Question

A strip of copper 150 µm thick and 4.50 mm wide is placed in a uniform magnetic field of magnitude B = 0.74 T, that is perpendicular to the strip. A current i = 18 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.)

Answer 1

Answer:

V = 6.55*10^{-6} v

Explanation:

The number density can be determined by using below formula:

$n = \frac{Bi}{Vle}$

where,

B  is uniform magnetic field 0.74

i is current 18 A

V is hall potential difference

l is thickness 150 MICRO METER

e is electron charge 1.6 *10^{-19} C

therefore V can be determined as

$V = \frac{iB}{nle}$

$V = \frac{18*0.74}{8.47*10^{28}*150*10^{-6}*1.6 *10^{-19}}$

V = 6.55*10^{-6} v