1) The relevant formula we need comes back in Chapter 28.1, where Knight tells us how current
is related to the motion of electrons, i = n A v(drift).
Note that i here is NOT the current (!!), it's the "electron current", the # of electrons passing by
each second. The last problem of last week's CAPA (or Knight 28.11) tells us that the regular
current I = i*e, it's the number of electrons passing each second times the charge per electron.
(And, I is in the opposite direction of i, but here we're dealing in magnitudes)
Putting this all together, I get v(drift) = (I/A) / (n e)
(Oh, hey, this also comes straight from Knight 28.12, which we just partly rederived!)
I is given, it's the current in Amps.
A is the cross sectional area, pi r^2. (Watch units, put r into meters BEFORE squaring!)
e is the charge of any charge carrier, but in this case, an electron, so e=1.6E-19 C. (I use a plus
sign because we're looking for the drift SPEED, which must be positive)
The only tricky one is "n" which is # of carrier/volume. I get this by the following logic:
We are given rho (kg/m^3), and MW (molecular weight: I first converted to metric by dividing
the given number by 1000, so MW is in kg/mole rather than g/mole) We know Na, Avogadro's
number, 6.022E23, is the number of atoms/mole (do you know what this means? If not, it's worth
learning (or re-learning) about Avogadros number, and moles**!) so we can combine these:
n = # atoms/m^3 = (# kg/m^3)*(#moles/kg)*(#atoms/mole)
= (kg/m^3) * (atoms/mol) / (kg/mol)
= rho * Na / MW.
These were all given in the problem (except Na, which is in the front flyleaf of your text), and
remember that you need to plug in MW in kg/mol, not g/mol, to keep the units consistent.
Now we know all the quantities in the equation at the top, just plug in.
**Here's a related joke: Do you know what Avocado's number is? It's a guaca-mole.
2) This one is pretty straightforward - use R = rho*L/A, where rho = resistivity (not density!),
and L=length, and A = cross sectional area. (Once again, use pi R^2, notice that they gave us
diameter, so R = D/2) Again, be sure to convert all quantities to MKS, before squaring or
anything. That's it. The numbers are all fairly realistic, the result (which was on the order of a
fraction of an Ohm) seems quite reasonable for real-life wires in your house. The more
interesting part is what we do with this in the next question:
3) CAPA asks for power. Now, there are 3 formulas, P=VI, V^2/R, or I^2R. In this case, they
told us I, and we just found R, so the easiest to use is I^2R. That's it. You're done with the
CAPA question, but let's *think* about this results for a second. With the numbers I was given
(which I claim are all perfectly reasonable household numbers) this came out to be 60W, as
much as a light bulb. Think about how hot a bulb feels! This is getting close to a fire risk
(although, granted, it's spread out over 30 meters, almost 100 feet), and is the reason why you