Find drift velocity of the current carriers in the copper strip

In summary, ABCD is a plane rectangular strip of conducting material with a steady current flowing from AD to BC. When a magnetic field is set up at right angles to the plane, a small potential difference appears between the mid-points of AB and CD. This phenomenon is known as the Hall effect and can be explained by the movement of charge carriers in response to the magnetic field. A practical application of this effect is the Hall sensor, which is used to measure magnetic fields.Using the given information about the copper strip, it can be determined that the drift velocity of the current carriers is 3.0 * 10-5 m/s, the number of charge carriers per unit volume is 1.0(4) * 1029 m
  • #1
moenste
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Homework Statement


ABCD is a plane rectangular strip of conducting material of uniform thickness, with a steady current flowing uniformly from AD to BC. The potential difference between E and F, the mid-points respectively of AB and CD, is zero, but when a magnetic field is set up at right angles to ABCD, into the plane of the diagram, a small, steady potential difference appears between E and F. Explain these observations and briefly describe a practical application of the phenomenon.

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When the rectangle is made of copper, of density 9 * 103 kg m-3, relative atomic mass 63, with thickness 1 mm, breadth 20 mm, carrying a current of 10 A and with an applied field of flux density 1.67 T, the potential difference between E and F is found to be 1 μV, F being positive with respect to E. Taking the electronic charge to be 1.6 * 10-19 C and the Avogadro constant to be 6 * 1023 mol-1, find:

(a) the drift velocity of the current carriers in the copper;

(b) the number of charge carriers per unit volume in the copper and the sign of their charge;

(c) the mobility (drift velocity per unit electric field) of these carriers, given the resistivity of copper to be 1.7 * 10-8 Ω m;

(d) the ratio of the number of atoms per unit volume to the number of charge carriers per unit volume in the copper.

Answers: (a) 3.0 * 10-5 m s-1, (b) 1.0(4) * 1029 m-3, (c) 3.5 * 10-3 m2 s-1 V-1, (d) 0.82.

2. The attempt at a solution
(a) I = n A v e → v = I / n A e where: I = 10 A, n = 9 * 103 kg m-3, A = (1 / 10 / 100) * (20 / 10 / 100) = 2 * 10-5 m2, e = 1.6 * 10-19 C.

v = 10 / (9 * 103) * (2 * 10-5) * (1.6 * 10-19) = 3.47 * 1020 m s-1.

Doesn't fit the answer. What am I missing?
 
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  • #2
hall effect !
 
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  • #3
lychette said:
hall effect !
I used the formula, I've put the numbers into it, still I get a wrong result. What's wrong?
 
  • #4
Any help please?
 
  • #5
moenste said:
(b) the number of charge carriers per unit volume in the copper and the sign of their charge;
I = n A v e

n = I / A v e = 10 / 2 * 10-5 * 3 * 10-5 * 1.6 * 10-19 = 1.042 * 1029 m-3.

But no idea how to approach (a), (c) or (d).
 
  • #6
Any help please?
 
  • #7
There are two useful formulas that pertain to the Hall effect. You've been using only one of them. @lychette gave you an excellent hint. Did you do a bit of research on the Hall Effect?

Take a look at: Hyperphysics Hall Effect then list the two useful formulas.
 
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  • #8
gneill said:
There are two useful formulas that pertain to the Hall effect. You've been using only one of them. @lychette gave you an excellent hint. Did you do a bit of research on the Hall Effect?

Take a look at: Hyperphysics Hall Effect then list the two useful formulas.
So, VH = B v D, v = VH / B d = 10-6 / (1.67 * 0.02) = 3 * 10-5.

I saw that formula before, but I didn't think that
moenste said:
the potential difference between E and F is found to be 1 μV
is the Hall Voltage.

moenste said:
(c) the mobility (drift velocity per unit electric field) of these carriers, given the resistivity of copper to be 1.7 * 10-8 Ω m;
What is mobility? VH = BI / net, where net = mobility?

Update: vd = μ E and σ = n e μe, where μ is mobility, σ is conductivity (or resistivity?), e is charge and n is number density.
 
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  • #9
Mobility is the drift velocity per unit of electric field that is causing that velocity. This one's a bit tricky because you need to first determine the electric field. This is not the field established by the Hall voltage which is transverse to the current flow, but the field due to the potential difference caused by the current and the resistance of the copper. So, very much like the potential drop across a resistor.

Note that they give you the resistivity of copper for this question. You know the current flowing, and you know the cross sectional area of the conductor... so what's the potential drop per unit length of the conductor?
 
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  • #10
gneill said:
Mobility is the drift velocity per unit of electric field that is causing that velocity. This one's a bit tricky because you need to first determine the electric field. This is not the field established by the Hall voltage which is transverse to the current flow, but the field due to the potential difference caused by the current and the resistance of the copper. So, very much like the potential drop across a resistor.

Note that they give you the resistivity of copper for this question. You know the current flowing, and you know the cross sectional area of the conductor... so what's the potential drop per unit length of the conductor?
moenste said:
Update: vd = μ E and σ = n e μe, where μ is mobility, σ is conductivity (or resistivity?), e is charge and n is number density.
I think I'll only need to determine n now and then calculate mobility.

Update: ρ = R A / L → R = ρ L / A = 1.7 * 10-8 * 0.02 / 2 * 10-5 = 1.7 * 10-5 Ω. V = I R = 10 * 1.7 * 10-5 = 1.7 * 10-4 V. vd = μ E → μ = vd / E = 3 * 10-5 / 1.7 * 10-4 = 0.176 m2 / s V.
 
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  • #11
moenste said:
I think I'll only need to determine n now and then calculate mobility.
That will work. It amounts to the same thing. ##\sigma## is conductivity.
 
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  • #12
gneill said:
That will work. It amounts to the same thing. ##\sigma## is conductivity.
moenste said:
Update: ρ = R A / L → R = ρ L / A = 1.7 * 10-8 * 0.02 / 2 * 10-5 = 1.7 * 10-5 Ω. V = I R = 10 * 1.7 * 10-5 = 1.7 * 10-4 V. vd = μ E → μ = vd / E = 3 * 10-5 / 1.7 * 10-4 = 0.176 m2 / s V.
What's missing here? I am mostly unsure in the L figure (length).
 
  • #13
You don't need the length (or at least not a particular value of length). You have the resistivity. Conductivity is the reciprocal of resistivity.
 
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  • #14
gneill said:
You don't need the length (or at least not a particular value of length). You have the resistivity. Conductivity is the reciprocal of resistivity.
σ = 1 / ρ

σ = n e μ → μ = σ / n e = (1 / (1.7 * 10-8)) / (1.042 * 1029) * ( 1.6 * 10-19) = 3.5 * 10-3 m2 V-1 s-1.

moenste said:
(d) the ratio of the number of atoms per unit volume to the number of charge carriers per unit volume in the copper.
How shall we address this last one?
 
  • #15
moenste said:
σ = 1 / ρ

σ = n e μ → μ = σ / n e = (1 / (1.7 * 10-8)) / (1.042 * 1029) * ( 1.6 * 10-19) = 3.5 * 10-3 m2 V-1 s-1.How shall we address this last one?
Well, what do you know already and what remains to found?
 
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  • #16
gneill said:
Well, what do you know already and what remains to found?
Number of atoms per unit volute and number of charge carriers per unit volume are unknown.

We have density 9 * 103 kg m-3, relative atomic mass 63 and NA = 6 * 1023 mol-1.
 
  • #17
moenste said:
Number of atoms per unit volute and number of charge carriers per unit volume are unknown.
What did you calculate for part (b)?
We have density 9 * 103 kg m-3, relative atomic mass 63 and NA = 6 * 1023 mol-1.
Yes. So work out the material's number density (atoms per unit volume).
 
  • #18
gneill said:
What did you calculate for part (b)?
moenste said:
(b) the number of charge carriers per unit volume in the copper and the sign of their charge;

So we only need the number of atoms per unit volume. Using this formula: (NA * density) / atomic mass = (6 * 1023) * (9 * 103) / 63 = 8.57 * 1025 m-3.

8.57 * 1025 / 1.0(4) * 1029 = 8.24 * 10-4.
 
  • #19
Make sure that your units match between given values.
 
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  • #20
gneill said:
Make sure that your units match between given values.
Indeed, 63 is g and not kg. The answer is (NA * density) / atomic mass = (6 *1023) * (9 * 103) / 0.063 = 8.57 * 1028 m-3.

8.57 * 1028 / 1.0(4) * 1029 = 0.82.

Thank you!
 

1. What is drift velocity?

Drift velocity is the average velocity of current carriers, such as electrons, in a material under the influence of an electric field. It is a result of collisions between the carriers and atoms in the material.

2. How is drift velocity calculated?

Drift velocity can be calculated by dividing the current density (J) by the charge carrier density (n) and the charge of the carrier (q). This can be represented by the equation v = J/(nq).

3. What factors affect the drift velocity of current carriers in a material?

The drift velocity of current carriers can be affected by the strength of the electric field, the charge and mass of the carriers, and the density and type of atoms in the material.

4. Why is copper commonly used in electrical wiring?

Copper is commonly used in electrical wiring because it has a high conductivity, which means it allows for efficient flow of current. It is also relatively low cost and abundant, making it a practical choice for many applications.

5. How does temperature affect the drift velocity of current carriers in copper?

Temperature can affect the drift velocity of current carriers in copper by increasing the collisions between the carriers and the atoms in the material. This can lead to a decrease in the overall drift velocity. However, at lower temperatures, the conductivity of copper actually increases, making it a more efficient conductor.

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