
The drift velocity of free electrons in a conductor is v, when a current. I is flowing in it if both the radius and current are doubled, then drift velocity will be.
A. \[\dfrac{v}{4}\]
B. \[\dfrac{v}{2}\]
C. \[2v\]
D. \[4v\]
Answer
163.8k+ views
Hint:The drift velocity is the rate of distance covered by the charge particle between two consecutive collisions from neighbouring atoms in an atom.
Formula used:
\[{v_d} = \dfrac{J}{{ne}}\]
where \[{v_d}\] is the drift velocity, \[J\] is the current density, n is the number of free electrons per unit volume and e is the charge on the electron.
Complete step by step solution:
When potential is applied across the wire then there is an electric field generated inside the wire which applies force on the free electrons. The applied force on the free electrons makes them move inside the wire. The average velocity gained by the free electron is called the drift velocity of the electron. Using the formula of the drift velocity,
\[{v_d} = \dfrac{J}{{ne}}\]
As the J is the current density of the wire, so it is equal to the electric current flowing per unit cross-sectional area of the wire,
\[J = \dfrac{i}{A}\]
On substituting the expression for the current density in the formula of drift velocity, we get
\[{v_d} = \dfrac{i}{{neA}}\]
\[\Rightarrow {v_d} = \dfrac{i}{{ne\pi {r^2}}}\]
It is given that both the current and the radius get doubled.
\[{i_2} = 2{i_1}\]
\[\Rightarrow {r_2} = 2{r_1}\]
The initial drift velocity is given as v,
\[v = \dfrac{{{i_1}}}{{ne\pi r_1^2}}\]
Let the final drift velocity is \[{v_2}\].
\[{v_2} = \dfrac{{{i_2}}}{{ne\pi r_2^2}} \\ \]
\[\Rightarrow {v_2} = \dfrac{{2{i_1}}}{{ne\pi {{\left( {2{r_1}} \right)}^2}}} \\ \]
\[\Rightarrow {v_2} = \dfrac{{2{i_1}}}{{4ne\pi r_1^2}} \\ \]
\[\Rightarrow {v_2} = \dfrac{1}{2}\left( {\dfrac{{{i_1}}}{{ne\pi r_1^2}}} \right) \\ \]
\[\therefore {v_2} = \dfrac{v}{2}\]
Hence, the final drift velocity is \[\dfrac{v}{2}\].
Therefore, the correct option is B.
Note: The drift velocity depends on the cross-section of the wire (A), the number of free electrons per unit volume (n) and the magnitude of the electric current. It does not depend upon the length of the wire.
Formula used:
\[{v_d} = \dfrac{J}{{ne}}\]
where \[{v_d}\] is the drift velocity, \[J\] is the current density, n is the number of free electrons per unit volume and e is the charge on the electron.
Complete step by step solution:
When potential is applied across the wire then there is an electric field generated inside the wire which applies force on the free electrons. The applied force on the free electrons makes them move inside the wire. The average velocity gained by the free electron is called the drift velocity of the electron. Using the formula of the drift velocity,
\[{v_d} = \dfrac{J}{{ne}}\]
As the J is the current density of the wire, so it is equal to the electric current flowing per unit cross-sectional area of the wire,
\[J = \dfrac{i}{A}\]
On substituting the expression for the current density in the formula of drift velocity, we get
\[{v_d} = \dfrac{i}{{neA}}\]
\[\Rightarrow {v_d} = \dfrac{i}{{ne\pi {r^2}}}\]
It is given that both the current and the radius get doubled.
\[{i_2} = 2{i_1}\]
\[\Rightarrow {r_2} = 2{r_1}\]
The initial drift velocity is given as v,
\[v = \dfrac{{{i_1}}}{{ne\pi r_1^2}}\]
Let the final drift velocity is \[{v_2}\].
\[{v_2} = \dfrac{{{i_2}}}{{ne\pi r_2^2}} \\ \]
\[\Rightarrow {v_2} = \dfrac{{2{i_1}}}{{ne\pi {{\left( {2{r_1}} \right)}^2}}} \\ \]
\[\Rightarrow {v_2} = \dfrac{{2{i_1}}}{{4ne\pi r_1^2}} \\ \]
\[\Rightarrow {v_2} = \dfrac{1}{2}\left( {\dfrac{{{i_1}}}{{ne\pi r_1^2}}} \right) \\ \]
\[\therefore {v_2} = \dfrac{v}{2}\]
Hence, the final drift velocity is \[\dfrac{v}{2}\].
Therefore, the correct option is B.
Note: The drift velocity depends on the cross-section of the wire (A), the number of free electrons per unit volume (n) and the magnitude of the electric current. It does not depend upon the length of the wire.
Recently Updated Pages
Uniform Acceleration - Definition, Equation, Examples, and FAQs

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Degree of Dissociation and Its Formula With Solved Example for JEE

Wheatstone Bridge for JEE Main Physics 2025

Charging and Discharging of Capacitor
