Question

In: Physics

How many conduction electrons are there in a 5.00 mmmm diameter gold wire that is 40.0...

How many conduction electrons are there in a 5.00 mmmm diameter gold wire that is 40.0 cmcm long?

How far must the sea of electrons in the wire move to deliver -28.0 nCnC of charge to an electrode?

Solutions

Expert Solution

Answer:- Given, Diameter (2r)= 5 mm

                                        r= 2.5 mm= 0.25 cm

Length of wire is equal to the height of cylindrical wire, h= 40 cm

Volume of cylindrical wire, V= r2 h

V= 3.14x(0.25)2x40

V= 7.85 cm3

No. of free electrons in gold, n= 5.9x1028 /m3

No. of free electrons (conduction electrons) in 1 m3 of gold= 5.9x1028

No. of free electrons in 1 cm3 of gold= 5.9x1022

No. of free electrons in 7.85 cm3of gold= 7.85x5.9x1022 = 4.6315x1023

Hence the No. of conduction electron in given wire of gold = 4.63x1023

Now, we are going to find that minimum distance which must be moved by sea of electrons to deliver the charge -28nC to an electrode.

The relation between drift velocity and electric current, i= n e A vd    .............(i)

Let time taken to deliver the charge (q= -28 nc) is 't'.

using i=q/t in equation (i),

q/t = n e A vd

q = n e A vd t                       (distance travelled by sea of electrons, s = vd t)

q = n e A s

s = q / (n e A) ....................(ii)         

= 28 nC =28x 10-9 C

n= 5.9x1028 /m3

Area of cross sectiono of gold wire ,    A= r2 = 3.14x(0.25)2 = 0.19625 cm3 = 1.9625x10-7 m3

from equation (ii),

s= (28x 10-9) / (5.9x1028x1.6x10-19x1.9625x10-7)

s= 1.511x 10-11 m


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