# Hall Effect

## The Attempt at a Solution

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

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mfb
Mentor
You can use WolframAlpha to check calculations like that.

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.

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

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