The Hall effect can be used to determine the density of mobile electrons in a conductor. A thin strip of the material being investigated is immersed in a magnetic field and oriented so that its surface is perpendicular to the field. In a particular measurement, the magnetic field strength was 0.767 T, the strip was 0.108 mm thick, the current along the strip was 2.87 A, and the Hall voltage between the strip's edges was 2.65 mV. Determine the number density n of mobile electrons (in the order of 1025) in the material. (e = 1.6 x 10-19 C) NOTE: You do not need to include 1025 in your answer. Type your answer...

University Physics Volume 2
18th Edition
ISBN:9781938168161
Author:OpenStax
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Chapter11: Magnetic Forces And Fields
Section: Chapter Questions
Problem 94AP: The Hall effect is to be used to find the density of charge carriers in an unknown material. A Hall...
Question
The Hall effect can be used to determine the density of mobile electrons in a conductor. A thin strip of the material being investigated is immersed in a magnetic field and
oriented so that its surface is perpendicular to the field. In a particular measurement, the magnetic field strength was 0.767 T, the strip was 0.108 mm thick, the current along
the strip was 2.87 A, and the Hall voltage between the strip's edges was 2.65 mV.
Determine the number density n of mobile electrons (in the order of 1025) in the material. (e = 1.6 x 10-19 C)
NOTE: You do not need to include 1025 in your answer.
Type your answer...
Transcribed Image Text:The Hall effect can be used to determine the density of mobile electrons in a conductor. A thin strip of the material being investigated is immersed in a magnetic field and oriented so that its surface is perpendicular to the field. In a particular measurement, the magnetic field strength was 0.767 T, the strip was 0.108 mm thick, the current along the strip was 2.87 A, and the Hall voltage between the strip's edges was 2.65 mV. Determine the number density n of mobile electrons (in the order of 1025) in the material. (e = 1.6 x 10-19 C) NOTE: You do not need to include 1025 in your answer. Type your answer...
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