Question

A silver wire of 1mm diameter has a charge of 90 coulmb flowing in 1hour 15mim. Silver contains 5.8× 10^28 free electrons per cm^3. Find current in the wire and drift velocity of the electrons

Answer 1

1 hour 15 minutes=4500 seconds

i=q/t

i=90/4500=0.02 amperes

j=i/a

j=0.02/(3.14*0.05*0.05)

j=2.55*10^4 amp/m^2

drift velocity=j/ne

v(d)=2.55*10000/((5.28*10^22)*(1.6*10^-19)

v(d)=2.69*10^-7

Answer 2

Answer:

The current in the wire and drift velocity of the electron is 0.02 A and $2.74\times10^{-6}\ m/s$

Explanation:

Given that,

Charge $Q= 90 C$

Diameter $d =1 mm$

The current is define as:

$I=\dfrac{Q}{t}$...(I)

Here, Q = charge

t = time

Put the value of charge and time in equation (I)

$I=\dfrac{90}{4500}$

$I=0.02\ A$

The drift velocity is define as:

$v_{d}=\dfrac{Q}{neAt}$

Here, Q = charge

n = number of free electron

A = area

t = time

Put the value of all element in equation (II)

$v_{d}=\dfrac{90}{5.8\times10^{28}\times1.6\times10^{-19}\times3.14\times(0.5\times10^{-3})^2\times4500}$

$v_{d}=2.74\times10^{-6}\ m/s$

Hence, The current in the wire and drift velocity of the electron is 0.02 A and $2.74\times10^{-6}\ m/s$