Name: Combination of Cells
Uploaded: 2019-07-19
Duration: 7 min 26 s
Description: Let’s say we have a collection of cells with which we need to achieve specific voltage and current values. Here, we discuss how we can combine them to get th...

Question

Prove that the current density of a metallic conductor is directly proportional to the drift speed of electrons.

Accepted Answer

Consider a conductor of length $l$ and area of cross-section $A$ having $n$ electrons per unit length, as shown in the figure.

Now volume of the conductor $V = A \times l$

Therefore, total number of electrons in the volume = $V o l u m e \times e l e c t r o n - d e n s i t y$ = $A \ln$

Now if 'e' is the charge of one electron, then the total charge of the conductor is $Q = c h a r g e o f o n e e^{-} \times t o t a l n o . o f e^{-} = A \ln \times e$

Now Electric field in the conductor , $E = \frac{V}{l}$

Now due to the presence of this electric field, the electrons start to drift in the direction opposite to the direction of electric field with a velocity $v_{d}$ also known as drift speed.

Therefore, time taken by the electrons to drift from one end to the other end of the conductor-

$t = \frac{l}{v_{d}}$

Now the amount of current in the conductor is -

$i = \frac{Q}{t} = \frac{A \ln e}{\frac{l}{v_{d}}} \Rightarrow i = n e A v_{d}$

$\frac{i}{A} = n e v_{d} \Rightarrow j = n e v_{d}$

Now since 'n' and 'e' are constant hence

$j = k \cdot v_{d} \Rightarrow j i s p r o p o r t i o n a l t o v_{d} H e n c e p r o v e d .$

Maharashtra Board Solutions for Chapter: Electric Current Through Conductors, Exercise 2: Exercises

Maharashtra Board Physics Solutions for Exercise - Maharashtra Board Solutions for Chapter: Electric Current Through Conductors, Exercise 2: Exercises

Questions from Maharashtra Board Solutions for Chapter: Electric Current Through Conductors, Exercise 2: Exercises with Hints & Solutions

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