Question

The mean time between collisions in iron is 4.6×10⁻¹⁵ s .

What electron current is driven through a 1.8-mm-diameter iron wire by a 0.062 V/m electric field?

Answer 1

Solution)

Mean time between collision is given by,

t = (m*Vd)/ (e*E)

Where

t = Mean time between collision = 4.6 * 10^-15 s

mass,m= 9.1 * 10^-31 Kg

We know,

Vd = i/n*A

e = 1.6 * 10^-19 C

E = 0.062 V/m

We know,

Electron density of iron= 8.5 * 10^28 m^-3

On Substituing Values , we get

4.6 * 10^-15 = ( 9.1 * 10^-31 * Vd) /(1.6*10^-19 * 0.062)

Vd = 5.014 * 10^-5

Also, we know

i/n*A = 5.014 * 10^-5

i = 5.014 * 10^-5 * 3.14* 1/4 * (1.8*10^-3)^2 * 8.5 * 10^28

i = 1.08 * 10^19 /s (Ans)

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