Nichrome and Copper Circuit w/ Electron Mobility

  • #1
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Homework Statement


In the circuit shown, two thick copper wires connect a 1.5 V battery to a Nichrome wire. Each copper wire has radius R = 0.009m and is L = 0.17m long. Copper has [itex]8.4 \times 10^{28}[/itex] mobile electrons/[itex]m^3[/itex] and an electron mobility of [itex]4.4 \times 10^{-3}[/itex](m/s)/(V/m). The Nichrome wire is L2=0.08m long and has radius R2 = 0.003m. Nichrome has [itex]9.0 \times 10^{28}[/itex] mobile electrons/[itex]m^3[/itex] and an electron mobility of [itex]7.0 \times 10^{-5}[/itex] (m/s)/(V/m).

a). What is the magnitude of the electric field in the thick copper wire?
b). What is the magnitude of the electric field in the thin Nichrome wire?

14e008af-1ee3-40de-a0f2-3a921331ba0d.jpe


Homework Equations


I = nAuE
v = uE

The Attempt at a Solution


First I just tried to analyze the copper wire section.
L = 0.17m
R = 0.009m
n = [itex]8.4 \times 10^{28}[/itex]
u = [itex]4.4 \times 10^{-3}[/itex]
A = [itex]2.54 \times 10^{-4}[/itex]

I began to plug this into I = nAuE to solve for E until I realized that I don't know I. So I tried substituting v/u into the equation so that I = nAu(v/u) = nAv, but then I realized that I don't know what the drift velocity (v) is..
 

Answers and Replies

  • #2
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Hmm, after reviewing more in the textbook, I see that [tex]I= \frac{V \sigma A}{L}[/tex]where [itex]\sigma=5.9 \times 10^7[/itex] as tabulated for the conductivity of copper in my textbook. V=1.5V, L = 0.17m and A = 2.54×10^−4
Does this work to find the proper I necessary to calculate E?
 
  • #3
gneill
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I'm not too familiar with this particular topic (sorry!) but if it were me I'd investigate the relationship between electron mobility and conductivity / resistivity. With resistivity in hand you would be in a position to calculate the resistance of each wire segment, the current, and potential drop across each. That's a direct path to the field...
 
  • #4
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I'm not too familiar with this particular topic (sorry!) but if it were me I'd investigate the relationship between electron mobility and conductivity / resistivity. With resistivity in hand you would be in a position to calculate the resistance of each wire segment, the current, and potential drop across each. That's a direct path to the field...
It's a very strange part of circuits that we haven't gone over, yet was included in the assignments, so it's kind of scaring me at the moment. I'm really not sure if I'm doing it correctly.

Well, aren't resistivity and conductivity inverses of each other? so [itex]\sigma = \frac{1}{\rho}[/itex]
I guess I could express everything in terms of resistivity, but that would seem to complicate it a little bit more.

EDIT: Oh, I think I see what you're saying.. But then I'm not really sure how to use resistance of the wire to calculate the field given the mobility and whatnot.
 
  • #5
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Speaking of which, I have another question that seems like you would understand:

A material, whose resistivity is [itex]\rho[/itex], is extruded into a length L of wire with area A. The material is then melted into a blob and re-extruded through a die whose opening is half the area of the original. Compare the new resistance of the wire with the old one. The resistivity remains the same.

[tex]R= \frac{\rho L}{A}[/tex]
So if I have half the area, that would mean that the resistance would double correct?
 
  • #6
gneill
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Suppose that you had the resistivity of each material and could calculate the resistance of each wire segment. Then you could determine the current and the potential drop across each. If you know the potential drop and the length of a wire, then you know the field (units are Volts/meter).
 
  • #7
gneill
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Speaking of which, I have another question that seems like you would understand:



[tex]R= \frac{\rho L}{A}[/tex]
So if I have half the area, that would mean that the resistance would double correct?
You need to investigate how the length of the wire changes when the cross-sectional area is altered. Clearly if the area is reduced the length of the wire that can be produced will be greater since the total amount of material remians the same. I'd expect the resistance to increase with a reduction in cross sectional area, but also increase because of the increase in length.
 
  • #8
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Suppose that you had the resistivity of each material and could calculate the resistance of each wire segment. Then you could determine the current and the potential drop across each. If you know the potential drop and the length of a wire, then you know the field (units are Volts/meter).
So let's say I want to find the voltage at the end of the copper wire on the left aka the new voltage for the start of the Nichrome wire on the left.
The current through the copper wire is:
[tex]I=V/R[/tex]
[tex]=\frac{1.5V}{\frac{2 \rho L}{A}}=\frac{1.5V}{\frac{2 (1.69 \times 10^{-8}) (0.17)}{2.54 \times 10^{-4}}}[/tex]


So then the voltage STARTING at the Nichrome wire would be that value times the resistance of Nichrome, right?

You need to investigate how the length of the wire changes when the cross-sectional area is altered. Clearly if the area is reduced the length of the wire that can be produced will be greater since the total amount of material remians the same. I'd expect the resistance to increase with a reduction in cross sectional area, but also increase because of the increase in length.

Yeah, that makes sense. thanks!

 
  • #9
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Oops -- ignore the 2 in the equation for resistance. Don't know why I put that there!
 
  • #10
gneill
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Calculate the resistance of each wire segment first. Then you have a simple series circuit of three resistances and a voltage source. The current can be found easily from the total resistance and total potential due to the voltage source, and then the individual potential drops can be determined.
 
  • #11
Mister T
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The voltage drop across the copper wire is not 1.5 V.
 
  • #12
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Calculate the resistance of each wire segment first. Then you have a simple series circuit of three resistances and a voltage source. The current can be found easily from the total resistance and total potential due to the voltage source, and then the individual potential drops can be determined.
Okay, I got that the resistance in the two copper wires are [itex]1.13 \times 10^{-5}[/itex] and the resistance in the Nichrome wire is 0.004.

So V=IR
I=V/R
I=1.5/.0040113?
 
  • #13
gneill
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Please indicate units on quantities.

I didn't check to see if your values are correct, but if there are two copper wires with resistance 1.13 x 10-5 Ω I'd expect that to be reflected in the total resistance...
 
  • #14
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Please indicate units on quantities.

I didn't check to see if your values are correct, but if there are two copper wires with resistance 1.13 x 10-5 Ω I'd expect that to be reflected in the total resistance...
True. But either way the total resistance rounds to 0.0040A anyway.

So I = 373.94 A. That seems very high doesn't it? I know that the resistivities are correct. As are the lengths and areas. So resistances is also correct..
 
  • #15
gneill
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I would expect the current value to be dominated by the Nichrome wire's resistance since its resistivity should be much higher than that of copper (after all, heating elements are made of Nichrome wire and power sources are delivered by copper wiring).

Now, the wire lengths are pretty short so I can imagine there being a high current, so I'm not too worried there. But I would like to see details of your calculation for the Nichrome wire's resistance.
 
  • #16
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[tex]R=\frac{\rho L}{A}=\frac{(1.49 \times 10^{-6}\Omega \cdot m)0.08m}{(2.83 \times 10^{-5}) m^2}=0.0042 \Omega[/tex]
 
  • #17
gneill
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Okay, you need to use the given values for electron mobility and mobile charge density to calculate the resistances. Note that Nichrome wire is an alloy whose properties vary with the concentration of its constituents, so using table values for a problem where particular characteristics are specified is not recommended.

Can you calculate the resistivity for this particular Nichrome wire given the provided electron density and mobility values?
 
  • #18
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I'm not really sure how...

I know that [itex]\sigma A V= I L[/itex] where sigma is the conductivity. After solving for [itex]\sigma[/itex], I could take the reciprocal to get resistivity. From there plug in resistivity into the equation for resistance..? Not sure where the mobility comes into to play though.. I'm really lost and confused about this problem now.
 
  • #19
gneill
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Do a web search on "electron mobility and resistivity" (that's what I did :smile:). You should find that:

##\sigma = n e \mu_e##

where ##\sigma## is the conductivity, n is the number density of mobile electrons, e the electric charge of an electron, and ##\mu_e## the electron mobility. All these values (except for the charge, which is a common constant) are given in the problem statement.
 
  • #20
Mister T
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Is there a connection between mobility and drift velocity?

Never mind, I just saw gneill's last post.
 
  • #21
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Do a web search on "electron mobility and resistivity" (that's what I did :smile:). You should find that:

##\sigma = n e \mu_e##

where ##\sigma## is the conductivity, n is the number density of mobile electrons, e the electric charge of an electron, and ##\mu_e## the electron mobility. All these values (except for the charge, which is a common constant) are given in the problem statement.
Ohh, interesting. I did not see that in my search!
So then the reciprocal would give me resistivity which would help me find resistance. So how would I find the new voltage in each piece of wire?
 
  • #22
gneill
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Is there a connection between mobility and drift velocity?
I'd say yes. If you tinker with the units of mobility you can mold them into ##\left(\frac{m}{s}\right) \left( \frac{V}{m} \right)^{-1}##.
 
  • #23
gneill
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Ohh, interesting. I did not see that in my search!
So then the reciprocal would give me resistivity which would help me find resistance. So how would I find the new voltage in each piece of wire?
It's a series circuit of three resistors...
Fig1.png

Find the potential drop across each.
 
  • #24
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It's a series circuit of three resistors...
View attachment 91194
Find the potential drop across each.
Okay, so I found the new total resistances. Copper stayed the same, but Nichrome's value did indeed change!
So then the total current is I=V/R, so 1.5/.002823, which gives me 531.42A. So now I can apply V=IR to each piece?
 
  • #25
gneill
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Okay, so I found the new total resistances. Copper stayed the same, but Nichrome's value did indeed change!
So then the total current is I=V/R, so 1.5/.002823, which gives me 531.42A. So now I can apply V=IR to each piece?
Yup. Standard circuit analysis.
 

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