Institutt for fysikk, NTNU TFY4155/FY1003 Elektrisitet og magnetisme Vår 2005

Solution to øving 10

Guidance Thursday March 17 and Friday March 18

Exercise 1

a) This system can be viewed as three resistances connected in series: the two 30 cm long Al wires and the resistance R = 10 Ω. The resistance of the two Al wires becomes

R_{A} = \frac{l}{σA} = ·\frac{0.60m}{3.54 · 10^{7} Ω^{−1}m^{−1} · 10^{−6}m^{2}} = 0.017 Ω

The same current I passes through the whole system. It is

I = \frac{V}{R + R_{A}} = \frac{1.5 V}{10.017 Ω} = 0.1497 A ≃ 0.15 A

according to Ohm’s law. Thus, we obtain the voltage drops

V_{R} = RI = 10 Ω · 0.1497 A = 1.497 V

over the resistance R and

V_{A} = R_{A}I = 0.017 Ω · 0.1497 A = 0.0025 V

over the two Al wires together.

b) We found the current I in a) above. The dissipated effect in the resistance R becomes

P = V_{R}I = 1.497 V · 0.1497 A = 0.224 W

(0.225 ≃ 0.23 W if we neglect the resistance of and the voltage drop across the two Al wires)

c) Here, we must first find the density of free electrons n. Next, we may use I = j · A = nevA in order to calculate the mean drift velocity v. In Al, we have a mass density 2700 kg pr m^{3}. This corresponds to 2700/0.02698 mol = 100074 mol = 100074 ·6.02 · 10^{23} atomer = 6.02 · 10^{28} atoms, and hence equally many free electrons, assuming one free electron pr Al atom. Mean drift velocity becomes

v = \frac{I}{neA} = \frac{0.15}{6.02 · 10^{28} · 1.6 · 10^{−19} · 10^{−6}} = 1.56 · 10^{−5} m/s = 15.6μm/s

Average thermal velocity for the electrons may be estimated by setting the kinetic energy equal to the thermal energy:

\frac{1}{2}mv^{2} = \frac{3}{2}k_{B}T
⇒ v = \sqrt{\frac{3k_{B}T}{m}} ≃ 10^{5} m/s

Here, k_{B} = 1.38 · 10^{−23} J/K is Boltzmann’s constant. We see that the mean drift velocity is roughly 10 orders of magnitude smaller than the average thermal velocity. In other words, it takes several hours for a given electron to get from one end to the other in our system!

Exercise 2

a) If 700 W corresponds to 90% of the total effect, the total effect becomes 700/0.9 W = 778 W. This effect is supposed to be delivered during a period of 0.005 s. Then, an energy (700/0.9) · 0.005 J = 3.89 J must be stored in the capacitor.

b) We have in the lectures derived an expression for the work required to charge a capacitor with capacitance C. Let us briefly remind ourselves: In order to increase the charge on the capacitor from q to q + dq, a work dW = v(q)dq is required. Here, v(q) is the voltage across the capacitor when the charge is q. Per definition, we have C = q/v, so that dW = C^{−1}q dq and the total work must be W = Q^{2}/2C, i.e., in order to increase the charge from 0 to Q. With Q = V C, this can also be written as W = CV^{2}/2. This means that the voltage required to store an energy 3.89 J in a capacitor with capacitance 0.80 mF is

V = \sqrt{\frac{2W}{C}} = \sqrt{\frac{2 · 3.89}{0.80 · 10^{−3}}} = 98.6 V

Exercise 3

Let us imagine we divide the conductor into cylindrical ”tubes” with inner radius r, outer radius r + dr, and therefore cross sections with area

dA = 2πr dr

The current in such a tube is

dI = j · dA = j_{0} (1 − \frac{r^{2}}{R^{2}} ) · 2πr dr

The total current I is found by integrating dI over the full cross section of the conductor, i.e., by letting r vary from 0 to R:

I = ∫ dI
= ∫_{0}^{R} j_{0} (1 − \frac{r^{2}}{R^{2}} ) · 2πr dr
= 2πj_{0}|_{0}^{R} (\frac{1}{2}r^{2} − \frac{r^{4}}{4R^{2}} )
= 2πj_{0} (\frac{1}{2}R^{2} − \frac{1}{4}R^{2})
= \frac{1}{2}j_{0}πR^{2}

Exercise 4

1. The full region between r = a and r = b may be viewed as many resistances dR connected in series, where each resistor is a thin spherical shell with radius r and thickness dr:

dR = \frac{ρ dr}{4πr^{2}}

The total resistance is found by summing up all these individual resistances, i.e., by integrating from r = a to r = b:

R = ∫ dR
= ∫_{a}^{b} \frac{ρ dr}{4πr^{2}}
= \frac{ρ}{4π} |_{a}^{b} (−\frac{1}{r})
= \frac{ρ}{4π} (\frac{1}{a} − \frac{1}{b})

2. The given expression for the current I shows that we may here use Gauss’ law for the electric field to determine I:

I = \frac{1}{ρ} · \frac{Q}{ε_{0}}

We must assume that the charge entering into the inner conductor immediately distributes itself over the spherical surface at r = a before is starts on its way through the material between r = a and r = b. The potential difference between the inner and outer conducting shell is easily determined, since we know the electric field E:

ΔV = V_{a} − V_{b}
= − ∫_{b}^{a} E(r) dr
= \frac{Q}{4πε_{0}} |_{b}^{a} \frac{1}{r}
= \frac{Q}{4πε_{0}} (\frac{1}{a} − \frac{1}{b})

From these expressions, it follows that the resistance is

R = \frac{ΔV}{I} = \frac{ρ}{4π} (\frac{1}{a} − \frac{1}{b})