ooo VIVA LTE 7:27 PM session.masteringphysics.com 66% Item 3 MasteringPhysics An

Question

ooo VIVA LTE 7:27 PM session.masteringphysics.com 66% Item 3 MasteringPhysics An electrical conductor designed to carry large currents has a circular cross section 2.70 mm in diameter and is 15.0 m long. The resistance between its ends is 0.110 Part A What is the resistivity of the material? Submit Give Up Part B If the electric field magnitude in the conductor is 1.23 V/m, what is the total current? Submit Give Up Part C If the material has 8.5 × 1028 free electrons per cubic meter. find the average drift speed under the conditions of part B. m/s

Explanation / Answer

1.

R = rho*L/A

rho = R*A/L

using given values:

rho = resistivity

rho = 0.110*pi*(1.35*10^-3)^2/15

rho = 4.20*10^-8 ohm-m

2.

E = V/d

V = E*d = 1.23*15 = 18.45 V

V = i*R

i = V/R = 18.45/0.11 = 167.73 Amp.

3.

Avg. drift speed = Vd = I/(n*e*A)

e = 1.6*10^-19

n = 8.5*10^28 electrons/m^3

A = pi*(1.35*10^-3)^2

So,

Vd = 267.73/(8.5*10^28*1.6*10^-19*pi*(1.35*10^-3)^2)

Vd = 3.44*10^-3 m/sec

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