Find drift velocity of the current carriers in the copper strip

[moenste]

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.

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?

 

[lychette]

hall effect !

 

[moenste]

hall effect !

I used the formula, I've put the numbers into it, still I get a wrong result. What's wrong?

 

[moenste]

Any help please?

 

[moenste]

(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).

 

[moenste]

Any help please?

 

[gneill]

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.

 

[moenste]

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, V_H = B v D, v = V_H / B d = 10^-6 / (1.67 * 0.02) = 3 * 10^-5.

I saw that formula before, but I didn't think that

the potential difference between E and F is found to be 1 μV

is the Hall Voltage.

(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? V_H = BI / net, where net = mobility?

Update: v_d = μ E and σ = n e μ_e, where μ is mobility, σ is conductivity (or resistivity?), e is charge and n is number density.

 

[gneill]

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]

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?

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. v_d = μ E → μ = v_d / E = 3 * 10^-5 / 1.7 * 10^-4 = 0.176 m² / s V.

 

[gneill]

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.

 

[moenste]

That will work. It amounts to the same thing. $\sigma$ is conductivity.

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).

 

[gneill]

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.

 

[moenste]

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.

(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?

 

[gneill]

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?

 

[moenste]

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 N_A = 6 * 10²³ mol^-1.

 

[gneill]

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 N_A = 6 * 10²³ mol^-1.

Yes. So work out the material's number density (atoms per unit volume).

 

[moenste]

What did you calculate for part (b)?

(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: (N_A * density) / atomic mass = (6 * 10²³) * (9 * 10³) / 63 = 8.57 * 10²⁵ m^-3.

8.57 * 10²⁵ / 1.0(4) * 10²⁹ = 8.24 * 10^-4.

 

[gneill]

Make sure that your units match between given values.

 

[moenste]

Make sure that your units match between given values.

Indeed, 63 is g and not kg. The answer is (N_A * 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!