What is the charge carrier density in the conductor

[uofcphysics]

The Hall voltage across a 0.800-mm-thick conductor in a 1.30 T magnetic field is 3.2 uV when the current is 12.0 A.

What is the charge carrier denisty in the conductor (in m^-3).



I was using the equation V_h=IB/ned
where V_h is the Hall voltage
I is the current
B is the magnetic field
n is the charge-carrier density
e is the charge of an electron
and d is the thickness



I rearranged to get n=IB/V_hed
and evaluated to get 3.806*10²⁵m^-3

But this came up wrong on the assignment and it looks like an unreasonable answer. I think I need additional values but basically this entire question is really confusing me.

Thanks for the help!

 

[nickjer]

Please show us what numbers you used, and try to keep everything in SI units. Because I get a different answer than you when plugging in the numbers you gave.

 

[uofcphysics]

I used the numbers I provided so:
I=12.0 A
B= 1.3 T
e= 1.6*10^(-19) C
V= 3.2*10^(-6) V
d= 0.8 mm (I left it because those are the units that the answer is in- but I'm guessing now that that is probably wrong).

What did I miss that is wrong in the calculation?

 

[nickjer]

The units the answer is in is meters, not millimeters. So d = 8*10^-4 m.

 

[paranormal]

The charge carrier density in the conductor can be calculated using the equation n = IB/Vhde, where I is the current, B is the magnetic field, Vh is the Hall voltage, d is the thickness of the conductor, and e is the charge of an electron.

In this case, we have values for I, B, Vh, and d, but we also need to know the value of e. The charge of an electron is a fundamental constant with a value of 1.602 x 10^-19 C. Without this value, we cannot accurately calculate the charge carrier density in the conductor.

I would recommend checking the given values and ensuring that all necessary information is provided before attempting to solve this problem. Without the value of e, the calculated charge carrier density will be incorrect.