Hall Effect

  • Thread starter Turion
  • Start date
  • #1
143
2

Homework Statement



OeP0E2B.png


Homework Equations





The Attempt at a Solution



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

Answers and Replies

  • #3
143
2
You can use WolframAlpha to check calculations like that.

I thought the method was wrong because it gives us d=1mm but we don't use it in our calculations.
 
  • #4
35,392
11,744
I'm not sure where the Hall voltage is measured, either 0.330mm or 1mm is not necessary.
 
  • #5
gneill
Mentor
20,925
2,866
Take a look at the Hyperphysics web page on the Hall Effect.

@Turion: Your calculation looks good.
 

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