Question

A conductor material has a free electrons density of 10^24 electrons per meter cube. when a voltage s applied a constant drift velocity of 1.5X10^-2 meter/second is attained by the electrons. of the cross sectional area of the material is 1 cm^2, calculate the magnitude of the current. Electronic charge is 1.6X10^-19 coulomb

Answer 1

Drift velocity is defined by:

v = I / nAq

v is the drift velocity
I is the current
n is the density of charge carriers
A is the cross-sectional area
q is the electric charge magnitude

Answer 2

after finding the drift velocity, what's next? how can I find the magnitude of the current?

Answer 3

You can solve this equation for the current:

v = I / nAq --> I = v * n * A * q

You have all 4 values needed, you just need to multiply them all together to find the current. Be careful with your units! Make sure everything is in meters (i notice the area A is in centimeters^2)

Answer 4

2400 is the answer I got, is this correct?

Answer 5

@Pompeii00

Answer 6

24?

Answer 7

I = (1.5 x 10^-2) * (10^24) * (0.0001) * (1.6 x 10^-19)

I don't have a calculator on me. But if you used the numbers above, then you should have the right answer.

Answer 8

in converting cm to m. shoukd I consider the square?

Answer 9

Yes. 1 cm = 0.01 m

When squared, you need to conver the square.

1 cm^2 = (0.01)^2 m^2 = 0.0001 m^2

Answer 10

ok, thank you very much :)