Rankers Physics
Topic: Current Electricity
Subtopic: Relation between Current and Drift Velocity

If the number of free electrons is \(5\times 10^{28} m^{-3}\) then the drift velocity of electron in a conductor of area of cross-section \(10^{-4} m^{2}\) for a current of 1.2 A is:  
\[1.5\times 10^{-2} m/s\]
\[1.5\times 10^{-3} m/s\]
\[1.5\times 10^{-4} m/s\]
\[1.5\times 10^{-6} m/s\]

Solution:

To find the drift velocity

vdv_d

of the electrons, we use the formula for current in terms of drift velocity:

 

I=nAevdI = n A e v_d

 

Where:


  • II
     

    is the current (1.2 A),


  • nn
     

    is the number of free electrons per unit volume ( 5×1028m35 \times 10^{28} \, \text{m}^{-3} 

    ),


  • AA
     

    is the cross-sectional area of the conductor ( 104m210^{-4} \, \text{m}^2 

    ),


  • ee
     

    is the charge of an electron ( 1.6×1019C1.6 \times 10^{-19} \, \text{C} 

    ),


  • vdv_d
     

    is the drift velocity of the electrons (which we need to calculate).

Step 1: Rearranging the formula to solve for vdv_d

 

 

vd=InAev_d = \frac{I}{n A e}

 

Step 2: Substituting the given values

 

vd=1.2(5×1028)×(104)×(1.6×1019)v_d = \frac{1.2}{(5 \times 10^{28}) \times (10^{-4}) \times (1.6 \times 10^{-19})}

 

Step 3: Performing the calculation

 

vd=1.2(5×1028)×(104)×(1.6×1019)v_d = \frac{1.2}{(5 \times 10^{28}) \times (10^{-4}) \times (1.6 \times 10^{-19})}

 

vd=1.28×106v_d = \frac{1.2}{8 \times 10^{6}}

 

vd=1.5×107m/sv_d = 1.5 \times 10^{-7} \, \text{m/s}

 

Final Answer:

The drift velocity of the electrons is

1.5×107m/s\boxed{1.5 \times 10^{-7} \, \text{m/s}}

.

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