is this where we use R= I/V question I dont even know how to start problem There

Question

is this where we use R= I/V question I dont even know how to start problem

There are 8.5 x 1028 conduction electrons per cubic meter in copper (n = 8.5 x 1028 m3). A piece of cooper wire which has a diameter of 0.5 mm carries a current of 0.26 amps. At what rate in Watts = J/s does an average electron gain kinetic energy in the copper wire described? If the wire is 0.6 m long, what is its volume? How many conduction electrons are in the wire? Determine the total rate at which the electrons in the copper wire gain kinetic energy between collisions. Note: The answer to b is also the rate at which thermal energy is generated in the wire and is equal to the Joule heating rate.

Explanation / Answer

Consider the wire has a volume of 1 m^3. The total number of electrons is

n =8.5*10^28 m^3

If the section of the wire is

S =pi*r^2 =pi*(0.25*10^-3)^2 =1.963*10^-7 m^2

The wire length is

L = V/S =1/1.963*10^-7 =5.093*10^6 m

The total resistance of this wire is (ro =1.68*10^-8 ohm*m)

R = ro*L/S = 1.68*10^-8*5.093*10^6/1.963*10^-7 =435872 ohm

The total power in the wire is

P = I^2*R = 0.26^2*435872 =29465 W

For one electron the individual power is

P0 = P/n =29465/8.5*10^28 =3.466*10^-25 W

b)

The volume in this case is

V = S*L =1.963*10^-7*0.6 =1.178*10^-8 m^3

The total number of conduction electrons in this wire is

N = n*V =8.5*10^28*1.178*10^-8 =1.001*10^22 electrons

c) the wire resistance in this case is

R = ro*L/S =1.68*10^-8*0.6/1.963*10^-7 =5.135*10^-2 Ohm

The Joule heating rate (J/s) or equivalent the rate with which the electrons gain kinetic energy is

P = I^2*R =0.26^2*5.135*10^-2 =3.471*10^-3 W =3.471 mW =3.471 (mJ/s)

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