Q1) a-The electrical conductivity and electron mobility for aluminum are .8X107 (2.m) and 0.0012 m/V.s, respectively. Calculate the Hall voltage for n aluminum specimen that is 15 mm thick for a current of 25 A and a nagnetic field of 0.6 tesla (imposed in a direction perpendicular to the current). -An Iron rod of length I and a cross section A has to be replaced by another od (where the length of the second rod is half of the first and its cross section is
Q1) a-The electrical conductivity and electron mobility for aluminum are .8X107 (2.m) and 0.0012 m/V.s, respectively. Calculate the Hall voltage for n aluminum specimen that is 15 mm thick for a current of 25 A and a nagnetic field of 0.6 tesla (imposed in a direction perpendicular to the current). -An Iron rod of length I and a cross section A has to be replaced by another od (where the length of the second rod is half of the first and its cross section is
Introductory Circuit Analysis (13th Edition)
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Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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![Q1) a-The electrical conductivity and electron mobility for aluminum are
* 3.8X10' (2.m)' and 0.0012 m?/V.s, respectively. Calculate the Hall voltage for
an aluminum specimen that is 15 mm thick for a current of 25 A and a
magnetic field of 0.6 tesla (imposed in a direction perpendicular to the current).
b-An Iron rod of length I and a cross section A has to be replaced by another
rod (where the length of the second rod is half of the first and its cross section is
one-third of the first). Find the material which has to be used in the second rod
to equalize the resistance of the first.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fedc48fdf-9f44-43ed-ad4d-c52c878d237d%2Ffbecfce3-7c2f-4eed-bf13-976482518f71%2Frvy7wub_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Q1) a-The electrical conductivity and electron mobility for aluminum are
* 3.8X10' (2.m)' and 0.0012 m?/V.s, respectively. Calculate the Hall voltage for
an aluminum specimen that is 15 mm thick for a current of 25 A and a
magnetic field of 0.6 tesla (imposed in a direction perpendicular to the current).
b-An Iron rod of length I and a cross section A has to be replaced by another
rod (where the length of the second rod is half of the first and its cross section is
one-third of the first). Find the material which has to be used in the second rod
to equalize the resistance of the first.
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