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A strip of copper 130 µm thick and 4.40 mm wide is placed in a uniform magnetic field of magnitude B = 0.79 T, that is perpendicular to the strip. A current i = 26 A is then sent through the strip such that a Hall potential difference V appears across the width. Calculate V. (The number of charge carriers per unit volume for copper is 8.47 × 1028 electrons/m3.)

Respuesta :

Answer:

V = 1.1658 × [tex]10^{-5}[/tex] V

Explanation:

given data

strip of copper thick = 130 µm

strip of copper wide = 4.40 mm

uniform magnetic field of magnitude B = 0.79 T

current i = 26 A

number of charge carriers per unit volume = 8.47 × [tex]10^{28}[/tex] electrons/m³

solution

we know that number density is express as

n = \frac{Bi}{Vle}      ...............1

B  is uniform magnetic field and i is current  and V is hall potential difference and l is thickness and e is electron charge 1.6 × [tex]10^{-19}[/tex]  C

so V will be as

V = \frac{iB}{nle}      .....................2

so put here value and we get V

V = [tex]\frac{26 \times 0.79}{8.47\times 10^{28}\times 130\times10^{-6}\times1.6 \times10^{-19}}[/tex]

V = 1.1658 × [tex]10^{-5}[/tex] V

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