February 10, 2009
PHY2054 Discussion-Spring ‘09
Quiz 3 (Chapter 17.1-17.8)
Name:
UFID:
*1. (2.5pts) In the Bohr model of the hydrogen atom, an electron (e = 1.60×10-19 C) moves around a
proton in a circular path. What is the effective current associated with the electron motion if the
electron moves with a speed of 1.50×106 m/s with a radius of 6.0×10-11 m?
The definition of current is the amount of charge that passes through a surface in one second. Thus
the effective current is the number of revolution per second times electronic charge. We have
I = ef = ev/(2πr) = 1.6×10-19×1.5×106/(2×3.14×6.0×10-11) = 6.37×10-4 A
**2. (2.5pts) A 100-m long aluminum wire 5.00 mm in diameter carries a steady current of 5.00 A.
Assuming one aluminum atom provides one electron, find how many days it takes one electron to
travel the full length of the wire? The density of aluminum is 2.70 g/cm3 and the atomic weight (or
molar weight) is 27.0 (g/mol).
According to our assumption, the number density of electron is the same as that of aluminum atoms,
which is given by
n = density/mass of each atom = density/(M/NA) = 2.7×103/(27×10-3/6.02×1023) = 6.02×1028 /m3
The drift velocity is calculated as
vd = I/enA = 5/[1.6×10-19×6.02×1028×π×(2.5×10-3)2] = 2.65×10-5 m/s
The time interval is given by the length of the wire divided by the velocity. We have
Δt = l/vd = 100/2.65×10-5 = 3.77×105 (s)×1(day)/(602×24) (s) = 43.6 days
Constants & Formulas
Electric current: I = ΔQ/Δt = qnvdA
Temperature Variation of resistance:
R = R0[1+α(T-T0)],where T0 = 20 ºC
Ohm’s law: ΔV = IR
Avogadro’s number: NA = 6.02×1023 /mol
Resistance & Resistivity: R = ρl/A
Power: P = IΔV = I2R = ΔV2/R
**3. (2.5pts) A platinum resistance thermometer (α = 3.92×10-3 /ºC) has resistances of 180. Ω when
placed in a 0 ºC ice bath and 320. Ω when immersed in a vessel containing melting unknown
material. Find the melting point of the material.
To find the melting point, we need to know the resistance of the thermometer at the reference
temperature. Using the resistance at 0 ºC, we get
R = R0[1+α(T-T0)] ⇒ R0 = R/[1+α(T-T0)] = 180/[1+3.92×10-3×(0-20)] = 195 Ω
Once we get the reference resistance, we can measure temperature by measuring resistance. The
melting point of the material is
R’ = R0[1+α(T’-T0)] ⇒ T’= (R’/ R0-1)/α+ T0 = [(320/195)-1]/3.92×10-3+20 = 184 ºC
**4. (5pts) Suppose that you wish to make a uniform wire out of 5.00 g of copper. If the copper wire
is designed to carry a current of 15.0 A with a power loss of 30.0 W, and if all the copper is to be
used, what will be the diameter of the wire? The resistivity of copper is 1.70×10-8 Ωm and the
density is 8.92×103 kg/m3.
The desired resistance of the wire is
P = I2R ⇒ R = P/I2 = 30/152 = 0.133 Ω
We make a wire with this resistance using 5 g of copper. The volume of the copper is
V = m/density = 5×10-3/8.92×103 = 5.61×10-7 m3
Since the volume of the wire does not change, the length of the wire is uniquely determined if we fix
the cross sectional area. Expressing the resistance in terms of the area, we have
R = ρl/A = ρ(V/A)/A ⇒ A = √(ρV/R) = √(1.7×10-8×5.61×10-7/0.133) = 2.68×10-7 m2
The diameter of the wire is
A = π(d/2)2 ⇒d = √(4A/π) = √(4×2.68×10-7 /3.14) = 5.84×10-4 m