Question

The accompanying table shows measurements of the Hall voltage and corresponding magnetic field for a probe used to measure magnetic fields. (b) If the measurements were taken with a current of 0.200 A and the sample is made from a material having a charge-carrier density of 1.00×10²⁶ carriers/m³, what is the thickness of the sample?

Answer 1

0.125 mm . is the thickness of the sample.

What do you mean by hall voltage ?

The Hall effect is the creation of a voltage difference (the Hall voltage) across an electrical conductor, which is transverse to an applied magnetic field perpendicular to the current and an electric current in the conductor. Edwin Hall made the discovery in 1879.

We need to know the material's current, magnetic field, length, number of charge carriers, and area in order to calculate the Hall voltage. The Hall voltage is computed using the formula: v=IBlneA=(100A)(1.5T)(1.0102m)(5.91028/m3)(1.61019C)(2.0105m2)=7.9106V.

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First we have to plot those point Then we can use some computer program to fit those point linearly to get slope

of that graph a and interception b. We already know, from theory, that Hall's voltage AVH and magnitude of

magnetic field B are connected as

Δ$V_{H} =\frac{I}{nqt} B$

where I is current trough probe, n is concentration of charge carriers, q = 1.6 • 10¯19 C is charge of charge

carries and t is thickness of the material. We have put the data from the problem on a graph and fitted linearly and

got

a = 100 μ$\frac{V}{T}$

b = —0.02  μV.

As we can see, our result are in agreement with theoretical assumptions because interception b is almost O, and a

is asked relation between Hall's voltage A VH and magnitude of magnetic field B. Then we can write

ΔVH =$100X10^{-6}$ V/TB

(4) Then we can use result (4) and numbers from the textbook to calculate the thickness of the sample as

$a=\frac{I}{nqt} \\t=\frac{I}{anq} \\t=\frac{.200A}{100X10^{-6}X 1.6 X10^{-19}X10^{26} } \\t=0.125mm$

To learn more about the hall voltage , Visit: brainly.com/question/19130911

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