# Calculating the magnetic field from the Hall Voltage

## 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 vd B = q EH

VH = EH d = vd B d

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

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

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

## The Attempt at a Solution

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

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

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

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

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
NA = 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
VH = 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 :)

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