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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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.
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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