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1 month ago

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The Hall effect can be used to calculate the charge-carrier number density in a conductor. If a conductor carrying a current of

2.0 A is 0.5 mm thick, and the Hall effect voltage is 4.5 µV when it is in a uniform magnetic field of 1.2 T, what is the density of charge carriers in the conductor?

1 answer:

kicyunya [3.2K]1 month ago

5 0

Answer:

$6.6*10^{27}e/m^3$

Explanation:

When calculating Hall voltage, it is crucial to have the current, magnetic field strength, length, area, and number of charge carriers available. The Hall voltage can be expressed using the equation:

$V_h = \frac{iB}{neL}$

Where:

i= the current

B= the magnetic field strength

L = the length

n = the number of charge carriers

e= charge of an electron

We need to replace values and solve for n:

$n= \frac{iB}{V_h e L}$

$n= \frac{2*1.2}{4.5*10^{-6}*5^10^{-3}*1.6*10^{-19}}$

$n= 6.6*10^{27}electron.m^{-3}$

As a result, the charge carrier density is $6.6*10^{27}e/m^3$

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