What is the charge carrier density in the conductor

uofcphysics
Messages
2
Reaction score
0
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 Vh=IB/ned
where Vh 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/Vhed
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!
 
Last edited by a moderator:
Physics news on Phys.org
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.
 
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?
 
The units the answer is in is meters, not millimeters. So d = 8*10^-4 m.
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top