Question

Assume this 1.20-mm-radius copper wire is electrically neutral in the Earth reference frame, in which it is at rest and carrying a current. Calculate the linear charge density λ′ of the copper wire in a reference frame moving along with the electrons if the electrons are moving at 5.20 ×10−4m/s .

Answer 1

Answer:

The charge density in the system is $4.25*10^4C/m$

Explanation:

To solve this problem it is necessary to keep in mind the concepts related to current and voltage through the density of electrons in a given area, considering their respective charge.

Our data given correspond to:

$r=1*10^{-3}m\\v = 5.2*10^{-4}m/s\\e= 1.6*10^{-19}C$

We need to asume here the number of free electrons in a copper conductor, at which is generally of $8.5 *10^{28}m^{-3}$

The equation to find the current is

$I = VenA$

Where

I =Current

V=Velocity

A = Cross-Section Area

e= Charge for a electron

n= Number of free electrons

Then replacing,

$I = (5.2*10^{-4})(1.6*10^{-19})(88.5 *10^{28})(\pi(1*10^{-3})^2)$

$I= 22.11a$

Now to find the linear charge density, we know that

$I = \frac{Q}{t} \rightarrow Q = It$

Where:

I: current intensity

Q: total electric charges

t: time in which electrical charges circulate through the conductor

And also that the velocity is given in proportion with length and time,

$V_d = \frac{l}{t} \rightarrow l = V_d t$

The charge density is defined as

$\lambda = \frac{Q}{l}\\\lambda = \frac{It}{V_d t}\\\lambda = \frac{I}{V_d}$

Replacing our values

$\lambda = \frac{22.11}{5.20*10{-4}}$

$\lambda= 4.25*10^4C/m$

Therefore the charge density in the system is $4.25*10^4C/m$