How do I find the average velocity of an electron?

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To find the average velocity of an electron in a bus bar carrying a current of 1600A, one must relate current to charge flow over time. The concentration of free electrons is given as 10^29 electrons per cubic meter, and the cross-sectional area of the bus bar is 0.4 cm by 16 cm. Using the equation that connects current, charge, and electron density, the average velocity can be calculated by considering the total charge and the number of electrons flowing. The provided values are essential for determining the average velocity of the electrons. Understanding these relationships is crucial for solving the problem effectively.
chueco
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Homework Statement



A current of 1600A exists in a rectangular (0.4-by-16cm) bus bar. the current is due to free electrons flowing the wire at an average velocity of v meters/second. If the concentration of free electrons is 10^29 electrons per cubic meter and if they are uniformly dispersed throughout the wire, then what is the average velocity of an electron?

Can someone please help?

Homework Equations



I don't know where to start.


The Attempt at a Solution

 
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chueco said:

Homework Statement



A current of 1600A exists in a rectangular (0.4-by-16cm) bus bar. the current is due to free electrons flowing the wire at an average velocity of v meters/second. If the concentration of free electrons is 10^29 electrons per cubic meter and if they are uniformly dispersed throughout the wire, then what is the average velocity of an electron?

Can someone please help?

Homework Equations



I don't know where to start.


The Attempt at a Solution


Welcome to the PF.

What is the equation that relates current I to the flow of charge Q per unit time T?

And if you start with that equation, then you need to take into account how much charge there is per electron, and how many electrons are flowing...

The numbers you are given are useful, because they tell you the density of the electrons, and the cross-sectional area of the conductor...
 

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