Calculating the magnetic field from the Hall Voltage

[dvd_247]

Homework Statement

You have built a sensor to detect the strength of unknown magnetic fields. You
use a rectangular sample of copper that is 14.2 cm wide and 0.5 cm thick. You apply a
current of 2.4 A to the copper. You know that there is a magnetic field perpendicular
to the current because you measure a Hall Voltage of 0.1μV. What was the magnitude
of the magnetic field that you detected?

Assume that one electron per atom is available for conduction. (2) Copper has a
density of 8.93 g/cm3 and a molar mass of 63.55 g/mol. (3) Remember that 1 mol of any
substance contains 6.02 x1023 atoms (Avogadro’s number).

Homework Equations

q v_d B = q E_H

V_H = E_H d = v_d B d

n = $$\frac{\rho N_{A}}{M}$$

B = $$\frac{E_{H}}{v_{d}}$$

E_H = $$\frac{V_{H}}{d}$$

The Attempt at a Solution

B = $$\frac{E_{H}}{v_{d}}$$

E_H = $$\frac{V_{H}}{d}$$

Therefore B = $$\frac{V_{H}}{v_{d} d}$$ ...

If v_d = $$\frac{I}{n q A}$$ ...

Then B = $$\frac{n q A V_{H}}{I d}$$

If n = $$\frac{\rho N_{A}}{M}$$

Then B = $$\frac{\rho N_{A} q A V_{H}}{M I d}$$

I worked out the following numbers (I don't know whether my error lies here or not)...

$$\rho$$ = 8.93 x 10^-9 kgm^-3
N_A = 6.02 x 10²³ atoms
q = 1.602 x 10^-19 C
A = 14.2 x 10^-2 x 0.5 x 10^-2 = 7.1 x 10^-4m²
V_H = 0.1 x 10^-6 v
M = 63.55 x 10^-3 kgmol^-1
I = 2.4 A
d = 0.5 x 10^-2 m

I have been stuck on this question for several hours now, and can't see where I'm going wrong. The answer I'm getting is 8.018 x 10^-11T, when the answer expected is between 2 and 4 T apparently. Any help would be much appreciated :)

 

[kuruman]

$$\rho$$ = 8.93 x 10^-9 kgm^-3

Check your conversions. Do you really believe that a cube of solid copper one meter on the side has a mass of 8.93 micrograms?

 

[dvd_247]

i've completed the question now, thanks for your help :) it was like you said, my conversion was completely wrong! also i was using the wrong measurement for d, it was in fact " 12.2 x 10^-2 and not 0.5 x 10^-2