Item I MasteringPhysics Item l An electrical conductor designed to carry large c

Question

Item I MasteringPhysics Item l An electrical conductor designed to carry large currents has a circular cross section 2.80 mm in diameter and is 13.6 m long. The resistance between its ends is 0.109 Part A What is the resistivity of the material? Submit Give Up Part B If the electric field magnitude in the conductor is 1.28 V/m, what is the total current? Submit Give U 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. u= m/s

Explanation / Answer

A)

In this situation, = R*A / L, where is the resistivity, R is the resistance, A is the area, and L is the length.

To find the area, we need to find the radius, which is simply half of the diameter.

2.80 mm / 2 = 1.40 mm = 0.00140 m

The area is *r^2 = *(0.00140)^2 = 6.15 x 10^-6 m^2

Now we can calculate the resistivity.

= (0.109 x (6.15 x 10^-6 m^2)) / 13.6 m

= 4.92 x 10^-8 *m

B)

We can find the total current by using Ohm's Law, which states that the current is equivalent to the ratio of voltage and resistance. To obtain the voltage, take the product of the electric field and the length of the conductor.

V = 1.28 V/m x 13.6 m = 17.40 V

I = V / R = 17.40 V / 0.109 = 160 A

C)

Drift velocity is equal to I / n*q*A, where is the charge carrier density, and q is the charge (1.60 x 10^-19 C).

n = 8.5*1028

v = 177 A / ((8.5*1028) x (1.60 x 10^-19 C) x (6.15 x 10^-6 m^2))

= 0.00211 m/s

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