 1 0^{- 3})^{2} m^{2} \approx 2.01 \cdot 1 0^{- 6} m^{2} = \frac{5.0}{2.01 \cdot 1 0 ^{- 6}} A/ m^{2} \approx 2.5 \cdot 1 0^{6} A/ m^{2} = \frac{2.49 \cdot 1 0 ^{6}}{8.5 \cdot 1 0 ^{28} \cdot 1.60 \cdot 1 0 ^{- 19}} m/s \approx 1.8 \cdot 1 0^{- 4} m/s = \frac{1.0}{1.83 \cdot 1 0 ^{- 4}} s \approx 5.5 \cdot 1 0^{3} s

The current is assumed to be distributed uniformly over the cross section, therefore the current density is the current divided by the area of the circular cross section, computed with the radius $0.80 mm$ . By the relation between current density and drift velocity, $J = n e v_{d}$ in magnitude, and $v_{d}$ is obtained. The electrons advance on average at constant velocity $v_{d}$ , hence the time to travel one meter is the ratio of length to velocity, about $1.5 h$ .

The result is very small compared with the thermal agitation speeds of the electrons. An appliance connected to the wire nevertheless switches on immediately, because the electric field is established along the whole circuit almost instantaneously and sets the electrons of every stretch in motion simultaneously.

Exercise 4

An aluminum wire has a stretch of diameter $d_{1} = 2.0 mm$ followed by a stretch of diameter $d_{2} = 1.0 mm$ and carries a steady current $I = 3.0 A$ . Aluminum has $n = 1.8 \cdot 1 0^{29}$ conduction electrons per cubic meter and resistivity $ρ = 2.8 \cdot 1 0^{- 8} Ω \cdot m$ . Compute, in each stretch, the current density, the drift velocity and the electric field.

I_{1} = I_{2} = I, J_{k} = \frac{4 I}{π d _{k}^{2}}, v_{d, k} = \frac{J _{k}}{n e}, E_{k} = ρ J_{k}

J_{1} J_{2} v_{d, 1} E_{1} = \frac{4 \cdot 3.0}{π \cdot ( 2.0 \cdot 1 0 ^{- 3} ) ^{2}} A/ m^{2} \approx 9.5 \cdot 1 0^{5} A/ m^{2} = 4 J_{1} \approx 3.8 \cdot 1 0^{6} A/ m^{2} = \frac{9.55 \cdot 1 0 ^{5}}{1.8 \cdot 1 0 ^{29} \cdot 1.60 \cdot 1 0 ^{- 19}} m/s \approx 3.3 \cdot 1 0^{- 5} m/s, v_{d, 2} = 4 v_{d, 1} \approx 1.3 \cdot 1 0^{- 4} m/s = 2.8 \cdot 1 0^{- 8} \cdot 9.55 \cdot 1 0^{5} V/m \approx 2.7 \cdot 1 0^{- 2} V/m, E_{2} = 4 E_{1} \approx 0.11 V/m

The current is steady and the wire has no branches, therefore by conservation of charge the current is the same in the two stretches. The cross-sectional area is proportional to the square of the diameter; since the diameter is halved, the area becomes one quarter and the current density quadruples. The drift velocity is proportional to $J$ , because $n$ and $e$ are the same in the two stretches, and it quadruples as well: the electrons speed up on entering the narrow stretch.

The electric field is obtained from Ohm’s law in local form, $E = ρ J$ , and in the narrow stretch it is four times stronger. In that stretch the voltage per unit length is larger, and a stretch of thin wire has a larger resistance than a stretch of thick wire of the same length.