8.0 x 10^15 electrons flow through a cross section of silver wire in 340 us with...

8.0 x 10^15 electrons flow through a cross section of silver wire in 340 us with a drift speed of 7.6 x 10^-4m/s .

What is the diameter of the wire?

Expert Solution

The number of electrons can be expressed as,

N_e = nAvt

Thus, the area of the cross-section of wire is,

A = N_e / nvt

(pi)r² = N_e /nvt

Thus, the radius of the wire is,

r = [Ne/ (pi)nvt)]1/2

= [(8.0 x 10^15) / (pi)(5.8x10²⁸ m^-3)(7.6 x 10^-4 m/s)(340x10^-6 s)]1/2

= 4.122x10^-4 m

Therefore, the diameter of the wire is,

d = 2r

= 2(4.122x10^-4 m)

= 8.244x10^-4 m

= 0.8244 mm

genius_generous answered 2 months ago

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