capa 7 - Phys 1120, CAPA #7 solutions. 1) The relevant...

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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. ... (Ouch!) 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
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capa 7 - Phys 1120, CAPA #7 solutions. 1) The relevant...

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