Hall Effect

1. Sep 30, 2013

Turion

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

$$Δ{ V }_{ H }=\frac { IB }{ nqt } \\ 9.6*{ 10 }^{ -6 }=\frac { 50*1.3 }{ n*e*3.3*{ 10 }^{ -4 } } \\ n=1.28*{ 10 }^{ 29 }\quad electrons/{ m }^{ 3 }$$

2. Sep 30, 2013

Staff: Mentor

You can use WolframAlpha to check calculations like that.

3. Oct 1, 2013

Turion

I thought the method was wrong because it gives us d=1mm but we don't use it in our calculations.

4. Oct 1, 2013

Staff: Mentor

I'm not sure where the Hall voltage is measured, either 0.330mm or 1mm is not necessary.

5. Oct 1, 2013

Staff: Mentor

Take a look at the Hyperphysics web page on the Hall Effect.