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An electrical conductor designed to carry large currents has a circular cross section 2.50 mm in diameter and is 14.0 m long. The resistance between its ends is 0.104Ω. (a) What is the resistivity of the material? (b) If the electric-field magnitude in the conductor is 1.28 V/m, what is the total current? (c) If the material has 8.5×1028free electrons per cubic meter, find the average drift speed under the conditions of part (b).

Short Answer

Expert verified

(a) The resistivity of the material is 3.65×10-8Ω·m.

b) If the electric-field magnitude in the conductor is 1.28 V/m , then the total current is 172.0 A .

(c) If the material has 8.5×1028free electrons per cubic meters, the average drift speed is2.58×10-3m/s

Step by step solution

01

Define the ohm’s law and resistance (R).

According to Ohm’s law the current flowing through conductor is directly proportional to the voltage across the two points.

V=IR

Where, Iis current in ampere localid="1655724078434" A, localid="1655724083468" Ris resistance in ohms localid="1655724088023" Ωand localid="1655724093311" Vis the potential difference volt localid="1655724097962" V.

Also, localid="1655724101438" V=EL

Where, localid="1655724105667" Eis the electric field across conductor and localid="1655724111023" Lis length of the conductor.

The ratio of V to I for a particular conductor is called its resistance R:

localid="1655724116244" R=VIorρLA

Where, localid="1655724026382" ρis resistivity localid="1655724030493" Ω·m,Lis length in m and localid="1655724036681" Ais area in localid="1655724042282" m2.

If the area of cross-section islocalid="1655724005712" πr2, concentration islocalid="1655724011647" nand localid="1655724017124" Iis current then drift velocitylocalid="1655724021912" Vdis given by equation:

localid="1655723999554" vd=lmn|q|r2

02

Determine the resistivity of material.

Given that,

R=0104Ωd=250mmL=14m

The resistivity of the material is:

ρ=RAL=R(πr2)L=0.104π2.52×11000214.0=3.65×10-8Ω·m

Hence, the resistivity of the material is 3.65×10-8Ω·m.

03

Determine the total current.

Given that,

E=1.28V/m

The current through conductor is

I=VR=ELR=1.25140.104=172.0A

Hence, if the electric-field magnitude in the conductor is 1.28 V/m , then the total current is 172.0 A .

04

Determine the drift speed.

The charge of electron q=1.6×10-19C.

The average drift speed is

role="math" localid="1655723824487" Vd=1πnqr2=172.0π8.5×10281.6×10-190.001252=2.58×10-3m/s

Hence, if the material has 8.5×1028free electrons per cubic meters, the average drift speed is2.58×10-3m/s

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