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Question: Assume that each atom of copper contributes one free electron. The density of copper is \(9gc{m^{ - ...

Assume that each atom of copper contributes one free electron. The density of copper is 9gcm39gc{m^{ - 3}} and the atomic weight of copper is 6363. If the current flowing through a copper wire of 1mm1mm diameter is 1.1ampere1.1ampere, the drift velocity of electrons will be:-
(A)0.01mms1(A)0.01mm{s^{ - 1}}
(B)0.02mms1(B)0.02mm{s^{ - 1}}
(C)0.2mms1(C)0.2mm{s^{ - 1}}
(D)0.1mms1(D)0.1mm{s^{ - 1}}

Explanation

Solution

In order to solve this question firstly we will write the expression for finding the drift velocity of the electrons. Then we will find the required quantities and convert them into SI units. On putting all these values in the expression of drift velocity, we will arrive at the solution.

Complete answer:
In this question we are given that,
Density of copper =9gcm3 = 9\dfrac{g}{{c{m^3}}}
Atomic weight =63 = 63
Current flowing =1.1A = 1.1A
Diameter of the wire =1mm = 1mm
The number of copper atoms =6.022×1023 = 6.022 \times {10^{23}} (the number of copper atoms is equal to the number of copper electrons)
We also know that,
density=massvolumedensity = \dfrac{{mass}}{{volume}}
On rearranging the above expression, we get,
volume=massdensityvolume = \dfrac{{mass}}{{density}}
On putting the required value of mass and density, we get,
volume=639volume = \dfrac{{63}}{9}
volume=7cm3volume = 7c{m^3}
On converting the above value into SI unit, we get,
volume=7×106m3volume = 7 \times {10^{ - 6}}{m^3}
Now, we can find the value of nn,
n=6.022×10237×106n = \dfrac{{6.022 \times {{10}^{23}}}}{{7 \times {{10}^{ - 6}}}}
n=0.86×1029n = 0.86 \times {10^{29}}
Area is given by the expression,
A=πr2A = \pi {r^2}
A=3.14×(0.5×103)2A = 3.14 \times {\left( {0.5 \times {{10}^{ - 3}}} \right)^2}
On solving the above value, we get,
A=0.78×106m3A = 0.78 \times {10^{ - 6}}{m^3}
We know that the formula of drift velocity is given by the expression,
vd=IneA{v_d} = \dfrac{I}{{neA}}
On substituting the required values in the above expression, we get,
vd=1.10.86×1029×1.6×1019×0.785×106{v_d} = \dfrac{{1.1}}{{0.86 \times {{10}^{29}} \times 1.6 \times {{10}^{ - 19}} \times 0.785 \times {{10}^{ - 6}}}}
On further solving the above value, we get,
vd=0.1mms{v_d} = 0.1\dfrac{{mm}}{s}

So, the correct answer is (D)0.1mms1(D)0.1mm{s^{ - 1}}.

Note:
Various types of subatomic particles like electrons move in any random direction all the time. When these electrons are subjected to an electric field, then they do continue to move randomly, but they slowly drift in one particular direction, in the direction of the electric field applied. This net velocity at which the electrons start to drift is known as drift velocity.